Intermediate Algebra: Functions and Graphs

Section8.5Equations with Fractions

In Example 8.2.27 of Section 8.2 Francine planned a 60-mile training run on her cycle-plane. The time required for the training run, in terms of the windspeed, $$x\text{,}$$ is given by:
\begin{equation*} t=f(x) = \dfrac{60}{15-x} \end{equation*}
If it takes Francine 9 hours to cover 60 miles, what is the speed of the wind? We can answer this question by reading values from the graph of $$f\text{,}$$ as shown at left. When $$t = 9\text{,}$$ the value of $$x$$ is between 8 and 9, so the windspeed is between 8 and 9 miles per hour.

Subsection8.5.1Solving Algebraically

If we need a more accurate value for the windspeed, we can solve the equation
\begin{equation*} \dfrac{60}{15-x}=9 \end{equation*}
To start, we multiply each side of the equation by the denominator of the fraction. This will clear the fraction and give us an equivalent equation without fractions.

Example8.5.1.

Solve the equation $$~~\dfrac{60}{15-x}=9$$
Solution.
We multiply both sides of the equation by $$\alert{15 - x}$$ to obtain
\begin{align*} \alert{(15 - x)}\frac{60}{15-x}\amp =9\alert{(15 - x)}\\ 60 \amp = 9(15 - x)\amp\amp \blert{\text{Apply the distributive law.}} \end{align*}
From here we can proceed as usual.
\begin{align*} 60 \amp = 135 - 9x\amp\amp \blert{\text{Subtract 135 from both sides.}}\\ -75 \amp = -9x\amp\amp \blert{\text{Divide by }-9.}\\ 8.\overline{3} \amp = x \end{align*}
The windspeed was $$8.\overline{3}\text{,}$$ or $$8\frac{1}{3}$$ miles per hour.

Checkpoint8.5.2.Practice 1.

Solve $$\dfrac{x^2}{x+4}=2$$
$$x=$$
$$-2, 4$$
Solution.
$$x = -2, x = 4$$

Checkpoint8.5.3.Practice 1.

Solve $$\dfrac{x^2}{x+4}=2$$
Solution.
$$x = -2, x = 4$$
If the equation contains more than one fraction, we can clear all the denominators at once by multiplying both sides by the LCD of the fractions.

Example8.5.4.

Solve $$~~\dfrac{3}{4} = 8- \dfrac{2x+11}{x-5}$$
Solution.
The LCD for the two fractions in the equation is $$\blert{4(x-5)}\text{.}$$ We multiply both sides of the equation by the LCD.
\begin{align*} {\blert{4(x-5)}}\left(\dfrac{3}{4}\right) \amp = \left(8- \dfrac{2x+11}{x-5}\right) \cdot {\blert{4(x-5)}} \amp \amp \blert{\stackrel{{\Large\text{Apply the}}} {\text{distributive law.}}}\\ {\blert{\cancel{4}(x-5)}}\left(\dfrac{3}{\cancel{4}}\right) \amp = {\blert{4(x-5)}}(8)- {\blert{4\cancel{(x-5)}}}\left(\dfrac{2x+11}{\cancel{x-5}}\right)\\ 3(x-5) \amp = 32(x-5)-4(2x+11) \end{align*}
We proceed as usual to complete the solution. First we use the distributive law to remove parentheses.
\begin{align*} 3x-15 \amp = 32x-160-8x-44 \amp \amp \blert{\text{Combine like terms.}}\\ 3x-15 \amp = 24x-204\\ -21x \amp = -189\\ x \amp = 9 \end{align*}

Caution8.5.5.

We must multiply each term of an equation by the LCD, whether or not the term is a fraction. In the previous Example we multiplied each term by the LCD, including the 8.

Checkpoint8.5.6.Practice 2.

Solve $$\quad\dfrac{1}{x-2}+\dfrac{2}{x} = 1$$
$$x=$$ Enter solutions separated by a comma.
$$4, 1$$
Solution.
$$1\text{,}$$ $$4$$

Checkpoint8.5.7.Practice 2.

