appendix-a.xml

<?xml version="1.0" encoding="UTF-8" ?>
<!-- <mathbook><book> -->

<appendix xml:id="appendix-a"   xmlns:xi="http://www.w3.org/2001/XInclude">
<title>Algebra Skills Refresher</title><introduction></introduction>
<section  xml:id="Numbers-and-Operations"><title>Numbers and Operations</title>
<subsection xml:id="Order-of-Operations"><title>Order of Operations</title>

<p>Numerical calculations often involve more than one operation. So that everyone agrees on how such expressions should be evaluated, we follow the order of operations.</p>

<assemblage><title>Order of Operations</title>
<p>
    <ol label="1">
        <li><p>
            Simplify any expressions within grouping symbols (parentheses, brackets, square root bars, or fraction bars). Start with the innermost grouping symbols and work outward.
        </p></li>
        <li><p>
            Evaluate all powers and roots.
        </p></li>
        <li><p>
            Perform multiplications and divisions in order from left to right.
        </p></li>
        <li><p>
            Perform additions and subtractions in order from left to right.
        </p></li>
    </ol>
</p>
</assemblage>
</subsection>




<subsection><title>Parentheses and Fraction Bars</title>
<p>
    We can use parentheses to override the multiplication-first rule. Compare the two expressions below.
    <md>
        <mrow>\amp\text{The sum of 4 times 6 and 10} \amp\amp 4 \cdot 6 + 10</mrow>
        <mrow>\amp\text{4 times the sum of 6 and 10} \amp\amp  4(6 + 10)</mrow>
    </md>
</p>
<p>
    In the first expression, we perform the multiplication <m>4\times 6</m> first, but in the second expression we perform the addition <m>6 + 10</m> first, because it is enclosed in parentheses.
</p>
<p>
    The location (or absence) of parentheses can drastically alter the meaning of an expression. In the following example, note how the location of the parentheses changes the value of the expression.
</p>

<example xml:id="example-parentheses">
    <p>
        <ol label="a">
            <li><p><m>
                \begin{aligned}5-3\cdot 4^2 \amp = 5-3\cdot 16\\
                \amp = 5-48=-43
                \end{aligned}                
            </m></p></li>
            <li><p><m>
                \begin{aligned}5-(3\cdot 4)^2 \amp = 5-12^2\\
                \amp= 5-144 = -139
                 \end{aligned}
           </m></p></li>
            <li><p><m>
                \begin{aligned}(5-3\cdot 4)^2 \amp = (5-12)^2\\
                \amp = (-7)^2 = 49
                \end{aligned}
            </m></p></li>
            <li><p><m>
                \begin{aligned}(5-3)\cdot 4^2 \amp = 2\cdot 16\\
                \amp = 32
                \end{aligned}
            </m></p></li>
        </ol>
    </p>
</example>

<warning>
    <p>
        In the expression <m>5 - 12^2</m>, which appears in <xref ref="example-parentheses" text="type-global"/>, the exponent <m>2</m> applies only to <m>12</m>, not to <m>-12</m>. Thus, <m>5 - 12^2 \ne 5 + 144</m>.
    </p>
</warning>

<p>
    The order of operations mentions other grouping devices besides parentheses: fraction bars and square root bars. Notice how the placement of the fraction bar affects the expressions in the next example.
</p>

<example>
    <p>
        <ol label="a">
            <li><p><m>
                \begin{aligned}
                \frac{1+2}{3\cdot 4} \amp = \frac{3}{12}\\
                \amp = \frac{1}{4}
                \end{aligned}
            </m></p></li>
            <li><p><m>
                \begin{aligned}
                1+\frac{2}{3\cdot 4} \amp = 1+ \frac{2}{12}\\
                \amp = 1+\frac{1}{6}=\frac{7}{6}
                \end{aligned}                
            </m></p></li>
            <li><p><m>
                \begin{aligned}
                \frac{1+2}{3}\cdot 4 \amp = \frac{3}{3}\cdot 4\\
                \amp = 1\cdot 4=4
                \end{aligned}
            </m></p></li>
            <li><p><m>
                \begin{aligned}
                1+\frac{2}{3}\cdot 4 \amp = 1+ \frac{8}{3}\\
                \amp = \frac{3}{3}+\frac{8}{3}=\frac{11}{3}
                \end{aligned}                
            </m></p></li>
        </ol>
    </p>
</example>

</subsection>




<subsection><title>Radicals</title>
<p>
    You are already familiar with square roots. Every nonnegative number has two square roots, defined as follows:
    <me>s ~~\text{ is a square root of }~~ n ~~\text{ if }~~s^2 = n</me>
    There are several other kinds of roots, one of which is called the <term>cube root</term><idx>cube root</idx>, denoted by <m>\sqrt[3]{n}</m>. We define the cube root as follows.
</p>
<assemblage><title>Cube Roots</title><idx>cube root</idx>
<p>
    <me>b ~~\text{ is a cube root of }~~~ n ~~~\text{ if }~~~ b ~\text{ cubed equals }~~n.</me>
    In symbols, we write
    <me>b=\sqrt[3]{n} ~~\text{ if } ~~ b^3=n</me>
</p></assemblage>

<p>
    Although we cannot take the square root of a <em>negative number</em>, we can take the <em>cube root</em> of <em>any</em> real number. For example,
    <me>\sqrt[3]{64}=4 ~~\text{ because }~~ 4^3=64</me>
    and
    <me>\sqrt[3]{-27}=-3 ~~\text{ because }~~ (-3)^3=-27</me>
</p>
<p>
    In the order of operations, simplifying radicals and powers comes after parentheses but before products and quotients.
</p>

<example>
    <p>
        Simplify each expression.
        <ol label="a">
            <li><m>3\sqrt[3]{-8}</m></li>
            <li><m>2-\sqrt[3]{-125}</m></li>
            <li><m>\displaystyle{\frac{6-\sqrt[3]{-27}}{2}}</m></li>
        </ol>
    </p>
<solution>
    <p>
        <ol label="a">
            <li><m>3\sqrt[3]{-8}=3(-2)=-6</m></li>
            <li><m>2-\sqrt[3]{-125}=2-(-5)=7</m></li>
            <li><m>\displaystyle{\frac{6-\sqrt[3]{-27}}{2}=\frac{6-(-3)}{2}=\frac{9}{2}}</m></li>
        </ol>
    </p>
</solution>
</example>

</subsection>



<subsection xml:id="Scientific-Notation"><title>Scientific Notation</title>
<p>
    Scientists and engineers regularly encounter very large numbers such as
    <me>5,980,000,000,000,000,000,000,000</me>
    (the mass of the Earth in kilograms) and very small numbers such as
    <me>0.000 \,000 \,000 \,000 \,000 \,000 \,000 \,001 \,67</me>
    (the mass of a hydrogen atom in grams). These numbers can be written in a more compact and useful form by using powers of <m>10</m>.
</p>
<p>
    In our base <m>10</m> number system, multiplying a number by a positive power of <m>10</m> has the effect of moving the decimal place <m>k</m> places to the right, where <m>k</m> is the exponent in the power of <m>10</m>. For example,
    <me>3.529 \times 10^2 = 352.9 ~~~\text{ and }~~~ 25 \times 104 = 250,000</me>
    Multiplying by a power of <m>10</m> with a negative exponent moves the decimal place to the left. For example,
    <me>1728 \times 10^{-3} = 1.728 ~~~\text{ and }~~~ 4.6 \times 10^{-5} = 0.000046</me>
</p>
<p>
    Using this property, we can write any number as the product of a number between <m>1</m> and <m>10</m> (including <m>1</m>) and a power of <m>10</m>. For example, the mass of the Earth and the mass of a hydrogen atom can be expressed as
    <me>5.98 \times 1024 \text{ kilograms}\hphantom{blank} \text{ and }\hphantom{blank}1.67 \times 10^{-24} \text{ gram}</me>
    respectively. A number written in this form is said to be expressed in <term>scientific notation</term><idx>scientific notation</idx>.
</p>

<assemblage><title>To Write a Number in Scientific Notation:</title>
<p>
    <ol label="1">
        <li><p>
            Locate the decimal point so that there is exactly one nonzero digit to its left.
        </p></li>
        <li><p>
            Count the number of places you moved the decimal point: This determines the power of <m>10</m>.
            <ol label="a">
                <li><p>
                    If the original number is greater than <m>10</m>, the exponent is positive.
                </p></li>
                <li><p>
                    If the original number is less than <m>1</m>, the exponent is negative.
                </p></li>
            </ol>
        </p></li>
    </ol>
</p>
</assemblage>

<example>
    <p>
        Write each number in scientific notation.
        <ol label="a">
            <li><p><m>
                \begin{aligned}
                478,000 \amp = 4.78000 \times 10^5\\
                \amp = 4.78 \times 10^5
                \end{aligned}
            </m></p></li>
            <li><p><m>
                \begin{aligned}
                0.00032 \amp= 00003.2 \times 10^{-4}\\
                \amp = 3.2 \times 10^{-4}
                \end{aligned}
            </m></p></li>
        </ol>
    </p>
</example>

<example>
    <p>
        The average American eats <m>110</m> kilograms of meat per year. It takes about <m>16</m> kilograms of grain to produce <m>1</m> kilogram of meat, and advanced farming techniques can produce about <m>6000</m> kilograms of grain on each hectare of arable land. (The hectare is <m>10,000</m> square meters, or just under <m>2 1⁄2</m> acres.) Now, the total land area of the Earth is about <m>13</m> billion hectares, but only about <m>11\%</m> of that land is arable. Is it possible for each of the <m>5.5</m> billion people on Earth to eat as much meat as Americans do?
    </p>
<solution>
    <p>
        First we will compute the amount of meat necessary to feed every person on Earth <m>110</m> kilograms per year. There are <m>5.5 \times 10^9</m> people on Earth.
        <me>(5.5\times 10^9 \text{ people})\times (110 \text{ kg/person}) = 6.05\times 10^{11} \text{ kg of meat}</me>
        Next we will compute the amount of grain needed to produce that much meat.
        <me>
            (16 \text{ kg of grain/kg of meat})\times (6.05\times 10^{11} \text{ kg of meat}) = 9.68\times 10^{12} \text{ kg of grain}
        </me>
        Next we will see how many hectares of land are needed to produce that much grain.
        <me>
            (9.68\times 10^{12} \text{ kg of grain})\div(6000 \text{ kg/hectare}) = 1.613\times 10^9 \text{ hectares}
        </me>
        Finally, we will compute the amount of arable land available for grain production.
        <me>0.11\times (13\times 10^9 \text{ hectares}) = 1.43\times 10^9 \text{ hectares}</me>
        Thus, even if we use every hectare of arable land to produce grain for livestock, we will not have enough to provide every person on Earth with <m>110</m> kilograms of meat per year.
    </p>
</solution>
</example>

</subsection>


<!--


-->
</section>

<section xml:id="appendix-Linear-Equations-and-Inequalities"><title>Linear Equations and Inequalities</title><introduction>
<p>An <term>equation</term> is just a mathematical statement that two expressions are equal. Equations relating two variables are particularly useful. If we know the value of one of the variables, we can find the corresponding value of the other variable by solving the equation.</p>

<example>
    <p>
        The equation <m>w = 6h</m> gives Loren's wages, <m>w</m>, in terms of the number of hours she works, <m>h</m>. How many hours does Loren need to work next week if she wants to earn <m>\$225</m>?
    </p>
<solution>
    <p>
        We know that <m>w = 225</m>, and we would like to know the value of <m>h</m>. We substitute the value for <m>w</m> into our equation and then solve for <m>h</m>.
        <md>
            <mrow>w \amp= 6h\amp\amp \text{Substitute 225 for }w.</mrow>
            <mrow>\alert{225} \amp= 6h\amp\amp \text{Divide both sides by } 6.</mrow>
            <mrow>\frac{225}{\alert{6}}\amp = \frac{6h}{\alert{6}}\amp\amp \text{Simplify.}</mrow>
            <mrow>375.5\amp = h</mrow>
        </md>
        Loren must work <m>37.5</m> hours in order to earn <m>\$225</m>. In reality, Loren will probably have to work for <m>38</m> hours, because most employers do not pay for portions of an hour's work. Thus, Loren needs to work for <m>38</m> hours.
    </p>
</solution></example>

<p>
    To solve an equation we can generate simpler equations that have the same solutions. Equations that have identical solutions are called <term>equivalent equations</term><idx>equivalent equations</idx>. For example,
    <me>3x - 5 = x + 3</me>
    and
    <me>2x = 8</me>
    are equivalent equations because the solution of each equation is <m>4</m>. Often we can find simpler equivalent equations by undoing in reverse order the operations performed on the variable.
</p></introduction>

<subsection><title>Solving Linear Equations</title>
<p>
    <term>Linear</term>, or first-degree, equations can be written so that every term is either a constant or a constant times the variable. The equations above are examples of linear equations. Recall the following rules for solving linear equations.
</p>

<assemblage><title>To Generate Equivalent Equations</title>
<p>
    <ol label="1">
        <li><p>
            We can add or subtract the <em>same</em> number on <em>both</em> sides of an equation.
        </p></li>
        <li><p>
            We can multiply or divide <m>both</m> sides of an equation by the <m>same</m> number (except zero).
        </p></li>
    </ol>
</p>
</assemblage>

<p>
    Applying either of these rules produces a new equation equivalent to the old one and thus preserves the solution.We use the rules to isolate the variable on one side of the equation.
</p>

<example><p>
    Solve the equation <m>3x - 5 = x + 3</m>.
</p>
<solution>
    <p>
        We first collect all the variable terms on one side of the equation, and the constant terms on the other side.
        <md>
            <mrow>
                3x - 5 \alert{{}- x} \amp = x + 3 \alert{{}- x}\amp\amp \text{Subtract }x \text{ from both sides.}
            </mrow>
            <mrow>
                2x - 5 \amp = 3\amp\amp \text{Simplify.}
            </mrow>
            <mrow>
                2x - 5 \alert{{}+ 5} \amp = 5 \alert{{}+ 5}\amp\amp \text{Add 5 to both sides.}
            </mrow>
            <mrow>
                2x \amp = 8 \amp\amp\text{Simplify.}
            </mrow>
            <mrow>
                \frac{2x}{\alert{2}}\amp =\frac{8}{\alert{2}}\amp\amp\text{Divide both sides by 2.}
            </mrow>
            <mrow>x\amp = 4\amp\amp\text{Simplify.}</mrow>
        </md>
        The solution is <m>4</m>. (You can check the solution by substituting <m>4</m> into the original equation to show that a true statement results.)
    </p>
</solution></example>

<p>
    The following steps should enable you to solve any linear equation. Of course, you may not need all the steps for a particular equation.
</p>

<assemblage><title>To Solve a Linear Equation:</title>
<p>
    <ol label="1">
        <li><p>
            Simplify each side of the equation separately.
            <ol label="a">
                <li><p>
                    Apply the distributive law to remove parentheses.
                </p></li>
                <li><p>
                    Collect like terms.
                </p></li>
            </ol>
        </p></li>
        <li><p>
            By adding or subtracting appropriate terms to both sides of the equation, get all the variable terms on one side and all the constant terms on the other.
        </p></li>
        <li><p>
            Divide both sides of the equation by the coefficient of the variable.
        </p></li>
    </ol>
</p>
</assemblage>

<example>
    <p>
        Solve <m>3(2x - 5) - 4x = 2x - (6 - 3x)</m>.
    </p>
<solution>
    <p>
        We begin by simplifying each side of the equation.
        <md>
            <mrow>
                3(2x - 5) - 4x \amp = 2x - (6 - 3x) \amp\amp\text{Apply the distributive law.}
            </mrow>
            <mrow>
                6x - 15 - 4x \amp= 2x - 6 + 3x \amp\amp\text{Combine like terms on each side.}
            </mrow>
            <mrow>
                2x - 15 \amp= 5x - 6
            </mrow>
        </md>
        Next, we collect all the variable terms on the left side of the equation, and all the constant terms on the right side.
        <md>
            <mrow>
                2x - 15 - 5x + 15 \amp= 5x - 6 - 5x + 15 \amp\amp\text{Add }-5x + 15 \text{ to both sides.}
            </mrow>
            <mrow>
                -3x \amp= 9
            </mrow>
        </md>
        Finally, we divide both sides of the equation by the coefficient of the variable.
        <md>
            <mrow>
                -3x \amp= 9 \amp\amp\text{Divide both sides by }-3.
            </mrow>
            <mrow>
                x \amp =-3
            </mrow>
        </md>
        The solution is <m>-3</m>.
    </p>
</solution></example>

</subsection>


<subsection><title>Formulas</title>
<p>
    A <term>formula</term><idx>formula</idx> is an equation that relates several variables. For example, the equation 
    <me>P = 2l + 2w</me>
    gives the perimeter of a rectangle in terms of its length and width.
</p>
<p>
    Suppose we have some wire fence to enclose an exercise area for rabbits, and we would like to see what dimensions are possible for different rectangles with that perimeter. In this case, it would be more useful to have a formula for the length of the rectangle in terms of its perimeter and its width. We can find such a formula by solving the perimeter formula for <m>l</m> in terms of <m>P</m> and <m>w</m>.
    <md>
        <mrow>2l + 2w \amp= P \amp\amp\text{Subtact }2 w \text{ from both sides.}</mrow>
        <mrow>2l \amp= P - 2w \amp\amp\text{Divide both sides by 2.}</mrow>
        <mrow>l \amp= \frac{P - 2w}{2}</mrow>
    </md>
        The result is a new formula that gives the length of a rectangle in terms of its perimeter and its width.
</p>

<example>
<p>
    The formula <m>5F = 9C + 160</m> relates the temperature in degrees Fahrenheit, <m>F</m>, to the temperature in degrees Celsius, <m>C</m>. Solve the formula for <m>C</m> in terms of <m>F</m>.
</p>
<solution>
    <p>
        We begin by isolating the term that contains <m>C</m>.
        <md>
            <mrow>
                5F \amp= 9C + 160 \amp\amp\text{Subtract 160 from both sides.}
            </mrow>
            <mrow>
                5F - 160 \amp = 9C \amp\amp\text{Divide both sides by 9.}
            </mrow>
            <mrow>
                \frac{5F - 160}{9} \amp =C
            </mrow>
        </md>
        We can also write the formula for <m>C</m> in terms of <m>F</m> as <m>C = \dfrac{5}{9}F - \dfrac{160}{9}</m>.
    </p>
</solution></example>

<example>
    <p>
        Solve <m>3x - 5y = 40</m> for <m>y</m> in terms of <m>x</m>.
    </p>
<solution>
    <p>
        We isolate <m>y</m> on one side of the equation.
        <md>
            <mrow>
                3x - 5y \amp = 40\amp\amp\text{Subtract }3x \text{ from both sides.}
            </mrow>
            <mrow>
                -5y \amp = 40 - 3x\amp\amp\text{Divide both sides by }-5.
            </mrow>
            <mrow>
                 \frac{-5y}{-5} \amp = \frac{40 - 3x}{-5}\amp\amp\text{Simplify both sides.}               
            </mrow>
            <mrow>
                y \amp = -8 + \frac{3}{5}x
            </mrow>
        </md>
    </p>
</solution></example>

</subsection>



<subsection><title>Linear Inequalities</title>
<p>
    The symbol <m>\gt</m> is called an <term>inequality symbol</term><idx>inequality symbol</idx>, and the statement <m>a\gt b</m> is called an <term>inequality</term><idx>inequality</idx>. There are four inequality symbols:
    <md>
        <mrow>\amp\gt \amp\amp\text{is greater than}</mrow>
        <mrow>\amp\lt \amp\amp\text{is less than}</mrow>
        <mrow>\amp\ge \amp\amp\text{is greater than or equal to}</mrow>
        <mrow>\amp\le \amp\amp\text{is less than or equal to}</mrow>
    </md>
</p>
<p>
    Inequalities that include the symbols <m>\gt</m> or <m>\le</m> are called <term>strict inequalities</term><idx>strict inequalities</idx>; those that include <m>\ge</m> or <m>\le</m> are called <term>nonstrict</term><idx>nonstrict</idx>.
</p>
<p>
    If we multiply or divide both sides of an inequality by a negative number, the direction of the inequality must be reversed. For example, if we multiply both sides of the inequality
    <me>2\lt 5</me>
    by <m>-3</m>, we get
    <md>
        <mrow>\alert{-3}(2) \amp\gt \alert{-3}(5)\amp\amp \text{Change inequality symbol from }\lt \text{ to }\gt.</mrow>
        <mrow>-6 \amp\gt -15.</mrow>
    </md>
    Because of this property, the rules for solving linear equations must be revised slightly for solving linear inequalities.
</p>