Solve $$\quad\dfrac{1}{x-2}+\dfrac{2}{x} = 1$$
Solution.
$$1\text{,}$$ $$4$$

Subsection8.5.2Proportions

A proportion is a statement that two ratios are equal. For example,
\begin{equation*} \dfrac{7}{5} = \dfrac{x}{6} \end{equation*}
To solve this proportion, we multiply both sides by the LCD, 30, to get
\begin{align*} \alert{30}\left(\dfrac{7}{5}\right) \amp = \left(\dfrac{x}{6}\right)\alert{30}\\ 42 \amp = 5x \amp \amp \blert{\text{Divide both sides by 5.}}\\ x = \dfrac{42}{5} = 8.4 \end{align*}
There is a short-cut we can use that avoids calculating an LCD. Observe that we can arrive at the equation $$42=5x$$ by cross-multiplying:
We then proceed as before to complete the solution.
The cross-multiplying shortcut is a fundamental property of proportions.

Property of Proportions.

\begin{equation*} \blert{\text{If}~~\dfrac{a}{b}=\dfrac{c}{d},~~~\text{then}~~~ad=bc,~~\text{as long as}~ b,d \ne 0} \end{equation*}

Example8.5.8.

The scale on a map of Fairfield County says that $$\dfrac{3}{4}$$ centimeter represents a distance of 10 kilometers. If Eastlake and Kenwood are 6 centimeters apart on the map, what is the distance between the two towns?
Solution.
The ratio of the two actual distances is the same as the ratio of the corresponding distances on the map. We let $$x$$ stand for the distance between Eastlake and Kenwood, and write a proportion.
We must be careful to keep the same order in both ratios. We choose to put the distance between towns in the numerators, and the distances on the scale in the denominator.
\begin{equation*} \dfrac{\text{distance between towns}}{\text{scale}}:~~~~\dfrac{x}{10} = \dfrac{6}{\dfrac{3}{4}} \end{equation*}
To solve the propotion, we cross-multiply.
\begin{align*} \dfrac{3}{4}x \amp = 10 \cdot 6\\ x \amp = \dfrac{60}{\dfrac{3}{4}} = 80 \end{align*}
The two towns are 80 kilometers apart.

Checkpoint8.5.9.Practice 3.

On a scale model of Fantasy Valley, $$1\dfrac{1}{2}$$ inches represents 50 yards. If the distance from the water slide to the bungee jump is 20 inches on the model, what is the distance between the two rides?
yards
$$666+{\frac{2}{3}}$$
Solution.
$$666\dfrac{2}{3}$$ yards

Checkpoint8.5.10.Practice 3.

On a scale model of Fantasy Valley, $$1\dfrac{1}{2}$$ inches represents 50 yards. If the distance from the water slide to the bungee jump is 20 inches on the model, what is the distance between the two rides?
Solution.
$$666\dfrac{2}{3}$$ yards

Caution8.5.11.

Do not try to use "cross-multiplying" on equations that are not proportions, or on any other operations involving fractions. The shortcut works only on proportions.

Checkpoint8.5.12.QuickCheck 1.

1. To clear fractions from an equation, we multiply each term by the
• LCD
• average
• reciprocal
of all the fractions.
2. True or False: When clearing fraction from an equation, we do not multiply terms that are not fractions.
• True
• False
3. A proportion is a statement that two
• equations
• factors
• ratios
are equal.
4. Cross-multiplying works only on
• reduced fractions
• improper fractions
• proportions
.
$$\text{LCD}$$
$$\text{False}$$
$$\text{ratios}$$
$$\text{proportions}$$
Solution.
1. LCD
2. False
3. ratios
4. proportions

Checkpoint8.5.13.QuickCheck 1.

1. To clear fractions from an equation, we multiply each term by the of all the fractions.
2. True or False: When clearing fraction from an equation, we do not multiply terms that are not fractions.
3. A proportion is a statement that two are equal.
4. Cross-multiplying works only on .