<assemblage><title>To Solve a Linear Inequality:</title>
<p>
    <ol label="1">
        <li><p>
            We may add or subtract the same number to both sides of an inequality without changing its solutions.
        </p></li>
        <li><p>
            We may multiply or divide both sides of an inequality by a <em>positive</em> number without changing its solutions.
        </p></li>
        <li><p>
            If we multiply or divide both sides of an inequality by a <em>negative</em> number, we must <em>reverse the direction of the inequality symbol</em>.
        </p></li>
    </ol>
</p>
</assemblage>

<example>
    <p>
        Solve the inequality <m>4 - 3x \ge -17</m>.
    </p>
<solution>
    <p>
        Use the rules above to isolate <m>x</m> on one side of the inequality.
        <md>
            <mrow>
                4 - 3x \amp\ge -17\amp\amp\text{Subtract 4 from both sides.}
            </mrow>
            <mrow>
                -3x \amp\ge -21\amp\amp\text{Divide both sides by }-3.
            </mrow>
            <mrow>x \amp\le 7</mrow>
        </md>
        Notice that we reversed the direction of the inequality when we divided by <m>-3</m>. Any number less than or equal to <m>7</m> is a solution of the inequality.
    </p>
</solution></example>

<p>
    A <term>compound inequality</term><idx>compound inequality</idx> involves two inequality symbols.
</p>

<example xml:id="example-compound-inequality">
    <p>
        Solve <m>4 \le 3x + 10 \le 16</m>.
    </p>
<solution>
    <p>
        We isolate <m>x</m> by performing the same operations on all three sides of the inequality.
        <md alignment="alignat">
            <mrow>
                4 \amp{}\le{} 3x\amp {}+{}\amp 10 \amp{}\le{} 16\hphantom{blank} \amp\text{Subtract }10.
            </mrow>
            <mrow>
                -6 \amp{}\le{} \amp 3x\amp {}{}\amp {}\le{} 6\hphantom{blank} \amp\text{Divide by }3.
            </mrow>
            <mrow>
                -2 \amp{}\le{} \amp x\amp {}{}\amp{}\le{} 2\hphantom{blank}\amp
            </mrow>
        </md>
        The solutions are all numbers between <m>-2</m> and <m>2</m>, inclusive.
    </p>
</solution></example>

</subsection>



<subsection><title>Interval Notation</title>
<p>
    The solutions of the inequality in <xref ref="example-compound-inequality" text="type-global"/> form an interval. An <term>interval</term><idx>interval</idx> is a set that consists of all the real numbers between two numbers <m>a</m> and <m>b</m>.
</p>
<p>
    The set <m>-2 \le x \le 2</m> includes its endpoints <m>-2</m> and <m>2</m>, so we call it a <term>closed interval</term><idx>closed interval</idx>, and we denote it by <m>[-2, 2]</m> (see <xref ref="fig-numline-closed-vs-open" text="type-global"/>a). The square brackets tell us that the endpoints are included in the interval. An interval that does not include its endpoints, such as <m>-2 \lt x \lt 2</m>, is called an <term>open interval</term><idx>open interval</idx>, and we denote it with round brackets, <m>(-2, 2)</m> (see <xref ref="fig-numline-closed-vs-open" text="type-global"/>b).
    <figure xml:id="fig-numline-closed-vs-open"><image source="images/fig-numline-closed-vs-open"><description>closed interval and open interva</description></image><caption>       
    </caption> </figure>
</p>

<warning>
    <p>
        Do not confuse the open interval <m>(-2, 2)</m> with the point <m>(-2, 2)</m>! The notation is the same, so you must decide from the context whether an interval or a point is being discussed.
    </p>
</warning>

<p>
    We can also discuss <term>infinite intervals</term><idx>infinite intervals</idx>, such as <m>x\lt 3</m> and <m>x\ge -1</m>, shown in <xref ref="fig-numline-infinite-intervals" text="type-global"/>. We denote the interval <m>x\lt 3</m> by <m>(-\infty, 3)</m>, and the interval <m>x\ge -1</m> by <m>[-1, \infty)</m>. The symbol <m>\infty</m>, for infinity, does not represent a specific real number; it indicates that the interval continues forever along the real line.
   <figure xml:id="fig-numline-infinite-intervals"><caption></caption><image source="images/fig-numline-infinite-intervals"  width="70%"><description>infinite intervals</description></image> </figure> 
</p>
<p>
    Finally, we can combine two or more intervals into a larger set. For example, the set consisting of <m>x\lt -1</m> or <m>x\gt 2</m>, shown in <xref ref="fig-numline-disjoint-infinite-intervals" text="type-global"/>, is the <term>union</term><idx>union</idx> of two intervals and is denoted by <m>(-\infty,-2) \cup (2,\infty)</m>.
    <figure xml:id="fig-numline-disjoint-infinite-intervals"><caption></caption><image source="images/fig-numline-disjoint-infinite-intervals"  width="70%"><description>number line with two disjoint infinite intervals</description></image> </figure>
</p>
<p>
    Many solutions of inequalities are intervals or unions of intervals.
</p>

<example>
    <p>
        Write each of the solution sets with interval notation and graph the solution set on a number line.
        <ol label="a">
            <li><p><m>3 \le x \lt 6</m></p></li>
            <li><p><m>x \ge -9</m></p></li>
            <li><p><m>x\le 1 ~~\text{ or }~~ x\gt 4</m></p></li>
            <li><p><m>-8 \lt x \le -5 ~~\text{ or }~~ -1 \le x \lt 3</m></p></li>
        </ol>
    </p>
<solution>
    <p>
        <ol label="a">
            <li><p>
                <m>[3, 6)</m>. This is called a <term>half-open</term><idx>half-open</idx> or <term>half-closed</term><idx>half-closed</idx> interval. (See <xref ref="fig-half-open-interval" text="type-global"/>.)
                <figure xml:id="fig-half-open-interval"><caption></caption><image source="images/fig-half-open-interval"  width="70%"><description>half-open interval</description></image> </figure>
            </p></li>
            <li><p>
                <m>[-9,\infty)</m>. We always use round brackets next to the symbol <m>\infty</m> because <m>\infty</m> is not a specific number and is not included in the set. (See <xref ref="fig-half-closed-infinite-interval" text="type-global"/>.)
                <figure xml:id="fig-half-closed-infinite-interval"><caption></caption><image source="images/fig-half-closed-infinite-interval"  width="70%"><description>half-closed infinite interval</description></image> </figure>
            </p></li>
            <li><p>
                <m>(-\infty, 1] \cup (4, \infty)</m>. The word <em>or</em> describes the union of two sets. (See <xref ref="fig-disjoint-infinite-intervals" text="type-global"/>.)
                <figure xml:id="fig-disjoint-infinite-intervals"><caption></caption><image source="images/fig-disjoint-infinite-intervals"  width="70%"><description>disjoint infinite intervals</description></image> </figure>
            </p></li>
            <li><p>
                <m>(-8,-5] \cup [-1, 3)</m>. (See <xref ref="fig-disjoint-finite-intervals" text="type-global"/>.)
                <figure xml:id="fig-disjoint-finite-intervals"><caption></caption><image source="images/fig-disjoint-finite-intervals"  width="70%"><description>disjoint finite intervals</description></image> </figure>
            </p></li>
        </ol>
        
    </p>
</solution>
</example>


</subsection>

</section>
<!-- 


    -->


<section xml:id="appendix-Algebraic-Expressions-and-Problem-Solving"><title>Algebraic Expressions and Problem Solving</title><introduction>
<p>You are familiar with the use of letters, or <term>variables</term>, to stand for unknown numbers in equations or formulas. Variables are also used to represent numerical quantities that change over time or in different situations. For example, <m>p</m> might stand for the atmospheric pressure at different heights above the Earth's surface. Or <m>N</m> might represent the number of people infected with cholera <m>t</m> days after the start of an epidemic.</p>
<p>
    An <term>algebraic expression</term><idx>algebraic expression</idx> is any meaningful combination of numbers, variables, and symbols of operation. Algebraic expressions are used to express relationships between variable quantities.
</p>

<example xml:id="example-Loren">
    <p>
        Loren makes <m>\$6</m> an hour working at the campus bookstore.
        <ol label="a">
            <li><p>
                Choose a variable for the number of hours Loren works per week.
            </p></li>
            <li><p>
                Write an algebraic expression for the amount of Loren's weekly earnings.
            </p></li>
        </ol>
    </p>
<solution>
    <p>
        <ol label="a">
            <li><p>
                Let <m>h</m> stand for the number of hours Loren works per week.
            </p></li>
            <li><p>
                The amount Loren earns is given by
                <me>6\times (\text{number of hours Loren worked})</me>
                or <m>6\cdot h</m>. Loren's weekly earnings can be expressed as <m>6h</m>.
            </p></li>
        </ol>
    </p>
</solution></example>

<p>
    The algebraic expression <m>6h</m> represents the amount of money Loren earns <em>in terms of</em> the number of hours she works. If we substitute a specific value for the variable in an expression, we find a numerical value for the expression. This is called <term>evaluating</term><idx>evaluating</idx> the expression.
</p>

<example>
    <p>
        If Loren from <xref ref="example-Loren" text="type-global"/> works for <m>16</m> hours in the bookstore this week, how much will she earn?
    </p>
<solution>
    <p>
        Evaluate the expression <m>6h</m> for <m>h=\alert{16}</m>.
        <me>
            6h = 6(\alert{16}) = 96
        </me>
        Loren will make <m>\$96</m>.
    </p>
</solution></example>

<example xml:id="example-April">
    <p>
        April sells environmentally friendly cleaning products. Her income consists of <m>\$200</m> per week plus a commission of <m>9\%</m> of her sales.
        <ol label="a">
            <li><p>
                Choose variables to represent the unknown quantities and write an algebraic expression for April's weekly income in terms of her sales.
            </p></li>
            <li><p>
                Find April's income for a week in which she sells <m>\$350</m> worth of cleaning products.
            </p></li>
        </ol>
    </p>
<solution>
    <p>
        <ol label="a">
            <li><p>
                Let <m>I</m> represent April's total income for the week, and let <m>S</m> represent the total amount of her sales. We translate the information from the problem into mathematical language as follows:
                <md>
                    <mrow>
                       \text{Her income consists of }\$200 . . .\text{ plus }. . . 9\% \text{ of her sales} 
                    </mrow>
                    <mrow>
                        I \hphantom{consists of}= \hphantom{of}200 \hphantom{plus+}+ \hphantom{....}0.09 \hphantom{of her}S
                    </mrow>
                </md>
                Thus, <m>I = 200 + 0.09S</m>.
            </p></li>
            <li><p>
                We want to evaluate our expression from part (a) with <m>S = 350</m>. We substitute <m>\alert{350}</m> for <m>S</m> to find
                <me>I = 200 + 0.09(\alert{350})</me>
                Following the order of operations, we perform the multiplication before the addition. Thus, we begin by computing <m>0.09(350)</m>.
                <md>
                    <mrow>
                        I \amp = 200 + 0.09(350)\amp\amp\text{Multiply }0.09 (350) \text{ first.}
                    </mrow>
                    <mrow>\amp = 200 + 31.5</mrow>
                    <mrow>\amp = 231.50</mrow>
                </md>
                April's income for the week is <m>\$231.50</m>.
            </p></li>
        </ol>
    </p>
</solution></example>

<remark><title><!--<image source="images/icon-GC.jpg"  width="8%"><description>Graphing Calculator</description></image>-->Calculator Tip</title>
<p>
    On a scientific or a graphing calculator, we can enter the expression from <xref ref="example-April" text="type-global"/> just as it is written:
    <me>200 + 0.09 ~ \boxed{\times}~ 350 ~\boxed{\text{ENTER}}</me>
    The calculator will perform the operations in the correct order—multiplication first.
</p></remark>

<example xml:id="example-parcel">
    <p>
        Economy Parcel Service charges <m>\$2.80</m> per pound to deliver a package from Pasadena to Cedar Rapids. Andrew wants to mail a painting that weighs <m>8.3</m> pounds, plus whatever packing material he uses.
        <ol label="a">
            <li><p>
                Choose variables to represent the unknown quantities and write an expression for the cost of shipping Andrew's painting.
            </p></li>
            <li><p>
                Find the shipping cost if Andrew uses <m>2.9</m> pounds of packing material.
            </p></li>
        </ol>
    </p>
<solution>
    <p>
        <ol label="a">
            <li><p>
                Let <m>C</m> stand for the shipping cost and let <m>w</m> stand for the weight of the packing material. Andrew must find the total weight of his package first, then multiply by the shipping charge. The total weight of the package is <m>8.3 + w</m> pounds. We use parentheses around this expression to show that it should be computed first, and the sum should be multiplied by the shipping charge of <m>\$2.80</m> per pound. Thus,
                <me>C = 2.80(8.3 + w)</me>
            </p></li>
            <li><p>
                Evaluate the formula from part (a) with <m>w = \alert{2.9}</m>.
                <md>
                    <mrow>
                        C \amp = 2.80(8.3 + \alert{2.9})\amp\amp\text{Add inside parentheses.}
                    </mrow>
                    <mrow>
                        \amp = 2.80(11.2)\amp\amp\text{Multiply.}
                    </mrow>
                    <mrow>
                        \amp = 31.36
                    </mrow>
                </md>
                The cost of shipping the painting is <m>\$31.36</m>.
            </p></li>
        </ol>
    </p>
</solution></example>

<remark><title><!--<image source="images/icon-GC.jpg"  width="8%"><description>Graphing Calculator</description></image>-->Calculator Tip</title>
<p>
    On a calculator, we enter the expression for <m>C</m> in the order it appears, including the parentheses. (Experiment to see whether your calculator requires you to enter the <c>×</c> symbol after 2.80.) The keying sequence
    <me>2.80 \times ( 8.3 + 2.9 ) \boxed{\text{ENTER}}</me>
    gives the correct result, <m>31.36</m>.
</p>
</remark>

<warning>
    <p>
        If we omit the parentheses, the calculator will perform the multiplication before the addition. Thus, the keying sequence
        <me>2.80 \times 8.3 + 2.9</me>
        gives an incorrect result for <xref ref="example-parcel" text="type-global"/>. (The sequence
        <me>8.3 + 2.9 \times 2.80</me>
        does not work either!)
    </p>
</warning></introduction>

<subsection><title>Problem Solving</title>
<p>
    Problem solving often involves translating a real-life problem into a computer programming language, or, in our case, into algebraic expressions. We can then use algebra to solve the mathematical problem and interpret the solution in the context of the original problem. Here are some guidelines for problem solving with algebraic equations.
</p>

<assemblage><title>Guidelines for Problem Solving</title>
<p>
    <ol>
        <li><p>
            Identify the unknown quantity and assign a variable to represent it.
        </p></li>
        <li><p>
            Find some quantity that can be expressed in two different ways and write an equation.
        </p></li>
        <li><p>
            Solve the equation.
        </p></li>
        <li><p>
            Interpret your solution to answer the question in the problem.
        </p></li>
    </ol>
</p>
</assemblage>

<p>
    In step 1, begin by writing an English phrase to describe the quantity you are looking for. Be as specific as possible—if you are going to write an equation about this quantity, you must understand its properties! Remember that your variable must represent a numerical quantity. For example, <m>x</m> can represent the <em>speed</em> of a train, but not just “the train.”
</p>
<p>
    Writing an equation is the hardest part of the problem. Note that the quantity mentioned in step 2 will probably <em>not</em> be the same unknown quantity you are looking for, but the algebraic expressions you write <em>will</em> involve your variable. For example, if your variable represents the <em>speed</em> of a train, your equation might be about the <em>distance</em> the train traveled.
</p>

</subsection>



<subsection><title>Supply and Demand</title>
<p>
    The law of supply and demand is fundamental in economics. If you increase the price of a product, the supply increases because its manufacturers are willing to provide more of the product, but the demand decreases because consumers are not willing to buy as much at a higher price. The price at which the demand for a product equals the supply is called the <term>equilibrium price</term><idx>equilibrium price</idx>.
</p>

<example>
    <p>
        The Coffee Connection finds that when it charges <m>p</m> dollars for a pound of coffee, it can sell <m>800 - 60p</m> pounds per month. On the other hand, at a price of <m>p</m> dollars a pound, International Food and Beverage will supply the Connection with <m>175 + 40p</m> pounds of coffee per month. What price should the Coffee Connection charge for a pound of coffee so that its monthly inventory will sell out?
    </p>
<solution>
    <p>
        <ol>
            <li><p>
                We are looking for the equilibrium price, <m>p</m>.
            </p></li>
            <li><p>
                The Coffee Connection would like the demand for its coffee to equal its supply. We equate the expressions for supply and for demand to obtain the equation
                <me>800 - 60p = 175 + 40p</me>
            </p></li>
            <li><p>
                Solve the equation. To get all terms containing the variable, <m>p</m>, on one side of the equation, we add <m>60p</m> to both sides and subtract <m>175</m> from both sides to obtain
                <md>
                    <mrow>
                        800 - 60p + \alert{60p - 175} \amp= 175 + 40p + \alert{60p - 175}
                    </mrow>
                    <mrow>
                        625 \amp = 100p \amp\amp\text{Divide both sides by }100.
                    </mrow>
                    <mrow>
                        6.25 \amp = p
                    </mrow>
                </md>
            </p></li>
            <li><p>
                The Coffee Connection should charge <m>\$6.25</m> per pound for its coffee.
            </p></li>
        </ol>
    </p>
</solution></example>
</subsection>

<subsection><title>Percent Problems</title>
<p>
    Recall the basic formula for computing percents.
</p>

<assemblage><title>Percent Formula</title>
<p>
    <me>P = rW</me>
    the <term>P</term>art (or percent) = the percentage <term>r</term>ate <m>\times</m> the <term>W</term>hole Amount
</p></assemblage>

<p>
    A <term>percent increase</term><idx>percent increase</idx> or <term>percent decrease</term><idx>percent decrease</idx> is calculated as a fraction of the <em>original</em> amount. For example, suppose you make <m>\$16.00</m> an hour now, but next month you are expecting a <m>5\%</m> raise. Your new salary should be
    <me>
        \stackrel{\text{Original salary}}{\$16.00} + \stackrel{\text{Increase}}{0.05(\$16.00)} = \stackrel{\text{New Salary}}{\$16.80}
    </me>
</p>
<example xml:id="example-percent-increase">
    <p>
        The price of housing in urban areas increased <m>4\%</m> over the past year. If a certain house costs <m>\$100,000</m> today, what was its price last year?
    </p>
<solution>
    <p>
        <ol>
            <li><p>
                Let <m>c</m> represent the cost of the house last year.
            </p></li>
            <li><p>
                Express the current price of the house in two different ways. During the past year, the price of the house increased by <m>4\%</m>, or <m>0.04c</m>. Its current price is thus
                <me>\stackrel{\text{Original cost}}{(1)c} + \stackrel{\text{Price increase}}{0.04c} = c(1 + 0.04) = 1.04c</me>
                This expression is equal to the value given for current price of the house:
                <me>1.04c = 100,000</me>
            </p></li>
            <li><p>
                To solve this equation, we divide both sides by <m>1.04</m> to find
                <me>c = \frac{100,000}{1.04}= 96,153.846</me>
            </p></li>
            <li><p>
                To the nearest cent, the cost of the house last year was <m>\$96,153.85</m>.
            </p></li>
        </ol>
    </p>
</solution></example>

<warning>
    <p>
        In <xref ref="example-percent-increase" text="type-global"/>, it would be incorrect to calculate last year's price by subtracting <m>4\%</m> of <m>\$100,000</m> from <m>\$100,000</m> to get <m>\$96,000</m>. (Do you see why?)
    </p>
</warning>
</subsection>