Subsection8.5.3Extraneous Solutions

An algebraic fraction is undefined for any values of $$x$$ that make its denominator equal zero. These values cannot be solutions to equations involving the fraction. Consider the equation
\begin{equation*} \dfrac{x}{x-3}=\frac{3}{x-3}+2 \end{equation*}
When we multiply both sides by the LCD, $$x - 3\text{,}$$ we obtain
\begin{align*} \alert{(x - 3)}\frac{x}{x-3} \amp =\alert{(x - 3)}\frac{3}{x-3}+\alert{(x - 3)}\cdot 2\\ x \amp = 3 + 2x - 6 \end{align*}
whose solution is $$x = 3\text{.}$$ However, $$x = 3$$ is not a solution of the original equation. Both sides of the equation are undefined at $$x = 3\text{.}$$ If you graph the two functions
\begin{equation*} Y_1=\frac{x}{x-3} \hphantom{space}\text{and}\hphantom{space}Y_2=\frac{3}{x-3}+2 \end{equation*}
you will find that the graphs never intersect, which means that there is no solution to the original equation.
What went wrong with our method of solution? We multiplied both sides of the equation by $$x - 3\text{,}$$ which is zero when $$x = 3\text{,}$$ so we really multiplied both sides of the equation by zero. Multiplying by zero does not produce an equivalent equation, and false solutions may be introduced.
An apparent solution that does not satisfy the original equation is called an extraneous solution. Whenever we multiply an equation by an expression containing the variable, we should check that the solution obtained does not cause any of the fractions to be undefined.

Example8.5.14.

Solve the equation $$~\dfrac{6}{x}+1=\dfrac{1}{x+2}\text{.}$$
Solution.
We multiply both sides by the LCD, $$x(x + 2)\text{.}$$ Notice that we multiply each term on the left side by the LCD, to get
or
\begin{equation*} 6(x + 2) + x(x + 2) = x \end{equation*}
We use the distributive law to remove the parentheses and write the result in standard form:
\begin{align*} 6x + 12 + x^2 + 2x\amp = x\\ x^2 + 7x + 12 \amp = 0 \end{align*}
This is a quadratic equation that we can solve by factoring.
\begin{equation*} (x + 3)(x + 4) = 0 \end{equation*}
so the solutions are $$x = -3$$ and $$x = -4\text{.}$$ Neither of these values causes either denominator to equal zero, so they are not extraneous solutions.

Caution8.5.15.

The following "solution" for the previous Example is incorrect. Do you see why?
\begin{align*} \alert{x(x + 2)}\frac{6}{x}+1 \amp =\alert{x(x + 2)}\frac{1}{x+2}\\ 6x+12+1 \amp = x\\ 5x \amp = -13\\ x \amp = \dfrac{-13}{5} \end{align*}

Checkpoint8.5.16.QuickCheck 2.

1. An algebraic fraction is undefined when
• the numerator equals zero
• the denominator equals zero
• it includes decimals
• there is a variable in the denominator
.
2. An apparent solution that does not satisfy the original equation is called
• an extraneous
• a double
• a principal
• a multiple
solution.
3. We may introduce extraneous solutions if we multiply both sides of an equation by
• a nonzero integer
• an expression containing the variable
• any irrational number
.
4. We must multiply each
• term
• denominator
• fraction only
of the equation by the LCD.
$$\text{the denominator equals zero}$$
$$\text{an extraneous}$$
$$\text{an expression containing the variable}$$
$$\text{term}$$
Solution.
1. the denominator equals zero
2. an extraneous
3. an expression containing the variable
4. term

Checkpoint8.5.17.QuickCheck 2.

1. An algebraic fraction is undefined when .
2. An apparent solution that does not satisfy the original equation is called solution.
3. We may introduce extraneous solutions if we multiply both sides of an equation by .
4. We must multiply each of the equation by the LCD.

Checkpoint8.5.18.Practice 4.

Solve $$\quad\dfrac{9}{x^2+x-2} + \dfrac{1}{x^2-x} = \dfrac{4}{x-1}$$
$$x=$$
$$\frac{-1}{2}$$
Solution.
$$\dfrac{-1}{2}$$

Checkpoint8.5.19.Practice 4.

Solve $$\quad\dfrac{9}{x^2+x-2} + \dfrac{1}{x^2-x} = \dfrac{4}{x-1}$$
Solution.
$$\dfrac{-1}{2}$$

Subsection8.5.4Solving Graphically

We can use graphs to solve the equation in the following Example.

Example8.5.20.

Solve the equation graphically: $$~\dfrac{6}{x}+1=\dfrac{1}{x+2}\text{.}$$
Solution.
We graph the two functions
\begin{equation*} Y_1=\frac{6}{x}+1 \hphantom{space}\text{and}\hphantom{space}Y_2=\frac{1}{x+2} \end{equation*}
in the window
\begin{align*} \text{Xmin} \amp = -4.7 \amp\amp \text{Xmax} = 4.7\\ \text{Ymin} \amp = -10 \amp\amp \text{Ymax} = 10 \end{align*}
as shown in figure (a).
It appears that the two graphs may intersect in the third quadrant, around $$x = -3\text{.}$$ To investigate further, we change the window settings to
\begin{align*} \text{Xmin} \amp = -4.55 \amp\amp \text{Xmax} = -2.2\\ \text{Ymin} \amp = -1.3 \amp\amp \text{Ymax} = -0.3 \end{align*}
and obtain the close-up view shown in figure (b). In this window, we can see that the graphs intersect in two distinct points, and by using the Trace we find that their $$x$$-coordinates are $$x = -3$$ and $$x = -4\text{.}$$

Checkpoint8.5.21.Practice 5.

The manager of a new health club kept track of the number of active members over the club’s first few months of operation. The number, $$N$$ of active members, in hundreds, $$t$$ months after the club opened is given by the equation
\begin{equation*} \begin{gathered} N=\dfrac{10t}{4+t^2} \end{gathered} \end{equation*}
1. Graph the equation in the window
\begin{equation*} \begin{aligned} \text{Xmin} \amp = 0 \amp\amp \text{Xmax} = 9.4\\ \text{Ymin} \amp = 0 \amp\amp \text{Ymax} = 3 \end{aligned} \end{equation*}
2. Use the graph to find out in which months the club had 200 active members.
Months: Enter month numbers separated by a comma.
$$1, 4$$
Solution.
1. A graph is below.
2. 1, 4
Graph for (a):

Checkpoint8.5.22.Practice 5.

The manager of a new health club kept track of the number of active members over the club’s first few months of operation. The number, $$N$$ of active members, in hundreds, $$t$$ months after the club opened is given by the equation
\begin{gather*} N=\dfrac{10t}{4+t^2} \end{gather*}
1. Graph the equation in the window
\begin{align*} \text{Xmin} \amp = 0 \amp\amp \text{Xmax} = 9.4\\ \text{Ymin} \amp = 0 \amp\amp \text{Ymax} = 3 \end{align*}
2. Use the graph to find out in which months the club had 200 active members.
Solution.
1. 1, 4

Subsection8.5.5Applications

Application problems may lead to equations with algebraic fractions.

Example8.5.23.

Rani times herself as she kayaks 30 miles down the Derwent River with the help of the current. Returning upstream against the current, she manages only 18 miles in the same amount of time. Rani knows that she can kayak at a rate of 12 miles per hour in still water. What is the speed of the current?
Solution.
If we let $$x$$ represent the speed of the current, we can use the formula
\begin{equation*} \text{time} = \dfrac{\text{distance}}{\text{rate}} \end{equation*}
to fill in the following table.
 Distance Rate Time Downstream $$30$$ $$12+x$$ $$\dfrac{30}{12+x}$$ Upstream $$18$$ $$12-x$$ $$\dfrac{18}{12-x}$$
Because Rani paddled for equal amounts of time upstream and downstream, we have the equation
\begin{equation*} \frac{30}{12+x}=\frac{18}{12-x} \end{equation*}
The LCD for the fractions in this equation is $$\alert{(12 + x)(12 - x)}\text{.}$$ We multiply both sides of the equation by the LCD to obtain
\begin{align*} \alert{\cancel{(12 + x)}(12 - x)}\dfrac{30}{\cancel{12+x}} \amp =\dfrac{18}{\cancel{12-x}}\alert{(12 + x)\cancel{(12 - x)}}\\ 30 (12 - x) \amp = 18 (12 + x) \end{align*}
Solving this equation, we find
\begin{align*} 360 - 30x \amp = 216 + 18x\\ 144 \amp = 48x\\ 3 \amp = x \end{align*}
The speed of the current is 3 miles per hour.
We can solve the equation in Example 8.5.23 graphically by considering two functions, one for each side of the equation. Graph the two functions
\begin{equation*} Y_1=\frac{30}{12+x} \hphantom{space}\text{and}\hphantom{space}Y_2=\frac{30}{12-x} \end{equation*}
in the window
\begin{align*} \text{Xmin} \amp = -9.4 \amp\amp \text{Xmax} = 9.4\\ \text{Ymin} \amp = 0 \amp\amp \text{Ymax} = 10 \end{align*}
to obtain the graph shown below.
The function $$Y_1$$ gives the time it takes Rani to paddle 30 miles downstream, and $$Y_2$$ gives the time it takes her to paddle 18 miles upstream. Both of these times depend on the speed of the current, $$x\text{.}$$
We are looking for a value of $$x$$ that makes $$Y_1$$ and $$Y_2$$ equal. This occurs at the intersection point of the two graphs, $$(3, 2)\text{.}$$ Thus, the speed of the current is 3 miles per hour, as we found in Example 8.5.23. The y-coordinate of the intersection point gives the time Rani paddled on each part of her trip: 2 hours each way.