<subsection><title>Weighted Averages</title>
<p>
    We find the <term>average</term><idx>average</idx>, or <term>mean</term><idx>mean</idx>, of a set of values by adding up the values and dividing the sum by the number of values. Thus, the average, <m>\overline{x}</m>, of the numbers <m>x_1, x_2, \ldots , x_n</m> is given by
    <me>
        \overline{x} = \frac{x_1 + x_2 + ·· ·+x_n}{n}
    </me>
    In a <term>weighted average</term><idx>weighted average</idx>, the numbers being averaged occur with different frequencies or are weighted differently in their contribution to the average value. For instance, suppose a biology class of 12 students takes a 10-point quiz. Of the 12 students, 2 receive 10s, three receive 9s, 5 receive 8s, and 2 receive scores of 6. The average score earned on the quiz is then
    <me>\overline{x} = \frac{\alert{2}(10) + \alert{3}(9) + \alert{5}(8) + \alert{2}(6)}{12}=8.25</me>
    The numbers in color are called the weights—in this example they represent the number of times each score was counted. Note that <m>n</m>, the total number of scores, is equal to the sum of the weights:
    <me>12 = 2 + 3 + 5 + 2</me>
</p>

<example xml:id="example-weighted-average">
    <p>
        Kwan's grade in his accounting class will be computed as follows: Tests count for <m>50\%</m> of the grade, homework counts for <m>20\%</m>, and the final exam counts for <m>30\%</m>. If Kwan has an average of <m>84</m> on tests and <m>92</m> on homework, what score does he need on the final exam to earn a grade of <m>90</m>?
    </p>
<solution>
    <p>
        <ol>
            <li><p>
                Let <m>x</m> represent the final exam score Kwan needs.
            </p></li>
            <li><p>
                Kwan's grade is the weighted average of his test, homework, and final exam scores.
                <me>\frac{\alert{0.50}(84) + \alert{0.20}(92) + \alert{0.30}x}{1.00}=90</me>
                (The sum of the weights is 1.00, or 100% of Kwan’s grade.) Multiply both sides of the equation by <m>1.00</m> to get
                <me>0.50(84) + 0.20(92) + 0.30x = 1.00(90)</me>
            </p></li>
            <li><p>
                Solve the equation. Simplify the left side first.
                <md>
                    <mrow>
                        60.4 + 0.30x \amp = 90\amp\amp\text{Subtract 60.4 from both sides.}
                    </mrow>
                    <mrow>
                        0.30x \amp = 29.6\amp\amp\text{Divide both sides by 0.30.}
                    </mrow>
                    <mrow>
                        x \amp = 98.7
                    </mrow>
                </md>
            </p></li>
            <li><p>
                Kwan needs a score of <m>98.7</m> on the final exam to earn a grade of <m>90</m>.
            </p></li>
        </ol>
    </p>
</solution></example>

<p>
    In step 2 of <xref ref="example-weighted-average" text="type-global"/>, we rewrote the formula for a weighted average in a simpler form.
</p>

<assemblage><title>Weighted Average</title><idx>weighted average</idx>
<p>
    The sum of the weighted values equals the sum of the weights times the average value. In symbols,
    <me>w_1x_1 + w_2x_2 + \cdots + w_nx_n = W\, \overline{x}</me>
    where <m>W</m> is the sum of the weights.
</p></assemblage>

<p>
    This form is particularly useful for solving problems involving mixtures.
</p>

<example>
    <p>
        The vet advised Delbert to feed his dog Rollo with kibble that is no more than <m>8\%</m> fat. Rollo likes JuicyBits, which are <m>15\%</m> fat. LeanMeal is more expensive, but it is only <m>5\%</m> fat. How much LeanMeal should Delbert mix with <m>50</m> pounds of JuicyBits to make amixture that is <m>8\%</m> fat?
    </p>
<solution>
    <p>
        <ol>
            <li><p>
                Let <m>p</m> represent the number of pounds of LeanMeal needed.
            </p></li>
            <li><p>
                In this problem, we want the weighted average of the fat contents in the two kibbles to be <m>8\%</m>. The weights are the number of pounds of each kibble we use. It is often useful to summarize the given information in a table.</p><p>
                <sidebyside><tabular left="minor" right="minor" top="minor" bottom="minor" halign="center">
                    <col halign="left"/>
                    <col halign="center"/>
                    <col halign="center"/>
                    <col halign="center"/>
                    <row>
                        <cell><m></m></cell>
                        <cell><m>\%</m> fat</cell>
                        <cell>Total pounds</cell>
                        <cell>Pounds of fat</cell>
                    </row>
                    <row>
                        <cell>Juicy Bits</cell>
                        <cell><m>15\%</m></cell>
                        <cell><m>50</m></cell>
                        <cell>0.15(50)</cell>
                    </row>
                    <row>
                        <cell>LeanMeal</cell>
                        <cell><m>5\%</m></cell>
                        <cell><m>p</m></cell>
                        <cell>0.05p</cell>
                    </row>
                    <row>
                        <cell>Mixture</cell>
                        <cell><m>8\%</m></cell>
                        <cell><m>50+p</m></cell>
                        <cell>0.08(50+p)</cell>
                    </row>
                </tabular></sidebyside>
                The amount of fat in the mixture must come from adding the amounts of fat in the two ingredients. This gives us an equation,
                <me>0.15(50) + 0.05p = 0.08(50 + p)</me>
                This equation is an example of the formula for weighted averages.
            </p></li>
            <li><p>
                Simplify each side of the equation, then solve.
                <md>
                    <mrow>
                        7.5 + 0.05p \amp = 4 + 0.08p
                    </mrow>
                    <mrow>
                        3.5 \amp = 0.03p
                    </mrow>
                    <mrow>
                        p \amp = 116.\overline{6}
                    </mrow>
                </md>
            </p></li>
            <li><p>
                Delbert should mix <m>116\frac{2}{3}</m> pounds of LeanMeal with <m>50</m> pounds of JuicyBits to make a mixture that is <m>8\%</m> fat.
            </p></li>
        </ol>
    </p>
</solution></example>
</subsection>


</section>
<!--


    -->


<section xml:id="appendix-Graphs-and-Equations"><title>Graphs and Equations</title><introduction>
<p>
    Graphs are useful tools for studying mathematical relationships. A graph provides an overview of a quantity of data, and it helps us identify trends or unexpected occurrences. Interpreting the graph can help us answer questions about the data.
</p>
<p>
    For example, here are some data showing the atmospheric pressure at different altitudes. Altitude is given in feet, and atmospheric pressure is given in inches of mercury.</p><p>
    <sidebyside><tabular halign="center" left="minor" right="minor" top="minor" bottom="minor">
        <col halign="left"/>
        <col halign="center"/>
        <col halign="center"/>
        <col halign="center"/>
        <col halign="center"/>
        <col halign="center"/>
        <col halign="center"/>
        <col halign="center"/>    
        <row>
            <cell>Altitude (ft)</cell>
            <cell><m>0</m></cell>
            <cell><m>5000</m></cell>
            <cell><m>10,000</m></cell>
            <cell><m>20,000</m></cell>
            <cell><m>30,000</m></cell>
            <cell><m>40,000</m></cell>
            <cell><m>50,000</m></cell>
        </row>
        <row>
            <cell>Pressure (in. Hg)</cell>
            <cell><m>29.7</m></cell>
            <cell><m>24.8</m></cell>
            <cell><m>20.5</m></cell>
            <cell><m>14.6</m></cell>
            <cell><m>10.6</m></cell>
            <cell><m>8.5</m></cell>
            <cell><m>7.3</m></cell>
        </row>
    </tabular></sidebyside>
</p>
<p>
    We observe a generally decreasing trend in pressure as the altitude increases, but it is difficult to say anything more precise about this relationship. A clearer picture emerges if we plot the data. To do this, we use two perpendicular number lines called axes. We use the horizontal axis for the values of the first variable, altitude, and the vertical axis for the values of the second variable, pressure.
</p>
<p>
    The entries in the table are called <term>ordered pairs</term>, in which the <term>first component</term> is the altitude and the <term>second component</term> is the atmospheric pressure measured at that altitude. For example, the first two entries can be represented by <m>(0, 29.7)</m> and <m>(5000, 24.8)</m>. We plot the points whose <term>coordinates</term> are given by the ordered pairs, as shown in <xref ref="fig-pressure-vs-altitude" text="type-global"/>a.
    <figure xml:id="fig-pressure-vs-altitude"><caption></caption><image source="images/fig-pressure-vs-altitude"  width="80%"><description>graphs of pressure vs altitude</description></image></figure>
</p>
<p>
    We can connect the data points with a smooth curve as shown in <xref ref="fig-pressure-vs-altitude" text="type-global"/>b. In doing this, we are assuming that one variable changes smoothly with respect to the other, and in fact this is true for many physical situations. Thus, a smooth curve will thus serve as a good model.
</p></introduction>

<subsection><title>Reading a Graph</title>
<p>
    Once we have constructed a graph, we can use it to estimate values of the variables between the known data points.
</p>

<example xml:id="example-pressure-vs-altitude">
    <p>
        From the graph in <xref ref="fig-pressure-vs-altitude" text="type-global"/>b, estimate the following: 
        <ol label="a">
            <li><p>
                The atmospheric pressure measured at an altitude of <m>15,000</m> feet 
            </p></li>
            <li><p>
                The altitude at which the pressure is <m>12</m> inches of mercury
            </p></li>
        </ol>
    </p>
<solution>
    <p>
        <ol label="a">
            <li><p>
                The point with first coordinate <m>15,000</m> on the graph in <xref ref="fig-pressure-vs-altitude2" text="type-global"/> has second coordinate approximately <m>17.4</m>. We estimate the pressure at <m>15,000</m> feet to be <m>17.4</m> inches of mercury.
                <figure xml:id="fig-pressure-vs-altitude2"><caption></caption><image source="images/fig-pressure-vs-altitude2"  width="50%"><description>graph of pressure vs altitude with two points highlighted</description></image> </figure>
         </p></li>
         <li><p>
             The point on the graph with second coordinate <m>12</m> has first coordinate approximately <m>25,000</m>, so an atmospheric pressure of <m>12</m> inches of mercury occurs at about <m>25,000</m> feet.
         </p></li>
    </ol>
    </p>
</solution></example>

<p>
    We can also use the graph to obtain information about the relationship between altitude and pressure that would be difficult to see from the data alone.
</p>

<example>
    <p>
        <ol label="a">
            <li><p>
                For what altitudes is the pressure less than <m>18</m> inches of mercury?
            </p></li>
            <li><p>
                How much does the pressure decrease as the altitude increases from <m>15,000</m> feet to <m>25,000</m> feet?
            </p></li>
            <li><p>
                For which <m>10,000</m>-foot increase in altitude does the pressure change most rapidly?
            </p></li>
        </ol>
    </p>
<solution>
    <p>
        <ol label="a">
            <li><p>
                From the graph in <xref ref="fig-pressure-vs-altitude" text="type-global"/>b, we see that the pressure has dropped to <m>18</m> inches of mercury at about <m>14,000</m> feet, and that it continues to decrease as the altitude increases. Therefore, the pressure is less than <m>18</m> inches of mercury for altitudes greater than <m>14,000</m> feet.
            </p></li>
            <li><p>
                The pressure at <m>15,000</m> feet is approximately <m>17.4</m> inches of mercury, and at <m>25,000</m> feet it is <m>12</m> inches. This represents a decrease in pressure of <m>17.4 - 12</m>, or <m>5.4</m>, inches of mercury.
            </p></li>
            <li><p>
                By studying the graph we see that the pressure decreases most rapidly at low altitudes, so we conclude that the greatest drop in pressure occurs between <m>0</m> and <m>10,000</m> feet.
            </p></li>
        </ol>
    </p>
</solution></example>

</subsection>



<subsection><title>Graphs of Equations</title>
<p>
    In <xref ref="example-pressure-vs-altitude" text="type-global"/>, we used a graph to illustrate data given in a table. Graphs can also help us analyze models given by equations. Let us first review some facts about solutions of equations in two variables.
</p>
<p>
    An equation in two variables, such as y = 2x + 3, is said to be satisfied if the variables are replaced by a pair of numbers that make the statement true. The pair of numbers is called a <term>solution</term><idx>solution</idx> of the equation and is usually written as an ordered pair <m>(x, y)</m>. (The first number in the pair is the value of <m>x</m> and the second number is the value of <m>y</m>.)
</p>
<p>
    To find a solution of a given equation, we can assign a number to one of the variables and then solve for the second variable.
</p>

<example>
    <p>
        Find solutions to the equation <m>y = 2x + 3</m>.
    </p>
<solution>
    <p>
        We choose some values for <m>x</m>, say, <m>-2</m>, <m>0</m>, and <m>1</m>. Substitute these <m>x</m>-values into the equation to find a corresponding <m>y</m>-value for each.
        <md>
            <mrow>
                \amp\text{When }x=-2,\amp\amp y = 2(\alert{-2}) + 3 =-1
            </mrow>
            <mrow>
                \amp\text{When }x=0,\amp\amp y = 2(\alert{0}) + 3 =3
            </mrow>
            <mrow>
                \amp\text{When }x=1,\amp\amp y = 2(\alert{1}) + 3 =5
            </mrow>
        </md>
        Thus, the ordered pairs <m>(-2, -1)</m>, <m>(0, 3)</m>, and <m>(1, 5)</m> are three solutions of <m>y = 2x + 3</m>. We can also substitute values for <m>y</m>. For example, if we let <m>y = \alert{10}</m>, we have
        <me>\alert{10} = 2x + 3</me>
        Solving this equation for <m>x</m>, we find <m>7 = 2x</m>, or <m>x = 3.5</m>. This means that the ordered pair <m>(3.5, 10)</m> is another solution of the equation <m>y = 2x + 3</m>.
    </p>
</solution></example>

<p>
    An equation in two variables may have infinitely many solutions, so we cannot list them all. However, we can display the solutions on a graph. For this we use a <term>Cartesian</term><idx>Cartesian</idx> (or <term>rectangular</term><idx>rectangular</idx>) <term>coordinate system</term><idx>coordinate system</idx>, as shown in <xref ref="fig-coordinate-plane" text="type-global"/>.
</p>
<sidebyside  widths="40% 35%" margins="auto">
    <figure xml:id="fig-coordinate-plane"><caption></caption><image source="images/fig-coordinate-plane"><description>coordinate plane</description></image></figure>
    <figure xml:id="fig-line"><caption></caption><image source="images/fig-line"><description>line in the coordinate plane</description></image></figure>

</sidebyside>
<p>
    The <term>graph of an equation</term><idx>graph of an equation</idx> is a picture of its solutions. A point is included in the graph if its coordinates satisfy the equation, and if the coordinates do not satisfy the equation, the point is not part of the graph. A graph of <m>y = 2x + 3</m> is shown in <xref ref="fig-line" text="type-global"/>.
</p>
<p>
    This graph does not display <em>all</em> the solutions of the equation, but it shows important features such as the intercepts on the <m>x</m>- and <m>y</m>-axes. Because there is a solution corresponding to every real number <m>x</m>, the graph extends infinitely in either direction, as indicated by the arrows.
</p>

<example>
        <sidebyside widths="45% 40%">
       <p> Use the graph of <m>y = 0.5x^2 - 2</m> in <xref ref="fig-parabola0" text="type-global"/> to decide whether the given ordered pairs are solutions of the equation. Verify your answers algebraically.
        <ol label="a">
            <li><p><m>(-4, 6)</m></p></li>
            <li><p><m>(3, 0)</m></p></li>
        </ol></p>
        <p><figure xml:id="fig-parabola0"><caption></caption><image source="images/fig-parabola0"  width="35%"><description>parabola in the coordinate plane</description></image></figure></p> 
    </sidebyside>
<solution>
    <p>
        <ol label="a">
            <li><p>
                Because the point <m>(-4, 6)</m> does lie on the graph, the ordered pair <m>x=-4, y = 6</m> is a solution of <m>y = 0.5x^2 - 2</m>. We can verify this by substituting <m>\alert{-4}</m> for <m>x</m> and <m>\blert{6}</m> for <m>y</m>:
                <md>
                    <mrow>0.5(\alert{-4})^2 - 2 \amp= 0.5(16) - 2</mrow>
                    <mrow>= 8 - 2 \amp = \blert{6}</mrow>
                </md>
            </p></li>
            <li><p>
                Because the point <m>(3, 0)</m> does not lie on the graph, the ordered pair <m>x = 3, y = 0</m> is not a solution of <m>y = 0.5x^2 - 2</m>. We substitute <m>\alert{3}</m> for <m>x</m> and <m>\blert{0}</m> for <m>y</m> to verify this.
                <md>
                    <mrow>0.5(\alert{3})^2 - 2 \amp= 0.5(9) - 2</mrow>
                    <mrow>= 4.5 - 2 \amp = 2.5 \ne \blert{0}</mrow>
                </md>
            </p></li>
        </ol>
    </p>
</solution>
</example>

</subsection>

</section>



<section xml:id="appendix-Linear-Systems-in-Two-Variables"><title>Linear Systems in Two Variables</title><idx>linear system</idx><introduction>
<p>A <m>2\times 2</m> <term>system</term> of equations is a set of <m>2</m> equations in the same <m>2</m> variables. A <term>solution</term><idx>solution</idx> of a <m>2\times 2</m> system is an ordered pair that makes each equation in the system true. In this section, we review two algebraic methods for solving <m>2\times 2</m> linear systems: substitution and elimination.</p></introduction>

<subsection><title>Solving Systems by Substitution</title>
<p>
    The basic strategy for the <term>substitution</term><idx>substitution</idx> method can be described as follows.
</p>

<assemblage><title>Steps for Solving a <n>2\tines 2</n> System by Substitution</title>
<p>
    <ol label="1">
        <li><p>
            Solve one of the equations for one of the variables in terms of the other.
        </p></li>
        <li><p>
            Substitute this expression into the second equation; doing so yields an equation in one variable.
        </p></li>
        <li><p>
            Solve the new equation.
        </p></li>
        <li><p>
            Use the result of step 1 to find the other variable.
        </p></li>
    </ol>
</p>
</assemblage>

<example>
    <p>
        Staci stocks two kinds of sleeping bags in her sporting goods store, a standard model and a down-filled model for colder temperatures. From past experience, she estimates that she will sell twice as many of the standard variety as of the down filled. She has room to stock 60 sleeping bags at a time. How many of each variety should Staci order?
    </p>
<solution>
    <p>
        <ol>
            <li><p>
                <md>
                    <mrow>\amp\text{Number of standard sleeping bags: }~x</mrow>
                    <mrow>\amp\text{Number of down-filled sleeping bags: }~y</mrow>
                </md>
            </p></li>
            <li><p>
                Write two equations about the variables. Staci needs twice as many standard model as down filled, so
                <md><mrow>x \amp = 2y\amp\amp\hphantom{blankety}(1)</mrow></md>
                Also, the total number of sleeping bags is 60, so
                <md><mrow>x+y \amp = 60\amp\amp\hphantom{blank}(2)</mrow></md>
            </p></li>
            <li><p>
                We will solve this system using substitution. Notice that Equation (1) is already solved for <m>x</m> in terms of <m>y</m>: <m>x = 2y</m>. Substitute <m>\alert{2y}</m> for <m>x</m> in Equation (2) to obtain
                <md>
                    <mrow>
                        \alert{2y} + y \amp = 60
                    </mrow>
                    <mrow>3y\amp = 60</mrow>
                </md>
                Solving for <m>y</m>, we find <m>y=\alert{20}</m>. Finally, substitute this value into Equation (1) to find
                <me>x = 2(\alert{20}) = 40</me>
                The solution to the system is <m>x = 40, y = 20</m>.
            </p></li>
            <li><p>
                Staci should order <m>40</m> standard sleeping bags and <m>20</m> down-filled bags.
            </p></li>
        </ol>
    </p>
</solution></example>