Checkpoint8.5.24.Practice 6.

A cruise boat travels 18 miles downstream and back in $$4\dfrac{1}{2}$$ hours. If the speed of the current is 3 miles per hour, what is the speed of the boat in still water?
1. Let $$x$$ represent the speed of the boat in still water, and fill in the table.
 Distance Rate Time Downstream Upstream
2. Write an equation to model the problem:
$$\text{Downstream time + Upstream time = Total trip time}$$
$$=$$
$$x=$$: The speed of the boat in still water is mph.
$$18$$
$$x+3$$
$$\frac{18}{x+3}$$
$$18$$
$$x-3$$
$$\frac{18}{x-3}$$
$$\frac{18}{x+3}+\frac{18}{x-3}$$
$$\frac{9}{2}$$
$$9$$
$$9$$
Solution.
1.  Distance Rate Time Downstream $$18$$ $$x+3$$ $$\dfrac{18}{x+3}$$ Upstream $$18$$ $$x-3$$ $$\dfrac{18}{x-3}$$
2. $$\displaystyle \dfrac{18}{x+3} + \dfrac{18}{x-3} = \dfrac{9}{2}$$
3. 9 mph

Checkpoint8.5.25.Practice 6.

A cruise boat travels 18 miles downstream and back in $$4\dfrac{1}{2}$$ hours. If the speed of the current is 3 miles per hour, what is the speed of the boat in still water?
1. Let $$x$$ represent the speed of the boat in still water, and fill in the table.
 Distance Rate Time Downstream $$\hphantom{00000}$$ $$\hphantom{00000}$$ $$\hphantom{00000}$$ Upstream $$\hphantom{00000}$$ $$\hphantom{00000}$$ $$\hphantom{00000}$$
2. Write an equation to model the problem:
$$\text{Downstream time + Upstream time = Total trip time}$$
Solution.
1.  Distance Rate Time Downstream $$18$$ $$x+3$$ $$\dfrac{18}{x+3}$$ Upstream $$18$$ $$x-3$$ $$\dfrac{18}{x-3}$$
2. $$\displaystyle \dfrac{18}{x+3} + \dfrac{18}{x-3} = \dfrac{9}{2}$$
3. 9 mph

Subsection8.5.6Formulas

Algebraic fractions may appear in formulas that relate several variables. If we want to solve for one variable in terms of the others, we may need to clear the fractions.

Example8.5.26.

Solve the formula $$~~p=\dfrac{v}{q+v}~~$$ for $$v\text{.}$$
Solution.
Because the variable we want appears in the denominator, we must first multiply both sides of the equation by that denominator, $$q+v\text{.}$$
We apply the distributive law on the left side, then collect all terms that involve $$v$$ on one side of the equation.
\begin{align*} qp + vp \amp = v \amp\amp \blert{\text{Subtract }vp \text{ from both sides.}}\\ qp = v - vp \end{align*}
We cannot combine the two terms containing $$v$$ because they are not like terms. However, we can factor out $$v\text{,}$$ so that the right side is written as a single term containing the variable $$v\text{.}$$ We can then complete the solution.
\begin{align*} qp \amp = v(1 - p) \amp\amp \blert{\text{Divide both sides by } 1- p.}\\ \frac{qp}{1-p} \amp= v \end{align*}

Checkpoint8.5.27.Practice 7.

Solve for $$a\text{:}$$ $$~\dfrac{2ab}{a+b}=H$$
$$a=$$
$$\frac{bH}{2b-H}$$
Solution.
$$a=\dfrac{bH}{2b-H}$$

Checkpoint8.5.28.Practice 7.

Solve for $$a\text{:}$$ $$~\dfrac{2ab}{a+b}=H$$
Solution.
$$a=\dfrac{bH}{2b-H}$$

Checkpoint8.5.29.QuickCheck 3.