</subsection>


<subsection><title>Solving Systems by Elimination</title>
<p>
    The method of substitution is convenient if one of the variables in the system has a coefficient of <m>1</m> or <m>-1</m>, because it is easy to solve for that variable. If none of the coefficients is <m>1</m> or <m>-1s</m>, then a second method, called <term>elimination</term><idx>elimination</idx>, is usually more efficient.
</p>
<p>
    The method of elimination is based on the following properties of linear equations.
</p>

<assemblage><title>Properties of Linear Systems</title>
<p>
    <ol label="1">
        <li><p>
            Multiplying a linear equation by a (nonzero) constant does not change its solutions. That is, any solution of the equation
            <me>ax + by = c</me>
            is also a solution of the equation
            <me>kax + kby = kc</me>
        </p></li>
        <li><p>
            Adding (or subtracting) two linear equations does not change their common solutions. That is, any solution of the system
            <md>
                <mrow>a_1x + b_1 y \amp = c_1</mrow>
                <mrow>a_2x + b_2 y \amp = c_2</mrow>
            </md>
            is also a solution of the equation
            <me>(a_1 + a_2) x + (b_1 + b_2) y = c_1 + c_2</me>
        </p></li>
    </ol>
</p>
</assemblage>

<example xml:id="example-linear-combinations">
    <p>
        Solve the system by the method of elimination.
        <md alignment="alignat">
            <mrow>2x\amp {}+{}\amp 3y\amp ={}\amp 8 \hphantom{blankblank}\amp(1)</mrow>
            <mrow>3x\amp {}-{}\amp 4y\amp ={}\amp -5 \hphantom{blankblank}\amp(2)</mrow>
        </md>
    </p>
<solution>
    <p>
        We first decide which variable to eliminate, <m>x</m> or <m>y</m>. We can choose whichever looks easiest. In this problem, we choose to eliminate <m>x</m>. We next look for the smallest number that both coefficients, <m>2</m> and <m>3</m>, divide into evenly. This number is <m>6</m>. We want the coefficients of <m>x</m> to become <m>6</m> and <m>-6</m>, so we will multiply Equation (1) by <m>3</m> and Equation (2) by <m>-2</m> to obtain
        <md alignment="alignat">
            <mrow>6x\amp {}+{}\amp 9y\amp ={}\amp 24 \hphantom{blankblank}\amp(1\text{a})</mrow>
            <mrow>-6x\amp {}+{}\amp 8y\amp ={}\amp 10 \hphantom{blankblank}\amp(2\text{a})</mrow>
        </md>
        Now add the corresponding terms of (1a) and (2a). The <m>x</m>-terms are eliminated, yielding an equation in one variable.
        <md alignment="alignat">
            <mrow>6x\amp {}+{}\amp 9y\amp ={}\amp 24 \hphantom{blankblank}\amp(1\text{a})</mrow>
            <mrow>-6x\amp {}+{}\amp 8y\amp ={}\amp 10 \hphantom{blankblank}\amp(2\text{a})</mrow>
            <mrow>{}\amp {}{}\amp 17y\amp ={}\amp 34 \hphantom{blankblank}\amp(3)</mrow>
        </md>
        Solve this equation for <m>y</m> to find <m>y=\alert{2}</m>. We can substitute this value of <m>y</m> into any of our equations involving both <m>x</m> and <m>y</m>. If we choose Equation (1), then
        <me>2x + 3 (\alert{2}) = 8</me>
        and solving this equation yields <m>x=1</m>. The ordered pair <m>(1, 2)</m> is a solution to the system. You should verify that these values satisfy both original equations.
    </p>
</solution></example>

<p>
    We summarize the strategy for solving a linear system by elimination.
</p>

<assemblage><title>Steps for Solving a <m>2\times 2</m> Linear System by Elimination</title>
<p>
    <ol label="1">
        <li><p>
            Choose one of the variables to eliminate. Multiply each equation by a suitable factor so that the coefficients of that variable are opposites.
        </p></li>
        <li><p>
            Add the two new equations termwise.
        </p></li>
        <li><p>
            Solve the resulting equation for the remaining variable.
        </p></li>
        <li><p>
            Substitute the value found in step 3 into either of the original equations and solve for the other variable.
        </p></li>
    </ol>
</p>
</assemblage>

<p>
    In <xref ref="example-linear-combinations" text="type-global"/>, we added <m>3</m> times the first equation to <m>-2</m> times the second equation. The result from adding a constant multiple of one equation to a constant multiple of another equation is called a <term>linear combination</term><idx>linear combination</idx> of the two equations. The method of elimination is also called the method of linear combinations<idx>linear combinations</idx>.
</p>
<p>
    If either equation in a system has fractional coefficients, it is helpful to clear the fractions before applying the method of linear combinations.
</p>

<example>
    <p>
        Solve the system by linear combinations.
        <md alignment="alignat">
            <mrow>\frac{2}{3}x\amp {}-{}\amp y\amp ={}\amp 2 \hphantom{blankblank}\amp(1)</mrow>
            <mrow>x\amp {}+{}\amp \frac{1}{2}y\amp ={}\amp 7 \hphantom{blankblank}\amp(2)</mrow>
        </md>
    </p>
<solution>
    <p>
        Multiply each side of Equation (1) by <m>3</m> and each side of Equation (2) by <m>2</m> to clear the fractions:
        <md alignment="alignat">
            <mrow>2x\amp {}-{}\amp 3y\amp ={}\amp 6 \hphantom{blankblank}\amp(1\text{a})</mrow>
            <mrow>2x\amp {}+{}\amp y\amp ={}\amp 14 \hphantom{blankblank}\amp(2\text{a})</mrow>
        </md>
        To eliminate the variable <m>x</m>, multiply Equation (2a) by <m>-1</m> and add the result to Equation (1a) to get
        <md>
            <mrow>-4y \amp = -8 \amp\amp\text{Divide both sides by -4.}</mrow>
            <mrow>y \amp = 2</mrow>
        </md>
        Substitute <m>\alert{2}</m> for <m>y</m> in one of the original equations and solve for <m>x</m>. We use Equation (2).
        <md>
            <mrow>x + \frac{1}{2}(\alert{2}) \amp = 7 \amp\amp\text{Subtract 1 from both sides.}</mrow>
            <mrow>x \amp = 6</mrow>
        </md>
        Verify that <m>x=6</m> and <m>y=2</m> satisfy both Equations (1) and (2). The solution to the system is the ordered pair <m>(6, 2)</m>.
    </p>
</solution></example>

</subsection>
</section>


<section xml:id="appendix-Laws-of-Exponents"><title>Laws of Exponents</title><idx>laws of exponents</idx><introduction>
<p>
    In this section, we review the rules for performing operations on powers.
</p></introduction>
<subsection><title>Product of Powers</title>
<p>
    Consider a product of two powers with the same base.
    <me>(a^3) (a^2) = a a a \cdot a a = a^5</me>
    because <m>a</m> occurs as a factor five times. The number of <m>a</m>'s in the product is the <m>sum</m> of the number of  <m>a</m>'s in each factor.
</p>

<assemblage><title>First Law of Exponents: Product of Powers</title>
<p>
    To multiply two powers with the same base, add the exponents and leave the base unchanged.
    <me>a^m \cdot a^n = a^{m+n}</me>
</p>
</assemblage>
<example xml:id="example-product-of-powers">
    <p>
        <sidebyside  width="80%"><image source="images/fig-product-of-powers">
          <description>
            figure showing that when powers of the same base are multiplied that base is retained with exponent equal to the sum of the original exponents
          </description>
        </image></sidebyside>
    </p>
</example>

<p>Here are some mistakes to avoid.</p>

<warning>
    <p>
        <ol label="1">
            <li><p>
                Note that we do not <em>multiply</em> the exponents when simplifying a product. For example,
                <me>b^4 \cdot b^2 \ne b^8</me>
                You can check this with your calculator by choosing a value for <m>b</m>, for instance, <m>b = 3</m>:
                <me>3^4 \cdot 3^2\ne 3^8</me>
            </p></li>
            <li><p>
                In order to apply the first law of exponents, the bases must be the same. For example,
                <me>2^3 \cdot 3^5 \ne 6^8</me>
                (Check this on your calculator.)
            </p></li>
            <li><p>
                We do not multiply the bases when simplifying a product. In <xref ref="example-product-of-powers" text="type-global"/>a, note that
                <me>5^3 \cdot 5^4\ne 25^7</me>
            </p></li>
            <li><p>
                Although we can simplify the product <m>x^2x^3</m> as <m>x^5</m>, we cannot simplify the sum <m>x^2 + x^3</m>, because <m>x^2</m> and <m>x^3</m> are not like terms.
            </p></li>
        </ol>
    </p>
</warning>

<example>
    <p>
        Multiply <m>(-3x^4z^2) (5x^3 z)</m>.
    </p>
<solution>
    <p>
        Rearrange the factors to group the numerical coefficients and the powers of each base. Apply the first law of exponents.
        <md>
            <mrow>(-3x^4z^2) (5x^3z) \amp = (-3) (5)x^4x^3z^2z</mrow>
            <mrow>\amp = -15x^7z^3</mrow>
        </md>
    </p>
</solution></example>

</subsection>




<subsection><title>Quotients of Powers</title>
<p>
    To reduce a fraction, we divide both numerator and denominator by any common factors.
    <me>
        \require{cancel} \require{color} \renewcommand{\CancelColor}{\blue}
        \frac{x^7}{x^4}=\frac{xxx\cancel{x}\cancel{x}\cancel{x}\cancel{x}}{\cancel{x}\cancel{x}\cancel{x}\cancel{x}}=\frac{x^3}{1}=x^3
    </me>
    We can obtain the same result more quickly by <em>subtracting</em> the exponent of the denominator from the exponent of the numerator.
    <me>\frac{x^7}{x^4}=x^{7-4}=x^3</me>
</p>
<p>
    What if the larger power occurs in the denominator of the fraction?
    <me>
        \frac{x^4}{x^7}=\frac{\cancel{x}\cancel{x}\cancel{x}\cancel{x}}{xxx\cancel{x}\cancel{x}\cancel{x}\cancel{x}}=\frac{1}{x^3}
    </me>
    In this case, we subtract the exponent of the numerator from the exponent of the denominator.
    <me>\frac{x^4}{x^7}=\frac{1}{x^{7-4}}=\frac{1}{x^3}</me>
    These examples suggest the following law.
</p>

<assemblage><title>Second Law of Exponents: Quotient of Powers</title>
<p>
    To divide two powers with the same base, subtract the smaller exponent from the larger one, keeping the same base.
    <ol labe="a">
        <li><p>
            If the larger exponent occurs in the numerator, put the power in the numerator.
            <me>\text{If }m\gt n,~\text{ then }~ \frac{a^m}{a^n}=a^{m-n} \hphantom{blank}(a \ne 0)</me>
        </p></li>
        <li><p>
            If the larger exponent occurs in the denominator, put the power in the denominator.
            <me>\text{If }m\lt n,~\text{ then }~ \frac{a^m}{a^n}=\frac{1}{a^{n-m}} \hphantom{blank}(a \ne 0)</me>
        </p></li>
    </ol>
</p>
</assemblage>

<example>
    <p>
        <ol>
            <li><p><m>
                \dfrac{3^8}{3^2}=3^{8-2}=3^6\hphantom{blankblankbla}\text{Subtract exponents: }~8 \gt 2.
            </m></p></li>
            <li><p><m>
                \displaystyle{\frac{w^3}{w^6}=\frac{1}{w^{6-3}}=\frac{1}{w^3}\hphantom{blankblank}\text{Subtract exponents: }3\lt 6.}
            </m></p></li>
        </ol>
    </p>
</example>
<example>
    <p>
        Divide
        <m>\dfrac{3x^2 y^4}{6x^3 y}.</m>
    </p>
<solution>
    <p>
        Consider the numerical coefficients and the powers of each variable separately. Use the second law of exponents to simplify each quotient of powers.
        <md>
            <mrow>
                \frac{3x^2 y^4}{6x^3 y} \amp = \frac{3}{6}\cdot \frac{x^2}{x^3} \cdot \frac{y^4}{y}\amp\amp\text{Subtract exponents.}
            </mrow>
            <mrow>\amp = \frac{1}{2}\cdot \frac{1}{x^{3-2}} \cdot y^{4-1}</mrow>
            <mrow>\amp = \frac{1}{2}\cdot \frac{1}{x} \cdot y^3=\frac{y^3}{2x}</mrow>
        </md>
    </p>
</solution></example>

</subsection>



<subsection><title>Power of a Power</title>
<p>
    Consider the expression <m>\left(a^4\right)^3</m>, the third power of <m>a^4</m>.
    <me>\left(a^4\right)^3=\left(a^4\right)\left(a^4\right)\left(a^4\right)
        =a^{4+4+4}=a^{12} \hphantom{blankblankblank}\text{Add exponents.}</me>
    We can obtain the same result by multiplying the exponents together.
    <me>\left(a^4\right)^3=a^{4 \,\cdot\, 3}=a^{12}</me>
</p>

<assemblage><title>Third Law of Exponents: Power of a Power</title>
<p>
    To raise a power to a power, keep the same base and multiply the exponents.
    <me>\left(a^m\right)^n = a^{mn}</me>
</p>
</assemblage>

<example>
    <p>
        <sidebyside  width="80%"><image source="images/fig-power-of-power">
          <description>
            examples of raising a power to a power by multiplying exponents
          </description>
        </image></sidebyside>
    </p>
</example>

<warning>
    <p>
        Notice the difference between the expressions
        <me>(x^3) (x^4) = x^{3+4} = x^7</me>
        and
        <me>\left(x^3\right)^4 = x^{3\, \cdot \, 4} = x^{12}</me>

The first expression is a product, so we add the exponents. The second expression raises a
power to a power, so we multiply the exponents.
    </p>
</warning>

</subsection>



<subsection><title>Power of a Product</title>
<p>
    To simplify the expression <m>(5a)^3</m>, we use the associative and commutative laws to regroup the factors as follows.
    <md>
        <mrow>
            (5a)^3 \amp = (5a) (5a) (5a)
        </mrow>
        <mrow>
            \amp =  5 \cdot 5 \cdot 5 \cdot a \cdot a \cdot a
        </mrow>
        <mrow>
            \amp = 5^3 a^3
        </mrow>
    </md>
    Thus, to raise a product to a power, we can simply raise each factor to the power.
</p>

<assemblage><title>Fourth Law of Exponents: Power of a Product</title>
<p>
    A power of a product is equal to the product of the powers of each of its factors.
    <me>(ab)^n = a^n b^n</me>
</p>
</assemblage>

<example>
    <p>
        <ol label="a">
            <li><p><m>(5a)^3 = 5^3a^3 = 125a^3\hphantom{blank}\text{Cube each factor}</m>.</p></li>
            <li><p><m>
                \begin{aligned}
                \left(-xy^2\right)^4\amp = (-x)^4\left(y^2\right)^4 \amp\amp\text{Raise each factor to the fourth power.}\\
                \amp =x^4y^8\amp\amp\text{Apply the third law of exponents.}
                \end{aligned}
            </m></p></li>
        </ol>
    </p>
</example>

<warning>
<p><ol label="1">
    <li><p>
        Compare the two expressions <m>3a^2</m> and <m>(3a)^2</m>; they are not the same. In the expression <m>3a^2</m>, only the factor <m>a</m> is squared. But in <m>(3a)^2</m>, both <m>3</m> and <m>a</m> are squared. Thus,
        <me>3a^2 ~~\text{  cannot be simplified}</me>
        but
        <me>(3a)^2 = 3^2a^2 = 9a^2</me>       
    </p></li>
    <li><p>
        Compare the two expressions <m>(3a)^2</m> and <m>(3 + a)^2</m>. The fourth law of exponents applies to the <em>product</em> <m>3a</m>, but not to the <em>sum</em> <m>3+a</m>. Thus,
        <me>(3 + a)^2 \ne 3^2 + a^2</me>
        In order to simplify <m>(3 + a)^2</m>, we must expand the binomial product:
        <me>(3 + a)^2 = (3 + a) (3 + a) = 9 + 6a + a^2</me>
    </p></li>
</ol></p>
</warning>

</subsection>


<subsection><title>Power of a Quotient</title>
<p>
    To simplify the expression <m>\displaystyle{\left(\frac{x}{3}\right)^4} </m>, we multiply together <m>4</m> copies of the fraction 
    <m>\dfrac{x}{3}</m>.
    <md>
        <mrow>
            \left(\frac{x}{3}\right)^4 \amp = \frac{x}{3}\cdot \frac{x}{3}\cdot \frac{x}{3}\cdot \frac{x}{3}
            = \frac{x\cdot x\cdot x\cdot x}{3\cdot 3\cdot 3\cdot 3}
        </mrow>
        <mrow>
            \amp = \frac{x^4}{3^4}=\frac{x^4}{81}
        </mrow>
    </md>
        In general, we have the following rule.
</p>

<assemblage><title>Fifth Law of Exponents: Power of a Quotient</title>
<p>
    To raise a quotient to a power, raise both the numerator and denominator to the power.
    <me>\left(\frac{a}{b}\right)^n=\frac{a^n}{b^n}</me>
</p>
</assemblage>


<p>
    For reference, we state all of the laws of exponents together. All the laws are valid when a and b are not equal to zero and when the exponents m and n are whole numbers.
</p>
<assemblage><title>Laws of Exponents</title><p>
    <ol label="*I*">
        <li><p><m>
            a^m\cdot a^n = a^{m+n}
        </m></p></li>
        <li><p>
            <ol label="a">
                <li><p><m>\dfrac{a^m}{a^n}=a^{m-n} \hphantom{blank}m\gt n</m></p></li>
                <li><p><m>\displaystyle{\frac{a^m}{a^n}=\frac{1}{a^{n-m}} \hphantom{blank}m\lt n}</m></p></li>

        </ol>
        </p></li>
        <li><p><m>
            \left(a^m\right)^n=a^{m+n}
        </m></p></li>
        <li><p><m>
            (ab)^n=a^n b^n
        </m></p></li>
        <li><p><m>
            \displaystyle{\left(\frac{a}{b}\right)^n = \frac{a^n}{b^n} }
        </m></p></li>
    </ol>
</p></assemblage>

<example>
    <p>
        Simplify <m>5x^2 y^3\left(2xy^2\right)^4</m>.
    </p>
<solution>
    <p>
        According to the order of operations, we should perform any powers before multiplications. Thus, we begin by simplifying <m>(2xy^2)^4</m>. We apply the fourth law.
        <md>
            <mrow>
                5x^2 y^3\left(2xy^2\right)^4 \amp = 5x^2 y^3 · 2^4x^4\left(y^2\right)^4 \amp\amp\text{Apply the fourth law.}
            </mrow>
            <mrow>
                \amp = 5x^2 y^3 · 2^4x^4 y^8
            </mrow>
        </md>
        Finally, multiply powers with the same base. Apply the first law.
        <me>5x^2 y^3 \cdot 2^4x^4 y^8 = 5 \cdot 2^4 x^2x^4 y^3 y^8 = 80x^6 y^{11}</me>
    </p>
</solution></example>

<example>
    <p>
        Simplify <m>\displaystyle{\left(\frac{2x}{z^2}\right)^3}</m>.
    </p>
<solution>
    <p>
        Begin by applying the fifth law.
        <md>
            <mrow>
                \require{\stackengine}
                \left(\frac{2x}{z^2}\right)^3 \amp = \frac{(2x)^3}{\left(z^2\right)^2}\amp\amp
                {\text{Apply the fourth law to the numerator }}{\text{and the third law to the denominator.}}
            </mrow>
            <mrow>
                \amp = \frac{2^3x^3}{z^6}=\frac{8x^3}{z^6}
            </mrow>
        </md>
    </p>
</solution>
</example>

</subsection>

</section>


<!-- **************************

-->

<section xml:id="appendix-Polynomials-and-Factoring"><title>Polynomials and Factoring</title><introduction>
<p>
    In <xref ref="appendix-Laws-of-Exponents" text="type-global"/>, we used the first law of exponents to multiply two or more monomials. In this section, we review techniques for multiplying and factoring polynomials of several terms.
</p></introduction>
<subsection><title>Polynomials</title>
<p>
    A <term>polynomial</term><idx>polynomial</idx> is a sum of terms in which all the exponents on the variables are whole numbers and no variables appear in the denominator or under a radical. The expressions
    <me>0.1 R^4, ~~d^2 + 32 d - 21, ~~\text{ and }~~128x^3 - 960x^2 + 8000</me>
    are all examples of polynomials in one variable.
</p>
<p>
    An algebraic expression consisting of one term of the form <m>cx^n</m>, where <m>c</m> is a constant and <m>n</m> is a whole number, is called a <term>monomial</term><idx>monomial</idx>. For example,
    <me>y3, ~~-3x^8, ~~\text{ and } ~~0.1R^4</me>
    are monomials. A polynomial is just a sum of one or more monomials.
</p>
<p>
    A polynomial with exactly two terms, such as <m>\dfrac{1}{2}n^2 + \dfrac{1}{2}n</m>, is called a <term>binomial</term><idx>binomial</idx>. A polynomial with exactly three terms, such as <m>d^2 + 32 d - 21</m> or <m>128x^3 - 960x^2 + 8000</m>, is called a <term>trinomial</term><idx>trinomial</idx>. We have no special names for polynomials with more than three terms.
</p>

<example>
    <p>
        Which of the following expressions are polynomials?
        <ol label="a">
            <li><p><m>\pi r^2</m></p></li>
            <li><p><m>23.4s^6 - 47.9s^4</m></p></li>
            <li><p><m>\dfrac{2}{3}w^3 - \dfrac{7}{3}w^2 + \dfrac{1}{3}w</m></p></li>
            <li><p><m>7 + m^{-2}</m></p></li>
            <li><p><m>\dfrac{x-2}{x+2}</m></p></li>
            <li><p><m>\sqrt[3]{4y}</m></p></li>
        </ol>
    </p>
<solution>
    <p>
        The first three are all polynomials. In fact, (a) is a monomial, (b) is a binomial, and (c) is a trinomial. The last three are not polynomials. The variable in (d) has a negative exponent, the variable in (e) occurs in the denominator, and the variable in (f) occurs under a radical.
    </p>
</solution></example>

<p>
    In a polynomial containing only one variable, the greatest exponent that appears on the variable is called the <term>degree</term><idx>degree</idx> of the polynomial. If there is no variable at all, then the polynomial is called a constant, and the degree of a constant is zero.
</p>

<example>
    <p>
        Give the degree of each polynomial.
        <ol label="a">
            <li><p>
                <m>b^3 - 3b^2 + 3b - 1</m>
            </p></li>
            <li><p>
                <m>10^{10}</m>
            </p></li>
            <li><p>
                <m>-4w^3</m>
            </p></li>
            <li><p>
                <m>s^2 - s^6</m>
            </p></li>
        </ol>
    </p>
<solution>
    <p>
        <ol label="a">
            <li><p>
                This is a polynomial in the variable <m>b</m>, and because the greatest exponent on <m>b</m> is <m>3</m>, the degree of this polynomial is <m>3</m>.
            </p></li>
            <li><p>
                This is a constant polynomial, so its degree is <m>0</m>. (The exponent on a constant does not affect the degree.)
            </p></li>
            <li><p>
                This monomial has degree <m>3</m>.
            </p></li>
            <li><p>
                This is a binomial of degree <m>6</m>.
            </p></li>
        </ol>
    </p>
</solution></example>

<p>
    We can evaluate a polynomial just as we evaluate any other algebraic expression: We replace the variable with a number and simplify the result.
</p>

<example>
    <p>
        Let <m>p(x) = -2x^2 + 3x - 1</m>. Evaluate each of the following.
        <ol label="a">
            <li><p><m>p(2)</m></p></li>
            <li><p><m>p(-1)</m></p></li>
            <li><p><m>p(t)</m></p></li>
            <li><p><m>p(t+3)</m></p></li>
        </ol>
    </p>
<solution>
    <p>
        In each case, we replace <m>x</m> by the given value.
        <ol label="a">
            <li><p>
                <m>p(\alert{2})=-2(\alert{2})^2 + 3(\alert{2}) - 1=-8+6-1=-3</m>
            </p></li>
            <li><p>
                <m>p(\alert{-1})=-2(\alert{-1})^2 + 3(\alert{-1}) - 1=-2+(-3)-1=-6</m>
            </p></li>
            <li><p>
                <m>p(\alert{t})=-2(\alert{t})^2 + 3(\alert{t}) - 1=--2t^2+3t-1</m>
            </p></li>
            <li><p>
                <m>
                    \begin{aligned}
                    \amp \\
                    p(\alert{t+3})\amp =-2(\alert{t+3})^2 + 3(\alert{t+3}) - 1\\
                    \amp = -2(t^2+6t+9)+3(t+3)-1\\
                    \amp = -2t^2-9t-10
                    \end{aligned}
                </m>
            </p></li>
        </ol>
    </p>
</solution></example>

</subsection>



<subsection><title>Products of Polynomials</title>
<p>
    To multiply polynomials, we use a generalized form of the distributive property:
    <me>a(b + c + d + \cdots) = ab + ac + ad + \cdots</me>
    To multiply a polynomial by a monomial, we multiply each term of the polynomial by the monomial.
</p>

<example>
    <p>
        <ol label="a">
            <li><p>
                <m>
                    \begin{aligned}
                    \amp\\
                    3x(x + y + z) \amp = 3x(x) + 3x(y) + 3x(z)\\
                    \amp = 3x^2+3xy+3xz
                    \end{aligned}
                </m>
            </p></li>
            <li><p>
                <m>
                    \begin{aligned}
                    \amp\\
                    -2ab^2(3a^2 - ab + 2b^2) \amp = -2ab^2(3a^2) - 2ab^2(-ab) - 2ab^2(2b^2)\\
                    \amp = -6a^3b^2+2a^2b^3-4ab^4
                    \end{aligned}
                      
                </m>
            </p></li>
        </ol>
    </p>
</example>

</subsection>




<subsection><title>Products of Binomials</title>
<p>
    Products of binomials occur so frequently that it is worthwhile to learn a shortcut for this type of multiplication. We can use the following scheme to perform the multiplication mentally. (See <xref ref="fig-FOIL" text="type-global"/>.)
    <figure xml:id="fig-FOIL"><image source="images/fig-FOIL" ALT="figure identifying the first, outer, inner, and last terms in a binomial product" width="50%"/><caption></caption> </figure>
</p>

<example>
    <p>
        <md>
            <mrow>(2x - 1)(x + 3) \amp = 2x^2 + 6x - x - 3</mrow>
            <mrow>\amp = 2x^2 + 5x - 3</mrow>
        </md>
    </p>
</example>

</subsection>



<subsection><title>Factoring</title>
<p>
    We sometimes find it useful to write a polynomial as a single <em>term</em> composed of two or more <em>factors</em>. This process is the reverse of multiplication and is called <term>factoring</term><idx>factoring</idx>. For example, observe that
    <me>3x^2 + 6x = 3x(x + 2)</me>
    We will only consider factorization in which the factors have integer coefficients.
</p>

</subsection>




<subsection><title>Common Factors</title>
<p>
    We can factor a common factor from a polynomial by using the distributive property in the form
    <me>ab + ac = a(b + c)</me>
    We first identify the common factor. For example, each term of the polynomial
    <me>6x^3 + 9x^2 - 3x</me>
    contains the monomial <m>3x</m> as a factor; therefore,
    <me>6x^3 + 9x^2 - 3x = 3x (<fillin characters="15"/>)</me>
    Next, we insert the proper polynomial factor within the parentheses. This factor can be determined by inspection. We ask ourselves for monomials that, when multiplied by <m>3x</m>, yield <m>6x^3</m>, <m>9x^2</m>, and <m>-3x</m>, respectively, and obtain
    <me>6x^3 + 9x^2 - 3x = 3x(2x^2 + 3x - 1)</me>
</p>
<p>
    We can check the result of factoring an expression by multiplying the factors. In the example above,
    <me>3x(2x^2 + 3x - 1) = 6x^3 + 9x^2 - 3x</me>
</p>

<example><p>
    <ol label = "a">
        <li><p><m>
            \begin{aligned} \\
            18x^2 y - 24xy2 \amp = 6xy(\text{?} - \text{?}) \\
            \amp = 6xy(3x - 4y)
            \end{aligned}
        </m></p>
        <p>
            because
            <me>6xy(3x - 4y) = 18x^2 y - 24xy^2</me>
    </p></li>
        <li><p><m>
            \begin{aligned} \\
            y(x - 2) + z(x - 2) \amp = (x - 2)(\text{?} - \text{?}) \\
            \amp = (x - 2)(y + z)
            \end{aligned}
        </m></p>
        <p>
            because
            <me>(x - 2)(y + z) = y(x - 2) + z(x - 2)</me>
    </p></li>
    </ol>
</p></example>

</subsection>

<subsection><title>Opposite of a Binomial</title>
<p>
    It is often useful to factor <m>-1</m> from the terms of a binomial.
    <md>
        <mrow>
            a - b \amp = (-1)(-a + b)
        </mrow>
        <mrow>\amp = (-1)(b - a) = -(b - a)</mrow>
    </md>
Hence, we have the following important relationship.
</p>

<assemblage><title>Opposite of a Binomial</title>
<p>
    <me>a-b=-(b-a)</me>
</p>
</assemblage>
<p>
    That is, <m>a-b</m> and <m>b-a</m> are opposites or negatives of each other.
</p>

<example>
    <p>
        <ol label="a">
            <li><p><m>3x - y = -(y - 3x)</m></p></li>
            <li><p><m>a - 2b = -(2b - a)</m></p></li>
        </ol>
    </p>
</example>

</subsection>

<subsection><title>Polynomial Division</title>
<p>
    We can divide one polynomial by a polynomial of lesser degree. The quotient will be the sum of a polynomial and a simpler algebraic fraction.
</p>
<p>
    If the divisor is a monomial, we can simply divide the monomial into each term of the numerator.
</p>

<example>
    <p>
        Divide <m>\dfrac{9x^3 - 6x^2 + 4}{3x}</m>
    </p>
<solution>
    <p>
        Divide <m>3x</m> into each term of the numerator.
        <md>
            <mrow>
               \frac{9x^3 - 6x^2 + 4}{3x} \amp = \frac{9x^3}{3x}- \frac{6x^2}{3x}+ \frac{4}{3x}
            </mrow>
            <mrow>
                \amp = 3x^2 - 2x + \frac{4}{3x}
            </mrow>
        </md>
        The quotient is the sum of a polynomial, <m>3x^2 - 2x</m>, and an algebraic fraction, <m>\dfrac{4}{3x}</m>.
    </p>
</solution></example>

<p>
    If the denominator is not a monomial, we can use a method similar to the long division algorithm used in arithmetic.
</p>

<example xml:id="example-long-division">
    <p>
        Divide <m>\dfrac{2x^2+x-7}{x+3}</m>        
    </p>
<solution>
    <p>
        First write</p><p>
        <sidebyside width="25%"><image source="images/fig-long-division.jpg"><description>long division box</description></image></sidebyside>
        and divide <m>2x^2</m> (the first term of the numerator) by <m>x</m> (the first term of the denominator) to obtain <m>2x</m>. (It may be helpful to write down the division: <m>\dfrac{2x^2}{2x}=x</m>.) Write <m>2x</m> above the quotient bar as the first term of the quotient, as shown below.
    </p>
    <p>
        Next, multiply <m>x+3</m> by <m>2x</m> to obtain <m>2x^2 + 6x</m>, and subtract this product from <m>2x^2 + x - 7</m>:</p><p>
        <sidebyside width="25%"><image source="images/fig-long-division2.jpg"><description>long division box</description></image></sidebyside>
    </p>
    <p>
        Repeating the process, divide <m>-5x</m> by <m>x</m> to obtain <m>-5</m>. Write <m>-5</m> as the second term of the quotient. Then multiply <m>x+3</m> by <m>-5</m> to obtain <m>-5x - 15</m>, and subtract:</p><p>
        <sidebyside width="25%"><image source="images/fig-long-division3.jpg"><description>long division box</description></image></sidebyside>
        Because the degree of <m>8</m> is less than the degree of <m>x + 3</m>, the division is finished. The quotient is <m>2x - 5</m>, with a remainder of <m>8</m>. We write the remainder as a fraction to obtain
        <me>
            \frac{2x^2 + x - 7}{x + 3}= 2x - 5 + \frac{8}{x + 3}
        </me>
    </p>
</solution></example>

<p>
    When using polynomial division, it helps to write the polynomials in descending powers of the variable. If the numerator is missing any terms, we can insert terms with zero coefficients so that like powers will be aligned. For example, to perform the division
    <me>\frac{3x - 1 + 4x^3}{2x-1}</me>
    we first write the numerator in descending powers as <m>4x^3 + 3x - 1</m>. We then insert <m>0x^2</m> between <m>4x^3</m> and <m>3x</m> and set up the quotient as</p><p>
    <sidebyside width="33%"><image source="images/fig-long-division4.jpg"><description>long division box</description></image></sidebyside>
    We then proceed as in <xref ref="example-long-division" text="type-global"/>. You can check that the quotient is
    <me>2x^2 + x + 2 + \frac{1}{2x-1}</me>
</p>

</subsection>

</section>

<!-- **************************

-->
<section xml:id="appendix-Factoring-Quadratic-Trinomials"><title>Factoring Quadratic Trinomials</title><introduction>
<p>Consider the trinomial
    <me>
        x^2 + 10x + 16
    </me>
    Can we find two binomial factors,
    <me>(x + a)(x + b)</me>
    whose product is the given trinomial? The product of the binomials is
    <me>(x + a)(x + b) = x^2 + (a + b)x + ab</me>
    Thus, we are looking for two numbers, a and b, that satisfy
    <me>(x + a)(x + b) = x^2 + (a + b)x + ab</me>
    <me>(x + a)(x + b) = x^2 + (a + b)x + ab= x^2 + 10x + 16</me>
    By comparing the coefficients of the terms in the two trinomials, we see that <m>a + b = 10</m> and <m>ab = 16</m>. That is, the sum of the two numbers is the coefficient of the linear term, <m>10</m>, and their product is the constant term, <m>16</m>.
</p>
<p>
    To find the numbers, we list all the possible integer factorizations of <m>16</m>:
    <me>1 \cdot 16, ~~~2 \cdot 8, ~~~\text{ and }~~~ 4 \cdot 4</me>
    We see that only one combination gives the correct linear term: <m>8</m> and <m>2</m>. These are the numbers <m>a</m> and <m>b</m>, so
    <me>x^2 + 10x + 16 = (x + 8)(x + 2)</me>
</p>
<p>
    In <xref ref="example-factor-trinomial" text="type-global"/> we factor quadratic trinomials in which one or more of the coefficients is negative.
</p>

<example xml:id="example-factor-trinomial">
    <p>
        Factor.
        <ol label="a">
            <li><p><m>x^2-7x+12</m></p></li>
            <li><p><m>x^2-x-12</m></p></li>
        </ol>
    </p>
<solution>
    <p>
        <ol label="a">
            <li><p>
                Find two numbers whose product is <m>12</m> and whose sum is <m>-7</m>. Because the product is positive and the sum is negative, the two numbers must both be negative. The possible factors of <m>12</m> are <m>-1</m> and <m>-12</m>, <m>-2</m> and <m>-6</m>, or <m>-3</m> and <m>-4</m>. Only <m>-4</m> and <m>-3</m> have the correct sum, <m>-7</m>. Hence,
                <me>x^2 - 7x + 12 = (x - 4) (x - 3)</me>
            </p></li>
            <li><p>
                Find two numbers whose product is <m>-12</m> and whose sum is <m>-1</m>. Because the product is negative, the two numbers must be of opposite sign and their sum must be <m>-1</m>. By listing the possible factors of <m>-12</m>, we find that the two numbers are <m>-4</m> and <m>3</m>. Hence,
                <me>x^2 - x - 12 = (x - 4) (x + 3)</me>
            </p></li>
        </ol>
    </p>
</solution></example>

<p>
    If the coefficient of the quadratic term is not <m>1</m>, we must also consider its factors.
</p>

<example><p>
    Factor <m>8x^2 - 9 - 21x</m>.
</p>
<solution>
    <p>
        <ol>
            <li><p>
                Write the trinomial in decreasing powers of <m>x</m>.
                <me>8x^2-21x-9</me>
            </p></li>
            <li><p>
                List the possible factors for the quadratic term.
                <md>
                    <mrow>(8x\hphantom{00000})\amp(x\hphantom{000000})</mrow>
                    <mrow>(4x\hphantom{00000})\amp(2x\hphantom{00000})</mrow>
                </md>
            </p></li>
            <li><p>
                Consider possible factors for the constant term: <m>9</m> may be factored as <m>9\cdot 1</m> or as <m>3\cdot 3</m>. Form all possible pairs of binomial factor using these factorizations. (See <xref ref="fig-factor-trinomial" text="type-global"/>.)
                <figure xml:id="fig-factor-trinomial"><caption></caption><image source="images/fig-factor-trinomial.jpg"  width="25%"><description>list of possible binomial factor pairs</description></image> </figure>
            </p></li>
            <li><p>
                Select the combinations of the products ❶ and ❷ whose sum or difference could be the linear term, <m>-21x</m>.
                <me>(8x \hphantom{000} 3) (x \hphantom{000} 3)</me>
            </p></li>
            <li><p>
                Insert the proper signs:
                <me>(8x + 3) (x - 3)</me>
            </p></li>
        </ol>
    </p>
</solution></example>

<p>
    With practice, you can usually factor trinomials of the form <m>Ax^2 + Bx + C</m> mentally. The following observations may help.
    <ol label="1">
        <li><p>
            If <m>A</m>, <m>B</m> and <m>C</m> are all positive, both signs in the factored form are positive. For example, as a first step in factoring <m>6x^2 + 11x + 4</m>, we could write
            <me>(\hphantom{000} + \hphantom{000} ) (\hphantom{000}  + \hphantom{000} )</me>

        </p></li>
        <li><p>
            If <m>A</m> and <m>C</m> are positive and <m>B</m> is negative, both signs in the factored form are negative. Thus as a first step in factoring <m>6x^2 - 11x + 4</m>, we could write
            <me>(\hphantom{000} - \hphantom{000} ) (\hphantom{000}  - \hphantom{000} )</me>
        </p></li>
        <li><p>
            If <m>C</m> is negative, the signs in the factored form are opposite. Thus as a first step in factoring <m>6x^2 - 5x - 4</m>, we could write
            <me>(\hphantom{000} + \hphantom{000} ) (\hphantom{000}  - \hphantom{000} )
                ~~\text{ or }~~  (\hphantom{000} - \hphantom{000} ) (\hphantom{000}  + \hphantom{000} )</me>
        </p></li>
    </ol>
</p>

<example>
    <p>
        <ol label="a">
            <li><p><m>
                \begin{aligned}
                6x^2 + 5x + 1 \amp = \hphantom{000} + \hphantom{000} ) (\hphantom{000}  + \hphantom{000} )\\
                \amp = (3x + 1) (2x + 1)
                \end{aligned}
            </m></p></li>
            <li><p><m>
                \begin{aligned}
                6x^2 - 5x + 1 \amp = (\hphantom{000} - \hphantom{000} ) (\hphantom{000}  - \hphantom{000} )\\
                \amp =  (3x - 1) (2x - 1)
                \end{aligned}
            </m></p></li>
            <li><p><m>
                \begin{aligned}
                6x^2 - x - 1 \amp = (\hphantom{000} + \hphantom{000} ) (\hphantom{000}  - \hphantom{000} )\\
                \amp =  (3x + 1) (2x - 1)
                \end{aligned}
               
            </m></p></li>
            <li><p><m>
                \begin{aligned}
                6x^2 - xy - y^2 \amp = (\hphantom{000} + \hphantom{000} ) (\hphantom{000}  - \hphantom{000} )\\
                \amp =  (3x + y) (2x - y)
                \end{aligned}
               
            </m></p></li>
        </ol>
    </p>
</example></introduction>

<subsection><title>Special Products and Factors</title>
<p>
    The products below are special cases of the multiplication of binomials. They occur so often that you should learn to recognize them on sight.
</p>

<assemblage><title>Special Products</title>
<p>
    <ol label="*I*">
        <li><p>
            <m>(a + b)^2 = (a + b) (a + b) = a^2 + 2ab + b^2</m>
        </p></li>
        <li><p>
            <m>(a - b)^2 = (a - b) (a - b) = a^2 - 2ab + b^2</m>
        </p></li>
        <li><p>
            <m>(a + b)(a-b) =  a^2 - b^2</m>
        </p></li>
    </ol>
</p>
</assemblage>

<warning><p>
    Notice that in (I) <m>(a + b)^2 \ne a^2 + b^2</m>, and that in (II) <m>(a - b)^2\ne a^2 - b^2</m>. For example,
    <md>
        <mrow>
            (x + 4)^2 \amp\ne x^2 + 16, \amp\amp\text{ instead } \amp (x + 4)^2 \amp = x^2 + 8x + 16
        </mrow>
        <mrow>
            (t-5)^2 \amp\ne t^2 - 16, \amp\amp\text{ instead } \amp (t -5)^2 \amp= t^2 -10t  + 25
        </mrow>
    </md>
</p></warning>

<example>
    <p>
        <ol label="a">
            <li><p><m>
                \begin{aligned}
                3(x + 4)^2 \amp = 3(x^2 + 2 \cdot 4x + 4^2) \\
                \amp = 3x^2 + 24x + 48
                \end{aligned}
            </m></p></li>
            <li><p><m>
                \begin{aligned}
                (y + 5) (y - 5) \amp = y^2-5^2 \\
                \amp = y^2-25
                \end{aligned}
            </m></p></li>
            <li><p><m>
                \begin{aligned}
                (3x - 2y)^2 \amp = (3x)^2 - 2(3x)(2y) + (2y)^2 \\
                \amp = 9x^2-12xy+4y^2
                \end{aligned}
            </m></p></li>
        </ol>
    </p>
</example>

<p>
    Each of the formulas for special products, when viewed from right to left, also represents a special case of factoring quadratic polynomials.
</p>

<assemblage><title>Special Factorizations</title>
<p>
    <ol label="*I*">
        <li><p>
            <m>a^2 + 2ab + b^2=(a + b)^2</m>
        </p></li>
        <li><p>
            <m>a^2 - 2ab + b^2=(a - b)^2 </m>
        </p></li>
        <li><p>
            <m>a^2 - b^2=(a + b)(a-b)</m>
        </p></li>
        <li><p><m>
            a^2+b^2 ~~\text{ cannot be factored}
        </m></p></li>
    </ol>
    
</p>
</assemblage>

<p>
    The trinomials in (I) and (II) are sometimes called <term>perfect-square trinomials</term><idx>perfect-square trinomials</idx> because they are squares of binomials. Note that the sum of two squares, <m>a^2 + b^2</m>, cannot be factored.
</p>

<example>
    <p>
        Factor.
        <ol label="a">
            <li><p>
                <m>x^2 + 8x + 16</m>
            </p></li>
            <li><p>
                <m>y^2 -10y + 25</m>
            </p></li>
            <li><p>
                <m>4a^2 - 12ab + 9b^2</m>
            </p></li>
            <li><p>
                <m>25m^2n^2 + 20mn + 4</m>
            </p></li>
        </ol>
    </p>
<solution>
    <p>
        <ol label="a">
            <li><p>
                Because <m>16</m> is equal to <m>4^2</m> and <m>8</m> is equal to <m>2\cdot 4</m>,
                <md>
                    <mrow>x^2 + 8x + 16 \amp = x^2 - 2 \cdot 4x + 4^2</mrow>
                    <mrow>\amp = (x + 4)^2</mrow>
                </md>
            </p></li>
            <li><p>
                Because <m>25=5^2</m>  and <m>10=2\cdot 5</m>,
                <md>
                    <mrow>y^2 - 10y + 25 \amp = y^2 - 2 \cdot 5y + 5^2</mrow>
                    <mrow>\amp = (y-5)^2</mrow>
                </md>
            </p></li>
            <li><p>
                Because <m>4a^2=(2a)^2</m>, <m>9b^2=(3b)^2</m>,  and <m>2ab=2(2a)(3b)</m>,
                <md>
                    <mrow>4a^2 - 12ab + 9b^2 \amp = (2a)^2 - 2(2a)(3b) + (3b)^2</mrow>
                    <mrow>\amp = (2a-3b)^2</mrow>
                </md>
            </p></li>
            <li><p>
                Because <m>25m^2n^2=(5mn)^2</m>, <m>4=2^2</m>,  and <m>20mn=2(5mn)(2)</m>,
                <md>
                    <mrow>25m^2n^2 + 20mn + 4 \amp = (5mn)^2 + 2(5mn)(2) + 2^2</mrow>
                    <mrow>\amp = (5mn+2)^2</mrow>
                </md>
            </p></li>
        </ol>
    </p>
</solution></example>

<p>
    Binomials of the form <m>a^2 - b^2</m> are often called the <term>difference of two squares</term><idx>difference of two squares</idx>.
</p>

<example xml:id="factor-difference-of-squares">
    <p>
        Factor if possible.
        <ol label="a">
            <li><p><m>x^2 - 81</m></p></li>
            <li><p><m>4x^2 - 9y^2</m></p></li>
            <li><p><m>x^2 + 81</m></p></li>
        </ol>
    </p>
<solution>
    <p>
        <ol label="a">
            <li><p>
                The expression <m>x^2 - 81</m> is the difference of two squares, <m>x^2 - 9^2</m>, and thus can be factored according to Special Factorization (III) above.
                <md>
                    <mrow>x^2-81 \amp = x^2-9^2</mrow>
                    <mrow>\amp = (x+9)(x-9)</mrow>
                </md>
            </p></li>
            <li><p>
                Because <m>4x^2 - 9y^2</m> can be written as <m>(2x)^2 -(3y)^2</m>,
                <md>
                    <mrow>4x^2 - 9y^2 \amp =(2x^2)-(3y)^2 </mrow>
                    <mrow>\amp =(2x+3y)(2x-3y) </mrow>
                </md>
            </p></li>
            <li><p>
                The expression <m>x^2 + 81</m>, or <m>x^2 + 0x + 81</m>, is <em>not</em> factorable, because no two real numbers have a product of <m>81</m> and a sum of <m>0</m>.
            </p></li>
        </ol>
    </p>
</solution></example>

<warning>
    <p>
        <m>x^2 + 81\ne (x + 9) (x + 9)</m>, which you can verify by multiplying 
            <me>(x + 9) (x + 9) = x^2 + 18x + 8</me>
    </p>
</warning>

<p>
    The factors <m>x + 9</m> and <m>x - 9</m> in <xref ref="factor-difference-of-squares" text="type-global"/>a are called <term>conjugates</term><idx>conjugates</idx> of each other. In general, any binomials of the form <m>a+b</m> and <m>a-b</m> are called a <term>conjugate pair</term><idx>conjugate pair</idx>.
</p>

</subsection>
</section>


<!-- **************************

-->
<section xml:id="appendix-Working-with-Algebraic-Fractions"><title>Working with Algebraic Fractions</title><introduction>
<p>
    A quotient of two polynomials is called a <term>rational expression</term><idx>rational expression</idx> or an <term>algebraic fraction</term><idx>algebraic fraction</idx>. Operations on algebraic fractions follow the same rules as operations on common fractions.
</p></introduction>

<subsection><title>Reducing Fractions</title>
<p>
    When we reduce an ordinary fraction such as <m>\dfrac{24}{36}</m>, we are using the fundamental principle of fractions.
</p>

<assemblage><title>Fundamental Principle of Fractions</title>
<p>
    If we multiply or divide the numerator and denominator of a fraction by the same (nonzero) number, the new fraction is equivalent to the old one. In symbols,
    <me>\frac{ac}{bc}=\frac{a}{b}, \hphantom{0000} (b, ~c \ne  0)</me>
</p>
</assemblage>

<p>
    Thus, for example,
    <me>\frac{24}{36}=\frac{2\cdot 12}{3\cdot 12}=\frac{2}{3}</me>
</p>
<p>
    We use the same procedure to reduce algebraic fractions: We look for common factors in the numerator and denominator and then apply the fundamental principle.
</p>

<example>
    <p>
        Reduce each algebraic fraction.
        <ol label = "a">
            <li><p><m>\dfrac{8x^3y}{6x^2y^3}</m></p></li>
            <li><p><m>\dfrac{6x-3}{3}</m></p></li>
        </ol>
    </p>
<solution>
    <p>
        Factor out any common factors from the numerator and denominator. Then divide numerator and denominator by the common factors.
        <ol label="a">
            <li><p><m>
                \begin{aligned}
                \frac{8x^3 y}{6x^2 y^3} \amp = \frac{4x \cdot 2x^2 y}{3y^2\cdot 2x^2 y} \\
                \amp = \frac{4x}{3y^2}
                \end{aligned}
            </m></p></li>
            <li><p><m>
                \begin{aligned}
                \frac{6x-3}{3} \amp = \frac{\cancel{3}(2x+1) }{\cancel{3}} \\
                \amp = 2x+1
                \end{aligned}
            </m></p></li>
        </ol>
    </p>
</solution></example>

<p>
    If the numerator or denominator of the fraction contains more than one term, it is especially important to <em>factor</em> before attempting to apply the fundamental principle. We can divide out common <em>factors</em> from the numerator and denominator of a fraction, but the fundamental principle does <em>not</em> apply to common <em>terms</em>.
</p>

<warning>
    <p>
        We can reduce
        <me>\frac{2xy}{3y}= \frac{2x}{3} </me>
        because <m>y</m> is a common factor in the numerator and denominator. However,
        <me>\frac{2x+y}{3+y} \ne \frac{2x}{3} </me>
        because <m>y</m> is a common term but is <em>not a common factor</em> of the numerator and denominator. Furthermore,
        <me>\frac{5x+3}{5y}\ne \frac{x+3}{y} </me>
        because <m>5</m> is not a factor of the <em>entire</em> numerator.
    </p>
</warning>

<example xml:id="example-reduce-fraction">
    <p>
        Reduce each fraction.
        <ol label="a">
            <li><p><m>
                \dfrac{4x+2}{4}
            </m></p></li>
            <li><p><m>
                \dfrac{9x^2+3}{6x+3}
            </m></p></li>
        </ol>
    </p>
<solution>
    <p>
        Factor the numerator and denominator. Then divide numerator and denominator by the common factors.
        <ol label="a">
            <li><p><m>
                \begin{aligned}
                \frac{4x + 2}{4} \amp = \frac{\cancel{2}(2x + 1)}{\cancel{2} (2)}\\
                \amp = \frac{2x+1}{2}
                \end{aligned}
            </m></p></li>
            <li><p><m>
                \begin{aligned}
                \frac{9x^2+3}{6x+3} \amp = \frac{\cancel{3}(3x^2 + 1)}{\cancel{3} (2x+1)}\\
                \amp = \frac{3x^2 + 1}{2x+1}
                \end{aligned}
            </m></p></li>
        </ol>
    </p>
</solution></example>

<warning>
    <p>
        Note that in <xref ref="example-reduce-fraction" text="type-global"/>a above,
        <me>\frac{4x + 2}{4} \ne x+2</me>
        and in <xref ref="example-reduce-fraction" text="type-global"/>b,
        <me>\frac{9x^2+3}{6x+3} \ne \frac{9x^2}{6x}</me>
    </p>
</warning>

<p>
    We summarize the procedure for reducing algebraic fractions as follows.
</p>

<assemblage><title>To Reduce an Algebraic Fraction:</title>
<p>
    <ol label="1">
        <li><p>
            Factor the numerator and denominator.
        </p></li>
        <li><p>
            Divide the numerator and denominator by any common factors.
        </p></li>
    </ol>
</p>
</assemblage>

<example>
    <p>
        Reduce each fraction.
        <ol label="a">
            <li><p><m>
                \dfrac{x^2 - 7x + 6}{36 - x^2}
            </m></p></li>
            <li><p><m>
                \dfrac{27x^3 - 1}{9 x^2-1}
            </m></p></li>
        </ol>
    </p>
<solution>
    <p>
        <ol label="a">
            <li><p>
                Factor numerator and denominator to obtain
                <me>\frac{(x - 6) (x - 1)}{(6 - x) (6 + x)} </me>
                The factor <m>x-6</m> in the numerator is the opposite of the factor <m>6-x</m> in the denominator. That is, <m>x - 6 = -1 (6 - x)</m>. Thus,
                <me>\frac{-1 \cancel{(6 - x )}(x - 1)}{\cancel{(6 - x )}(6 + x)}=\frac{-1 (x - 1)}{6 + x}=\frac{1-x}{6+x} </me>
            </p></li>
            <li><p>
                The numerator of the fraction is a difference of two cubes, and the denominator is a difference of two squares. Factor each to obtain
                <me>\frac{\cancel{(3x-1)}(9x^2+3x+1)}{\cancel{(3x-1)}(3x+1)}=\frac{9x^2+3x+1}{3x+1} </me>
            </p></li>
        </ol>
    </p>
</solution></example>

</subsection>



<subsection><title>Products of Fractions</title>
<p>
    To multiply two or more common fractions together, we multiply their numerators together and multiply their denominators together. The same is true for a product of algebraic fractions. For example, xy
    <md>
        <mrow>\frac{6x^2}{y}\cdot\frac{xy}{2}\amp =\frac{6x^2\cdot}{y\cdot 2}=\frac{6x^3y}{2y}\amp\amp\text{Reduce.} </mrow>
        <mrow>\amp =\frac{3x^3(2y)}{2y}=3x^3 </mrow>
    </md>
    We can simplify the process by first factoring each numerator and denominator and dividing out any common factors.
    <me>\frac{6x^2}{y}\cdot\frac{\cancel{2}\cdot 3x^2}{\bcancel{y}}\cdot\frac{x\bcancel{y}}{\cancel{2}}=3x^3 </me>
</p>
<p>
    In general, we have the following procedure for finding the product of algebraic fractions.
</p>

<assemblage><title>To Multiply Algebraic Fractions:</title>
<p>
    <ol label="1">
        <li><p>
            Factor each numerator and denominator.
        </p></li>
        <li><p>
            Divide out any factors that appear in both a numerator and a denominator.
        </p></li>
        <li><p>
            Multiply together the numerators; multiply together the denominators.
        </p></li>
    </ol>
</p>
</assemblage>

<example>
    <p>
        Find each product.
        <ol label="a">
            <li><p><m>
                \dfrac{5}{x^2-1}\cdot\dfrac{x+2}{x}
            </m></p></li>
            <li><p><m>
                \dfrac{4y^2-1}{4-y^2}\cdot\dfrac{y^2-2y}{4y+2}
            </m></p></li>
        </ol>
    </p>
<solution>
    <p>
        <ol label="a">
            <li><p>
                The denominator of the first fraction factors into <m>(x + 1) (x - 1)</m>. There are no common factors to divide out, so we multiply the numerators together and multiply the denominators together.
                <me>\frac{5}{x^2-1}\cdot\frac{x+2}{x}=\frac{5(x+2)}{x(x^2-1)}=\frac{5x+10}{x^3-x} </me>
            </p></li>
            <li><p>
                Factor each numerator and each denominator. Look for common factors.
                <md>
                    <mrow>
                        \frac{4y^2-1}{4-y^2}\cdot\frac{y^2-2y}{4y+2} \amp = \frac{(2y-1)(\bcancel{2y+1})}{\cancel{(2-y)}(2+y)} \cdot \frac{y\cancelto{\alert{-1}}{(y-2)}}{2(\bcancel{y+1})}\amp\amp\text{Divide out common factors.}
                    </mrow>
                    <mrow>
                        \amp =\frac{-y(2y-1)}{y(y+2)} \amp\amp\text{Note: }y-2=-(2-y)
                    </mrow>
                </md>
            </p></li>
        </ol>
    </p>
</solution></example>

</subsection>


<subsection><title>Quotients of Fractions</title>
<p>
    To divide two algebraic fractions we multiply the first fraction by the reciprocal of the second fraction. For example,
    <md>
        <mrow>
            \frac{2x^3}{3y}\div \frac{4x}{5y^2}\amp = \frac{2x^3}{3y}\cdot \frac{5y^2}{4x}
        </mrow>
        <mrow>
            \amp = \frac{\bcancel{2x}\cdot x^2}{3\cancel{y}}\cdot \frac{5y\cdot\cancel{y}}{2\cdot \bcancel{2x}}=\frac{5x^2y}{6}
        </mrow>
    </md>
</p>
<p>
    If the fractions involve polynomials of more than one term, we may need to factor each numerator and denominator in order to recognize any common factors. This suggests the following procedure for dividing algebraic fractions.
</p>

<assemblage><title>To Divide Algebraic Fractions:</title>
<p>
    <ol label="1">
        <li><p>
            Multiply the first fraction by the reciprocal of the second fraction.
        </p></li>
        <li><p>
            Factor each numerator and denominator.
        </p></li>
        <li><p>
            Divide out any factors that appear in both a numerator and a denominator.
        </p></li>
        <li><p>
            Multiply together the numerators; multiply together the denominators.
        </p></li>
    </ol>
</p>
</assemblage>

<example>
    <p>
        Find each quotient.
        <ol label="a">
            <li><p><m>
                \dfrac{x^2-1}{x+3}\div\dfrac{x^2-x-2}{x^2+5x+6}
            </m></p></li>
            <li><p><m>
                \dfrac{6ab}{2a+b}\div(4a^2b)
            </m></p></li>
        </ol>
    </p>
<solution>
    <p>
        <ol label="a">
            <li><p>
                Multiply the first fraction by the reciprocal of the second fraction.
                <md>
                    <mrow>
                        \frac{x^2-1}{x+3}\div\frac{x^2-x-2}{x^2+5x+6}
                        \amp =\frac{x^2-1}{x+3}\cdot\frac{x^2+5x+6}{x^2-x-2} \amp\amp\text{Factor.}
                    </mrow>
                    <mrow>
                        \amp = \frac{(x-1)(\cancel{x+1})}{\bcancel{x+3}}\cdot\frac{(\bcancel{x+3})(x+2)}{(\cancel{x+1})(x-2)}
                    </mrow>
                    <mrow>
                        \amp = \frac{(x-1)(x+2)}{x-2}
                    </mrow>
                </md>
            </p></li>
            <li><p>
                Multiply the first fraction by the reciprocal of the second fraction.
                <md>
                    <mrow>
                        \frac{6ab}{2a+b}\div(4a^2b) \amp =\frac{\cancelto{\alert{3}}{6}\bcancel{ab}}{2a+b}\cdot\frac{1}{\cancelto{\alert{2}}{4}a\cdot\bcancel{ab}}
                        \amp\amp\text{Divide out common factors.}
                    </mrow>
                    <mrow>
                        \amp = \frac{3}{2a(2a+b)}=\frac{3}{4a^2+2ab}
                    </mrow>
                </md>
            </p></li>
        </ol>
    </p>
</solution></example>

</subsection>


<subsection><title>Sums and Differences of Like Fractions</title>
<p>
    Algebraic fractions with the same denominator are called <term>like fractions</term><idx>like fractions</idx>. To add or subtract like fractions, we combine their numerators and keep the same denominator for the sum or difference. This method is an application of the distributive law.
</p>

<example>
    <p>
        Find each sum or difference.
        <ol label="a">
            <li><p><m>
                \dfrac{2x}{9z^2}+\dfrac{5x}{9z^2}
            </m></p></li>
            <li><p><m>
                \dfrac{2x-1}{x+3}-\dfrac{5x-3}{x+3}
            </m></p></li>
        </ol>
    </p>
<solution>
    <p>
        <ol label="a">
            <li><p>
                Because these are like fractions, we add their numerators and keep the same denominator.
                <me>
                    \frac{2x}{9z^2}+\frac{5x}{9z^2}=\frac{2x+5x}{9z^2}=\frac{7x}{9z^2}
                </me>
            </p></li>
            <li><p>
                Be careful to subtract the <em>entire</em> numerator of the second fraction: Use parentheses to show that the subtraction applies to both terms of <m>5x - 3</m>.
                <md>
                    <mrow>
                        \frac{2x-1}{x+3}-\frac{5x-3}{x+3}\amp = \frac{2x-1-(5x-3)}{x+3}
                    </mrow>
                    <mrow>
                        \amp =\frac{2x-1-5x+3}{x+3}=\frac{-3x+2}{x+3}
                    </mrow>
                </md>
            </p></li>
        </ol>
    </p>
</solution></example>

</subsection>



<subsection><title>Lowest Common Denominator</title>
<p>
    To add or subtract fractions with different denominators, we must first find a <term>common denominator</term><idx>common denominator</idx>. For arithmetic fractions, we use the smallest natural number that is exactly divisible by each of the given denominators. For example, to add the fractions <m>\dfrac{1}{6}</m> and <m>\dfrac{3}{8}</m>, we use <m>24</m> as the common denominator because <m>24</m> is the smallest natural number that both <m>6</m> and <m>8</m> divide into evenly.
</p>
<p>
    We define the <term>lowest common denominator</term><idx>lowest common denominator</idx> (<term>LCD</term><idx>LCD</idx>) of two or more algebraic fractions as the polynomial of least degree that is exactly divisible by each of the given denominators.
</p>

<example xml:id="example-LCD">
    <p>
        Find the LCD for the fractions <m>\dfrac{3x}{x+2}</m> and <m>\dfrac{2x}{x-3}</m>
    </p>
<solution>
    <p>
        The LCD is a polynomial that has as factors both <m>x + 2</m> and <m>x-3</m>. The simplest such polynomial is <m>(x + 2) (x - 3)</m>, or <m>x^2 - x - 6</m>. For our purposes, it will be more convenient to leave the LCD in factored form, so the LCD is <m>(x + 2) (x - 3)</m>.
    </p>
</solution></example>
<p>
    The LCD in <xref ref="example-LCD" text="type-global"/> was easy to find because each original denominator consisted of a single factor; that is, neither denominator could be factored. In that case, the LCD is just the product of the original denominators. We can always find a common denominator by multiplying together all the denominators in the given fractions, but this may not give us the <em>simplest</em> or <em>lowest</em> common denominator. Using anything other than the simplest possible common denominator will complicate our work needlessly.
</p>
<p>
    If any of the denominators in the given fractions can be factored, we factor them before looking for the LCD.
</p>

<assemblage><title>To Find the LCD of Algebraic Fractions:</title>
<p>
    <ol label="1">
        <li><p>
            Factor each denominator completely.
        </p></li>
        <li><p>
            Include each different factor in the LCD as many times as it occurs in any <em>one</em> of the given denominators.
        </p></li>
    </ol>
</p>
</assemblage>

<example xml:id="example-LCD2">
    <p>
         Find the LCD for the fractions <m>\dfrac{2x}{x^2-1}</m> and <m>\dfrac{x+3}{x^2+x}</m>.
    </p>
<solution>
    <p>
        Factor the denominators of each of the given fractions.
        <me>x^2 - 1 = (x - 1) (x + 1)~~~\text{ and }~~~x^2 + x = x (x + 1)</me>
        The factor <m>(x - 1)</m> occurs once in the first denominator, the factor <m>x</m> occurs once in the second denominator, and the factor <m>(x + 1)</m> occurs once in each denominator. Therefore, we include in our LCD one copy of each of these factors. The LCD is <m>x (x + 1) (x - 1)</m>.
    </p>
</solution></example>

<warning>
    <p>
        In <xref ref="example-LCD2" text="type-global"/>, we do not include two factors of <m>(x + 1)</m> in the LCD. We need only one factor of <m>(x + 1)</m> because <m>(x + 1)</m> occurs only once in either denominator. You should check that each original denominator divides evenly into our LCD, <m>x (x + 1)(x - 1)</m>.
    </p>
</warning>

</subsection>



<subsection><title>Building Fractions</title>
<p>
    After finding the LCD, we <term>build</term><idx>build</idx> each fraction to an equivalent one with the LCD as its denominator. The new fractions will be like fractions, and we can combine them as explained above.
</p>
<p>
    Building a fraction is the opposite of reducing a fraction, in the sense that we multiply, rather than divide, the numerator and denominator by an appropriate factor. To find the <term>building factor</term><idx>building factor</idx>, we compare the factors of the original denominator with those of the desired common denominator.
</p>

<example xml:id="example-build-fractions">
    <p>
        Build each of the fractions <m>\dfrac{3x}{x+2} </m> and <m>\dfrac{2x}{x-3} </m> to equivalent fractions with the LCD <m>(x + 2)(x - 3)</m> as denominator.
    </p>
<solution>
    <p>
        Compare the denominator of the given fraction to the LCD. We see that the fraction <m>\dfrac{3x}{x+2} </m> needs a factor of <m>(x - 3)</m> in its denominator, so <m>(x - 3)</m> is the building factor for the first fraction. We multiply the numerator and denominator of the first fraction by <m>(x - 3)</m> to obtain an equivalent fraction:
        <me>\frac{3x}{x+2}=\frac{3x\alert{(x-3)}}{(x+2)\alert{(x-3)}}=\frac{3x^2-9x}{x^2-x-6} </me>
        The fraction <m>\dfrac{2x}{x-3} </m> needs a factor of <m>(x + 2)</m> in the denominator, so we multiply numerator and denominator by <m>(x + 2)</m>:
        <me>
            \frac{2x}{x-3}=\frac{2x\alert{(x+2)}}{(x-3)\alert{(x+2)}}= \frac{2x^2 + 4x}{x^2 - x - 6}
        </me>
    </p>
</solution></example>
<p>
    The two new fractions we obtained in <xref ref="example-build-fractions" text="type-global"/> are like fractions; they have the same denominator.
</p>

</subsection>

<subsection><title>Sums and Differences of Unlike Fractions</title>
<p>
    We are now ready to add or subtract algebraic fractions with unlike denominators. We will do this in four steps.
</p>
<assemblage><title>To Add or Subtract Fractions with Unlike Denominators:</title>
<p>
    <ol label="1">
        <li><p>
            Find the LCD for the given fractions.
        </p></li>
        <li><p>
            Build each fraction to an equivalent fraction with the LCD as its denominator.
        </p></li>
        <li><p>
            Add or subtract the numerators of the resulting like fractions. Use the LCD as the denominator of the sum or difference.
        </p></li>
        <li><p>
            Reduce the sum or difference, if possible.
        </p></li>
    </ol>
</p>
</assemblage>

<example>
    <p>
        Subtract <m>\dfrac{3x}{x+2}-\dfrac{2x}{x-3} </m>.
    </p>
<solution>
    <p>
        <ol>
            <li><p>
                The LCD for these fractions is <m>(x + 2) (x - 3)</m>.
            </p></li>
            <li><p>
                We build each fraction to an equivalent one with the LCD, as we did in <xref ref="example-build-fractions" text="type-global"/>.
                <me>
                    \frac{3x}{x+2}=\frac{3x^2-9x}{x^2-x-6} ~~\text{ and }~~ \frac{2x}{x-3}=\frac{2x^2+4x}{x^2-x-6}
                </me>
            </p></li>
            <li><p>
                Combine the numerators over the same denominator.
                <md>
                    <mrow>
                        \frac{3x}{x+2}-\frac{2x}{x-3}\amp =\frac{3x^2-9x}{x^2-x-6}-\frac{2x^2+4x}{x^2-x-6}
                        \amp\amp\text{Subtract the numerators.}
                    </mrow>
                    <mrow>
                        \amp = \frac{(3x^2-9x)-(2x^2+4x)}{x^2-x-6}
                    </mrow>
                    <mrow>
                        \amp = \frac{x^2-13x}{x^2-x-6}
                    </mrow>
                </md>
            </p></li>
            <li><p>
                Reduce the result, if possible. If we factor both numerator and denominator, we find
                <me>\frac{x(x-13)}{(x-3)(x+2)} </me>
                The fraction cannot be reduced.
            </p></li>
        </ol>
    </p>
</solution></example>

<example>
    <p>
        Write as a single fraction: <m>\displaystyle{1 + \frac{2}{a}-\frac{a^2 + 2}{a^2 + a}}</m>.
    </p>
<solution>
    <p>
        <ol>
            <li><title>Step 1</title><p>
                To find the LCD, factor each denominator:
                <md>
                    <mrow>a\amp =a</mrow>
                    <mrow>a^2 + a\amp = a (a + 1)</mrow>
                </md>
                The LCD is <m>a(a + 1)</m>.
            </p></li>
            <li><title>Step 2</title><p>
                Build each term to an equivalent fraction with the LCD as denominator. (The building factors for each fraction are shown in color.) The third fraction already has the LCD for its denominator.
                <md>
                    <mrow>
                        1 = \frac{1 \cdot \alert{a (a + 1)}}{1 \cdot\alert{a (a + 1)}}\amp =\frac{a^2+a} {a (a + 1)}
                    </mrow>
                    <mrow>
                        \frac{2}{a}=\frac{2\cdot\alert{(a+1)} }{a\cdot\alert{(a+1)}}\amp =\frac{2a+2}{a(a+1)}
                    </mrow>
                    <mrow>
                        \frac{a^2 + 2}{a^2 + a} \amp = \frac{a^2 + 2}{a(a+1)}
                    </mrow>
                </md>
            </p></li>
            <li><title>Step 3</title><p>
                Combine the numerators over the LCD.
                <md>
                    <mrow>
                        1 + \frac{2}{a}-\frac{a^2 + 2}{a^2 + a} \amp = \frac{a^2+a} {a (a + 1)} + \frac{2a+2}{a(a+1)}- \frac{a^2 + 2}{a(a+1)}
                    </mrow>
                    <mrow>
                        \amp = \frac{a^2+a +(2a+2)-(a^2+2) } {a (a + 1)}
                    </mrow>
                    <mrow>
                        \amp = \frac{3a} {a (a + 1)}
                    </mrow>
                </md>
            </p></li>
            <li><title>Step 4</title><p>
                Reduce the fraction to find
                <me>
                    \frac{3\cancel{a}}{\cancel{a}(a+1)}=\frac{3}{a+1}
                </me>
            </p></li>
        </ol>
    </p>
</solution>
</example>

</subsection>



<subsection><title>Complex Fractions</title>
<p>
    A fraction that contains one or more fractions in either its numerator or its denominator or both is called a complex fraction. For example,
    <me>
        \dfrac{\dfrac{2}{3}}{\dfrac{5}{6}} ~~~\text{ and }~~~ \dfrac{x+\dfrac{3}{4}}{x-\dfrac{1}{2}}
    </me>
    are complex fractions. Like simple fractions, complex fractions represent quotients. For the examples above,
    <me>
        \dfrac{\dfrac{2}{3}}{\dfrac{5}{6}}=\dfrac{2}{3}\div\dfrac{5}{6} ~~~\text{ and }~~~ 
        \dfrac{x+\dfrac{3}{4}}{x-\dfrac{1}{2}} = \left(x+\dfrac{3}{4}\right)\div\left(x-\dfrac{1}{2}\right)
    </me>
    We can always simplify a complex fraction into a standard algebraic fraction. If the denominator of the complex fraction is a single term, we can treat the fraction as a division problem and multiply the numerator by the reciprocal of the denominator. Thus,
    <me>
        \dfrac{\dfrac{2}{3}}{\dfrac{5}{6}}=\dfrac{2}{3}\div\dfrac{5}{6} =\dfrac{2}{3}\cdot\dfrac{6}{5} = \dfrac{4}{5}
    </me>
</p>
<p>
    If the numerator or denominator of the complex fraction contains more than one term, it is easier to use the fundamental principle of fractions to simplify the expression.
</p>

<example>
    <p>
        Simplify <m>\displaystyle{\frac{x+\dfrac{3}{4}}{x-\dfrac{1}{2}} }</m>
    </p>
<solution>
    <p>
        Consider all of the simple fractions that appear in the complex fraction; in this example <m>\dfrac{1}{2} </m> and <m>\dfrac{3}{4} </m>. The LCD of these fractions is <m>4</m>. If we multiply the numerator and denominator of the complex fraction by <m>4</m>, we will eliminate the fractions within the fraction. Be sure to multiply <em>each</em> term of the numerator and <em>each</em> term of the denominator by <m>4</m>.
        <me>
            \frac{\alert{4}\left(x+\dfrac{3}{4}\right)}{\alert{4}\left(x-\dfrac{1}{2}\right)}
            = \frac{\alert{4}(x)+\alert{4}\left(\dfrac{3}{4}\right)}{\alert{4}(x)-\alert{4}\left(\dfrac{1}{2}\right)}
            =\frac{4x+3}{4x-2}
        </me>
        Thus, the original complex fraction is equivalent to the simple fraction <m>\displaystyle{\frac{4x+3}{4x-2}} </m>.
    </p>
</solution></example>

<p>
    We summarize the method for simplifying complex fractions as follows.
</p>
<assemblage><title>To Simplify a Complex Fraction:</title>
<p>
    <ol label="1">
        <li><p>
            Find the LCD of all the fractions contained in the complex fraction.
        </p></li>
        <li><p>
            Multiply each term in the numerator and the denominator of the complex fraction by the LCD.
        </p></li>
        <li><p>
            Reduce the resulting simple fraction, if possible.
        </p></li>
    </ol>
</p>
</assemblage>
</subsection>



<subsection><title>Negative Exponents</title>
<p>
    Algebraic fractions are sometimes written using negative exponents. (You can review negative exponents in <xref ref="Variation" auto="yes"/>.)
</p>

<example xml:id="example-negative-exponents-complex-fractions">
    <p>
        Write each expression as a single algebraic fraction.
        <ol label="a">
            <li><p><m>
                x^{-1}-y^{-1}
            </m></p></li>
            <li><p><m>
                \left(x^{-2}+y^{-2}\right)^{-1}
            </m></p></li>
        </ol>
    </p>
<solution>
    <p>
        <ol label="a">
            <li><p><m>
                \displaystyle{x^{-1}-y^{-1}=\frac{1}{x}-\frac{1}{y} ~~~\text{ or } ~~~\frac{y-x}{xy}}
            </m></p></li>
            <li><p><m>
                \displaystyle{\left(x^{-2}+y^{-2}\right)^{-1}=\left(\frac{1}{x^{2}}+\frac{1}{y^{2}}\right)^{-1}
                =\left(\frac{y^{2}+x^{2}}{x^2y^2}\right)^{-1}=\frac{x^2y^2}{y^2+x^2}}
            </m></p></li>
        </ol>
    </p>
</solution></example>

<p>
    When working with fractions and exponents, it is important to avoid some tempting but <em>incorrect</em> algebraic operations.
</p>

<warning>
    <p>
        <ol label="1">
            <li><p>
                In <xref ref="example-negative-exponents-complex-fractions" text="type-global"/>a, note that
                <me>\frac{1}{x}-\frac{1}{y}\ne\frac{1}{x-y} </me>
                For example, you can check that for <m>x=2</m> and <m>y=3</m>,
                <me>\frac{1}{2}-\frac{1}{3}\ne\frac{1}{2-3}=-1 </me>
            </p></li>
            <li><p>
                 In <xref ref="example-negative-exponents-complex-fractions" text="type-global"/>b, note that
                 <me>\left(x^{-2}+y^{-2}\right)^{-1}\ne x^2+y^2</me>
            </p></li>
        </ol>
    </p>
</warning>

<p>
    In general, the fourth law of exponents does <em>not</em> apply to sums and differences; that is,
    <me>(a+b)^n\ne a^n + b^n </me>
</p>

</subsection>

</section>




<!-- **************************

-->
<section xml:id="appendix-Working-with-Radicals"><title>Working with Radicals</title><introduction>
<p>In some situations, radical notation is more convenient to use than exponents. In these cases, we usually simplify radical expressions algebraically as much as possible before using a calculator to obtain decimal approximations.</p></introduction>

<subsection><title>Properties of Radicals</title>
<p>
    Because <m>\sqrt[n]{a}=a^{1/n}</m>, we can use the laws of exponents to derive two important properties that are useful in simplifying radicals.
</p>

<assemblage><title>Properties of Radicals</title>
<p>
    <ol label="1">
        <li><p><m>\displaystyle{
            \sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b}\text{, }\hphantom{blank000}\text{for } a,  b \ge 0
        }
        </m></p></li>
        <li><p><m>\displaystyle{
            \sqrt[n]{\frac{a}{b}}=\frac{\sqrt[n]{a}}{\sqrt[n]{b}}\text{, }\hphantom{blankblank}\text{for } a\ge 0,~~  b \gt 0
        }
        </m></p></li>
    </ol>
</p>
</assemblage>
<p>
    As examples, you can verify that
    <me>
        \sqrt{36}=\sqrt{4}\sqrt{9} ~~~\text{ and }~~~ \sqrt[3]{\frac{1}{8}}=\frac{\sqrt[3]{1}}{\sqrt[3]{8}}
    </me>
</p>

<example>
    <p>
        Which of the following are true?
        <ol label="a">
            <li><p>
                Is <m>\sqrt{36+64}= \sqrt{36}+\sqrt{64}\text{?}
            </m></p></li>
            <li><p>
                Is <m>\sqrt[3]{8(64)}= \sqrt[3]{8}\sqrt[3]{64}\text{?}
            </m></p></li>
            <li><p>
                Is <m>\sqrt{x^2+4}= x+2\text{?}
            </m></p></li>
            <li><p>
                Is <m>\sqrt[3]{8x^3}= 2x\text{?}
            </m></p></li>
        </ol>
    </p>
<solution>
    <p>
        The statements in (b) and (d) are true, and both are examples of the first property of radicals. Statements (a) and (c) are false. In general, <m>\sqrt[n]{a+b}</m> is not equal to <m>\sqrt[n]{a}+\sqrt[n]{b}</m>, and <m>\sqrt[n]{a-b}</m> is not equal to <m>\sqrt[n]{a}-\sqrt[n]{b}</m>.
    </p>
</solution></example>
</subsection>


<subsection><title>Simplifying Radicals</title>
<p>
    We use Property (1) to simplify radical expressions by factoring the radicand. For example, to simplify <m>\sqrt[3]{108}</m>, we look for perfect cubes that divide evenly into <m>108</m>. The easiest way to do this is to try the perfect cubes in order: <m>1, 8, 27, 64, 125, \ldots</m> and so on, until we find one that is a factor. For this example, we find that <m>108 = 27 · 4</m>. Using Property (1), we write
    <me>\sqrt[3]{108}=\sqrt[3]{27}\sqrt[3]{4} </me>
    Simplify the first factor to find
    <me>\sqrt[3]{108}= 3 \sqrt[3]{4} </me>
    This expression is considered simpler than the original radical because the new radicand, <m>4</m>, is smaller than the original, <m>108</m>.
</p>
<p>
    We can also simplify radicals containing variables. If the exponent on the variable is a multiple of the index, we can extract the variable from the radical. For instance,
    <me>\sqrt[3]{12}=x^{12/3}=x^4 </me>
    (You can verify this by noting that <m>(x^4)^3 = x^{12}</m>.) If the exponent on the variable is not a multiple of the index, we factor out the highest power that is a multiple. For example,
    <md>
        <mrow>
            \sqrt[3]{x^{11}} \amp = \sqrt[3]{x^9 \cdot x^2}\amp\amp\text{Apply Property (1).}
        </mrow>
        <mrow>\amp = \sqrt[3]{x^9}\cdot\sqrt[3]{2} \amp\amp\text{Simplify }\sqrt[3] {x^9}=x^{9/3}. </mrow>
    </md>
    <md>
        \amp = x^3 \sqrt[3]{x^2}
    </md>
</p>

<example>
    <p>
        Simplify each radical.
        <ol label="a">
            <li><p><m>
                \sqrt{18x^5}
            </m></p></li>
            <li><p><m>\sqrt[3]{24x^6y^8} </m></p></li>
        </ol>
    </p>
<solution>
    <p>
        <ol label = "a">
            <li><p>
                The index of the radical is <m>2</m>, so we look for perfect square factors of <m>18x^5</m>. The factor <m>9</m> is a perfect square, and <m>x^4</m> has an exponent divisible by <m>2</m>. Thus,
                <md>
                    <mrow>
                        \sqrt{18x^5} \amp = \sqrt{9x^4\cdot 2x}\amp\amp\text{Apply Property (1).}
                    </mrow>
                    <mrow>
                        \amp = \sqrt{9x^4} \sqrt{2x}\amp\amp\text{Take square roots.}
                    </mrow>
                    <mrow>
                        \amp = 3x^2\sqrt{2x}
                    </mrow>
                </md>
            </p></li>
            <li><p>
                The index of the radical is <m>3</m>, so we look for perfect cube factors of <m>24x^6 y^8</m>. The factor <m>8</m> is a perfect cube, and <m>x^6</m> and <m>y^6</m> have exponents divisible by <m>3</m>. Thus,
                <md>
                    <mrow>
                        \sqrt[3]{24x^6y^8} \amp = \sqrt[3]{8x^6 y^6 \cdot 3 y^2}\amp\amp\text{Apply Property (1).}
                    </mrow>
                    <mrow>
                        \amp = \sqrt[3]{8x^6 y^6} \sqrt[3]{3 y^2}\amp\amp\text{Take cube roots.}
                    </mrow>
                    <mrow>
                        \amp = 2x^2 y^2\sqrt[3]{3y^2}
                    </mrow>
                </md>
            </p></li>
        </ol>
    </p>
</solution></example>

<warning>
    <p>
        Property (1) applies only to products under the radical, not to sums or differences. Thus, for example,
        <me>
            \sqrt{4\cdot 9}=\sqrt{4}\sqrt{9}=2\cdot 3, ~~~\text{ but }~~~\sqrt{4+ 9}\ne \sqrt{4}+\sqrt{9}
        </me>
        and
        <me>
            \sqrt[3]{x^3y^6}=\sqrt[3]{x^3}\sqrt[3]{y^6}=xy^2, ~~~\text{ but }~~~\sqrt[3]{x^3-y^6}\ne \sqrt[3]{x^3}-\sqrt[3]{y^6}
        </me>
    </p>
</warning>

<p>
    To simplify roots of fractions, we use Property (2), which allows us to write the expression as a quotient of two radicals.
</p>

<example>
    <p>
        <ol label="a">
            <li><p><m>\displaystyle{
                \sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{\sqrt{4}}=\frac{\sqrt{3}}{2}
            }
            </m></p></li>
            <li><p><m>
                \displaystyle{\sqrt[3]{\frac{5}{8}} =\frac{\sqrt[3]{5}}{\sqrt[3]{8}}=\frac{\sqrt[3]{5}}{2} }
            </m></p></li>
        </ol>
    </p>
</example>

<p>
    We can also use Properties (1) and (2) to simplify products and quotients of radicals.
</p>

<example>
    <p>
        Simplify.
        <ol label="a">
            <li><p><m>
                \sqrt[4]{6x^2} \sqrt[4]{8x^3}
            </m></p></li>
            <li><p><m>\displaystyle{
                \frac{\sqrt[3]{16y^5}}{\sqrt[3]{y^2}}
            }
            </m></p></li>
        </ol>
    </p>
<solution>
    <p>
        <ol label="a">
            <li><p>
                First apply Property (1) to write the product as a single radical, then simplify.
                <md>
                    <mrow>
                        \sqrt[4]{6x^2} \sqrt[4]{8x^3}\amp =\sqrt[4]{48x^5}\amp\amp\text{Factor out perfect fourth powers.} 
                    </mrow>
                    <mrow>
                        \amp = \sqrt[4]{16x^4}\sqrt[4]{3x}\amp\amp\text{Simplify.}
                    </mrow>
                    <mrow>
                        \amp = 2x \sqrt[4]{3x}
                    </mrow>
                </md>
            </p></li>
            <li><p>
                Apply Property (2) to write the quotient as a single radical.
                <md>
                    <mrow>
                       \frac{\sqrt[3]{16y^5}}{\sqrt[3]{y^2}}\amp =\sqrt[3]{\frac{16y^5}{y^2}}\amp\amp\text{Reduce.} 
                    </mrow>
                    <mrow>
                        \amp = \sqrt[3]{16y^3}\amp\amp\text{Simplify: factor out perfect cubes.}
                    </mrow>
                    <mrow>
                        \amp = \sqrt[3]{8y^3}\sqrt[3]{2}
                    </mrow>
                    <mrow>
                        \amp = 2y \sqrt[3]{2}
                    </mrow>
                </md>
            </p></li>
        </ol>
    </p>
</solution>
</example>
</subsection>


<subsection><title>Sums and Differences of Radicals</title>
<p>
    You know that sums or differences of like terms can be combined by adding or subtracting their coefficients:
    <me>
        3xy + 5xy = (3 + 5)xy = 8xy
    </me>
    Like radicals, that is, radicals of the same index and radicand, can be combined in the same way.
</p>

<example xml:id="example-like-radicals">
    <p>
        <ol label="a">
            <li><p><m>
                \begin{aligned}
                3\sqrt{3}+4\sqrt{3} \amp = (3+4)\sqrt{3}\\
                \amp = 7\sqrt{3}
                \end{aligned}
            </m></p></li>y
            <li><p><m>
                \begin{aligned}
                4\sqrt[3]{y}-6\sqrt[3]{y} \amp = (4-6)\sqrt[3]{y}\\
                \amp = -2\sqrt[3]{y}
                \end{aligned}
            </m></p></li>
        </ol>
    </p>
</example>

<warning>
    <p>
        <ol label="1">
            <li><p>
                In <xref ref="example-like-radicals" text="type-global"/>a, <m>3\sqrt{3}+4\sqrt{3}\ne 7\sqrt{6} </m>. Only the coefficients are added; the radicand does not change.
            </p></li>
            <li><p>
                Sums of radicals with different radicands or different indices cannot be combined. Thus, 
                <me>\sqrt{11}+\sqrt{5}\ne\sqrt{16} </me>
                <me>\sqrt[3]{10x}-\sqrt[3]{2x}\ne\sqrt[3]{8x} </me>
                and
                <me>\sqrt[3]{7}+\sqrt{7}\ne\sqrt[5]{7} </me>
                None of the expressions above can be simplified.
            </p></li>
        </ol>
    </p>
</warning>
</subsection>


<subsection><title>Products of Radicals</title>
<p>
    According to Property (1), radicals of the same index can be multiplied together.
</p>
<assemblage><title>Product of Radicals</title>
<p>
    <me>\sqrt[n]{a} \sqrt[n]{b}=\sqrt[n]{ab} \hphantom{blankblank} (a, b \ge 0) </me>
</p>
</assemblage>

<p>
    Thus, for example,
    <me>
        \sqrt{2}\sqrt{18}\sqrt{36}=6 ~~~\text{ and }~~~\sqrt[3]{2x}\sqrt[3]{4x^2}=\sqrt[3]{8x^3}=2x
    </me>
</p>
<p>
    For products involving binomials, we can apply the distributive law.
</p>

<example><p>
<ol label="a">
    <li><p>
        <m>
            \begin{aligned}
            \sqrt{3} \left(\sqrt{2x}+\sqrt{6} \right) \amp = \sqrt{3\cdot 2x}+ \sqrt{3\cdot 6}\\
            \amp = \sqrt{6x}+\sqrt{18}=\sqrt{6x}+3\sqrt{2}
            \end{aligned}
        </m>
    </p></li>
    <li><p>
        <m>
            \begin{aligned}
            \left(\sqrt{x}-\sqrt{y} \right)\left(\sqrt{x}+\sqrt{y} \right) \amp = \sqrt{x^2}+\sqrt{xy}-\sqrt{xy}-\sqrt{y^2}\\
            \amp = x-y
            \end{aligned}
        </m>
    </p></li>
</ol></p>
</example>
</subsection>



<subsection><title>Rationalizing the Denominator</title>
<p>
    It is easier to work with radicals if there are no roots in the denominators of fractions. We can use the fundamental principle of fractions to remove radicals from the denominator. This process is called <term>rationalizing the denominator</term><idx>rationalizing the denominator</idx>. For square roots, we multiply the numerator and denominator of the fraction by the radical in the denominator.
</p>
<example>
    <p>
        Rationalize the denominator of each fraction.
        <ol label="a">
            <li><p><m>
                \displaystyle{\sqrt{\frac{1}{3}}}
            </m></p></li>
            <li><p><m>
                \displaystyle{\frac{\sqrt{2}}{\sqrt{50x}} }
            </m></p></li>
        </ol>
    </p>
<solution>
    <p>
        <ol label="a">
            <li><p>
                Apply Property (2) to write the radical as a quotient.
                <m>
                    \begin{aligned}
                    \sqrt{\frac{1}{3}}\amp = \frac{\sqrt{1}}{\sqrt{3}} \\
                    \amp =\frac{1}{\sqrt{3}}
                    \amp\amp\text{Multiply numerator and denominator by } \sqrt{3}.\\
                    \amp = \frac{1\cdot \alert{\sqrt{3}}}{\sqrt{3}\cdot \alert{\sqrt{3}}}\\
                    \amp = \frac{\sqrt{3}}{3}
                    \end{aligned}
                </m>
            </p></li>
            <li><p>
                It is always best to simplify the denominator before rationalizing.
                <m>
                    \begin{aligned}
                    \frac{\sqrt{2}}{\sqrt{50x}}\amp = \frac{\sqrt{2}}{5\sqrt{2x}} 
                    \amp\amp\text{Multiply numerator and denominator by } \sqrt{2x}.\\
                    \amp = \frac{\sqrt{2}\cdot \alert{\sqrt{2x}}}{5\sqrt{2x}\cdot \alert{\sqrt{2x}}}
                    \amp\amp\text{Simplify.} \\
                    \amp = \frac{\sqrt{4x}}{5(2x)}\\
                    \amp =\frac{2\sqrt{x}}{10x}
                    \end{aligned}
                </m>
            </p></li>
        </ol>
    </p>
</solution></example>
<p>
    If the denominator of a fraction is a <em>binomial</em> in which one or both terms is a radical, we can use a special building factor to rationalize it. First, recall that
    <me>(p - q) (p + q) = p^2 - q^2</me>
    where the product consists of perfect squares only. Each of the two factors <m>p - q</m> and <m>p + q</m> is said to be the <term>conjugate</term><idx>conjugate</idx> of the other.
</p>
<p>
    Now consider a fraction of the form
    <me>
        \frac{a}{b+\sqrt{c}}
    </me>
    If we multiply the numerator and denominator of this fraction by the conjugate of the denominator, we get
    <me>
        \frac{a \alert{(b-\sqrt{c})}}{(b+\sqrt{c})\alert{(b-\sqrt{c})}}=\frac{ab-a\sqrt{c}}{b^2-(\sqrt{c})^2}
        =\frac{ab-a\sqrt{c}}{b^2-c}
    </me>
    The denominator of the fraction no longer contains any radicals—it has been rationalized.
</p>
<p>
    Multiplying numerator and denominator by the conjugate of the denominator also works on fractions of the form
    <me>\frac{a}{\sqrt{b}+c} ~~~\text{ and }~~~\frac{a}{\sqrt{b}+\sqrt{c}} </me>
    We leave the verification of these cases as exercises.
</p>

<example>
    <p>
        Rationalize the denominator: <m>\displaystyle{\frac{x}{\sqrt{2}+\sqrt{x}}} </m>.
    </p>
<solution>
    <p>
        Multiply numerator and denominator by the conjugate of the denominator, <m>\sqrt{2}-\sqrt{x} </m>.
        <me>
            \frac{x (\alert{\sqrt{2}-\sqrt{x}}) }{(\sqrt{2}+\sqrt{x})(\alert{\sqrt{2}-\sqrt{x}} )}
            =\frac{x(\sqrt{2}-\sqrt{x})}{2-x}
        </me>
    </p>
</solution></example>
</subsection>



<subsection><title>Simplifying <m>\sqrt[n]{x^n} </m></title>
<p>
    Raising to a power is the inverse operation for extracting roots; that is,
    <me>\boxed{\hphantom{0000} \left(\sqrt[n]{a} \right)^n =a \hphantom{0000}}</me>
    as long as <m>\sqrt[n]{a} </m> is a real number. For example,
    <me>
        \left(\sqrt[4]{16} \right)^4= 2^4 = 16,~~~\text{ and }~~~\left(\sqrt[3]{-125} \right)^3= (-5)^3 = -125
    </me>
</p>
<p>
    Now consider the power and root operations in the opposite order; is it true that <m>\sqrt[n]{a^n}=a </m>? If the index <m>n</m> is an odd number, then the statement is always true. For example,
    <me>
        \sqrt[3]{2^3}=\sqrt[3]{8}=2 ~~~\text{ and }~~~\sqrt[3]{(-2)^3}=\sqrt[3]{-8}=-2
    </me>
    However, if <m>n</m> is even, we must be careful. Recall that the principal root <m>\sqrt[n]{x} </m> is always positive, so if <m>a</m> is a negative number, it cannot be true that <m>\sqrt[n]{a^n}=a </m>. For example, if <m>a = -3</m>, then
    <me>
        \sqrt{(-3)^2}=\sqrt{9}=3
    </me>
    Instead, we see that, for even roots, <m>\sqrt[n]{a^n}=\abs{a} </m>.
</p>
<p>
    We summarize our results in below.
</p>

<assemblage><title>Roots of Powers</title>
<p>
    <ol label="1">
        <li><p>
            If <m>n</m> is odd, <m>\hphantom{blankblank}\sqrt[n]{a^n}=a </m>
        </p></li>
        <li><p>
            If <m>n</m> is even, <m>\hphantom{blankblank}\sqrt[n]{a^n}=\abs{a} </m></p>
        <p>
            In particular, <m>\hphantom{blankblank}\sqrt{a^2}=\abs{a} </m>
        </p></li>
    </ol>
</p>
</assemblage>

<example>
    <p>
        <ol label="a">
            <li><p>
                <m>\sqrt{16x^2}= 4\abs{x} </m>
            </p></li>
            <li><p>
                <m>\sqrt{(x-1)^2}= \abs{x-1} </m>
            </p></li>
        </ol>
    </p>
</example>
</subsection>



<subsection><title>Extraneous Solutions to Radical Equations</title>
<p>
    It is important to check the solution to a radical equation, because it is possible to introduce false, or <term>extraneous</term><idx>extraneous</idx>, solutions when we square both sides of the equation. For example, the equation 
    <me>\sqrt{x}=-5 </me>
    has no solution, because <m>\sqrt{x}</m> is never a negative number. However, if we try to solve the equation by squaring both sides, we find
    <md>
        <mrow>
            (\sqrt{x})^\alert{2}\amp = (-5)^\alert{2}
        </mrow>
        <mrow>
            x\amp = 25
        </mrow>
    </md>
    You can check that <m>25</m> is <em>not</em> a solution to the original equation, <m>\sqrt{x}=-5 </m>, because <m>\sqrt{25} </m> does not equal <m>-5</m>.
</p>
<p>
    If each side of an equation is raised to an odd power, extraneous solutions will not be introduced. However, if we raise both sides to an even power, we should check each solution in the original equation.
</p>

<example xml:id="example-radical-equationS">
    <p>
        Solve the equation <m>\sqrt{x+2}+4=x </m>.
    </p>
<solution>
    <p>
        First, isolate the radical expression on one side of the equation. (This will make it easier to square both sides.)
        <md>
            <mrow>
                \sqrt{x+2}\amp = x-4\amp\amp\text{Square both sides of the equation.}
            </mrow>
            <mrow>
                \left(\sqrt{x+2}\right)^\alert{2}\amp = (x-4)^\alert{2}
            </mrow>
            <mrow>
                x+2\amp = x^2-8x+16 \amp\amp\text{Subtract }x + 2 \text{ from both sides.}
            </mrow>
            <mrow>
                x^2-9x+14\amp = 0\amp\amp\text{Factor the left side.}
            </mrow>
            <mrow>
                (x-2)(x-7)\amp = 0 \amp\amp\text{Set each factor to zero.}
            </mrow>
            <mrow>
                x=2\hphantom{000}\amp\text{or}\hphantom{000}x=7
            </mrow>
        </md>
    </p>

    <p><em>Check</em></p><sidebyside margins="5%"><paragraphs>
    <p>Does <m>\sqrt{\alert{2}+2}+4=\alert{2} </m>? <m>\hphantom{blank}{\text{No; }2\text{ is not a solution.}} </m></p>
    <p>Does <m>\sqrt{\alert{7}+2}+4=\alert{7} </m>? <m>\hphantom{blank}{\text{Yes; }7\text{ is  a solution.}} </m>
    </p></paragraphs></sidebyside>
        
    <p>
        The apparent solution <m>2</m> is extraneous. The only solution to the original equation is <m>7</m>. We can verify the solution by graphing the equations
        <me>y_1=\sqrt{x+2} ~~~\text{ and }~~~ y_2=x-4</me>
        as shown in <xref ref="fig-radical-equation" text="type-global"/>. The graphs intersect in only one point, <m>(7, 3)</m>, so there is only one solution, <m>x=7</m>.
    </p>
    <figure xml:id="fig-radical-equation"><caption></caption><image source="images/fig-radical-equation"  width="40%"><description>graphs of both sides of radical equation</description></image> </figure>

</solution></example>

<warning>
    <p>
        When we square both sides of an equation, it is <em>not</em> correct to square each term of the equation separately. Thus, in <xref ref="example-radical-equationS" text="type-global"/>, the original equation is not equivalent to
        <me>(\sqrt{x+2})^2+4^2=x^2 </me>
        This is because <m>(a + b)^2 \ne a^2 + b^2</m>. Instead, we must square the <em>entire</em> left side of the equation as a binomial, like this,
        <me>(\sqrt{x+2}+4)^2=x^2 </me>
        or we may proceed as shown in <xref ref="example-radical-equationS" text="type-global"/>.
    </p>
</warning>
</subsection>



<subsection><title>Equations with More than One Radical</title>
<p>
    Sometimes it is necessary to square both sides of an equation more than once in order to eliminate all the radicals.
</p>

<example xml:id="example-radical-equation-with-3-terms">
    <p>
        Solve <m>\sqrt{x-7}+\sqrt{x}=7 </m>.
    </p>
<solution>
    <p>
        First, isolate the more complicated radical on one side of the equation. (This will make it easier to square both sides.) We will subtract <m>\sqrt{x} </m> from both sides.
        <me>\sqrt{x-7}=7-\sqrt{x} </me>
        Now square each side to remove one radical. Be careful when squaring the binomial <m>7-\sqrt{x} </m>.
        <md>
            <mrow>
                (\sqrt{x-7})^\alert{2}\amp =(7-\sqrt{x})^\alert{2}
            </mrow>
            <mrow>
                x-7\amp = 49-14\sqrt{x}+x
            </mrow>
        </md>
        Collect like terms, and isolate the radical on one side of the equation.
        <md>
            <mrow>
                -56 \amp = -14\sqrt{x}\amp\amp\text{Divide both sides by }-14.
            </mrow>
            <mrow>
                4 \amp = \sqrt{x}
            </mrow>
        </md>
        Now square again to obtain
        <md>
            <mrow>
                (4)^\alert{2} \amp = (\sqrt{x})^\alert{2}
            </mrow>
            <mrow>
                16 \amp = x
            </mrow>
        </md></p>
    <p><em>Check</em></p>
    <p><sidebyside margins="5%"><paragraphs><p>
                Does <m>\sqrt{\alert{16}-7}+\sqrt{\alert{16}}=7</m>? <m>\hphantom{blank}\text{Yes. The solution is } 16</m>.
            </p></paragraphs></sidebyside>
    </p>
</solution>
</example>

<warning><p>Recall that we cannot solve a radical equation by squaring each term separately. In other words, it is <em>incorrect</em> to begin <xref ref="example-radical-equation-with-3-terms" text="type-hybrid"/> by writing
<me>(\sqrt{x-7})^2+(\sqrt{x})^2=7^2</me>
We must square the <em>entire expression</em> on each side of the equal sign as one piece.
</p></warning>

</subsection>

</section>



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<xi:include href="./section-a-11.xml" />        <!--   Facts from Geometry  -->
<xi:include href="./section-a-12.xml" />        <!--   Real Number System  -->

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</appendix>


<!-- </book>  </mathbook> -->