True or False.
1. To solve an equation graphically, we graph two functions, $$y=$$ (each side of the equation.)
• True
• False
2. The solutions are the $$y$$-coordinates of the intersection points of the two graphs.
• True
• False
3. To solve a formula that is linear in the desired variable, we must get all terms including that variable on one side of the equation.
• True
• False
4. If two or more terms on one side of the equation include the desired variable, we factor it out.
• True
• False
$$\text{True}$$
$$\text{False}$$
$$\text{True}$$
$$\text{True}$$
Solution.
1. True
2. False
3. True
4. True

Checkpoint8.5.30.QuickCheck 3.

True or False.
1. To solve an equation graphically, we graph two functions, $$y=$$ (each side of the equation.)
2. The solutions are the $$y$$-coordinates of the intersection points of the two graphs.
3. To solve a formula that is linear in the desired variable, we must get all terms including that variable on one side of the equation.
4. If two or more terms on one side of the equation include the desired variable, we factor it out.

Exercises8.5.7Problem Set 8.5

Warm Up

Exercise Group.
For Problems 1–8, solve.
1.
$$\dfrac{6}{w+2}=4$$
$$\dfrac{-1}{2}$$
2.
$$\dfrac{12}{r-7}=3$$
3.
$$9=\dfrac{h-5}{h-2}$$
$$\dfrac{13}{8}$$
4.
$$-3=\dfrac{v+1}{v-6}$$
5.
$$\dfrac{15}{s^2}=8$$
$$\pm\sqrt{\dfrac{15}{8}}$$
6.
$$\dfrac{3}{m^2}=5$$
7.
$$4.3=\sqrt{\dfrac{18}{y}}$$
$$\dfrac{1800}{1849}\approx 0.97$$
8.
$$6.5=\dfrac{52}{\sqrt{z}}$$

Skills Practice

Exercise Group.
For Problems 9–16, solve.
9.
$$\dfrac{4}{x} - 3 = \dfrac{5}{2x+3}$$
$$-2,~1$$
10.
$$\dfrac{4}{x-1} - \dfrac{4}{x+2} = \dfrac{3}{7}$$
11.
$$\dfrac{2}{n^2-2n} + \dfrac{1}{2n} = \dfrac{-1}{n^2+2n}$$
$$-6$$
12.
$$\dfrac{3}{x-2}=\dfrac{1}{2}+\dfrac{2x-7}{2x-4}$$
13.
$$\dfrac{4}{x+2}-\dfrac{1}{x}=\dfrac{2x-1}{x^2+2x}$$
$$1$$
14.
$$\dfrac{1}{x-1} +\dfrac{2}{x+1}=\dfrac{x-2}{x^2-1}$$
15.
$$\dfrac{x}{x+2} -\dfrac{3}{x-2}=\dfrac{x^2+8}{x^2-4}$$
$$\dfrac{-14}{5}$$
16.
$$-3= \dfrac{-10}{x+2} +\dfrac{10}{x+5}$$
17.
What is wrong with the solution to the following addition problem:
Solution:
\begin{align*} \alert{8} \cdot \dfrac{3}{8} + \alert{8} \cdot \dfrac{1}{2} \amp = 3+4 \amp \amp \blert{\text{Multiply by the LCD, 8.}}\\ \amp = 7 \amp \amp \blert{\text{The answer is 7.}} \end{align*}
We don’t multiply by the LCD in addition problems.
18.
Compare the methods in each calculation with fractions. Explain how the LCD is used in each operation.
1. Add: $$~\dfrac{3}{x} + \dfrac{1}{x+3}$$
2. Divide: $$~\dfrac{3}{x} \div \dfrac{1}{x+3}$$
3. Solve: $$~\dfrac{3}{x} + \dfrac{1}{x+3} = 3$$
4. Simplify: $$~\dfrac{1+\dfrac{3}{x}}{3+\dfrac{1}{x+3}}$$
Exercise Group.
For Problems 19–24, solve the formula for the specified variable.
19.
$$S=\dfrac{a}{1-r},~~~~$$ for $$r$$
$$r=\dfrac{S-a}{a}$$
20.
$$m=\dfrac{y-k}{x-h},~~~~$$ for $$x$$
21.
$$\dfrac{x}{a} + \dfrac{y}{b} = 1,~~~~$$ for $$x$$
$$x=a-\dfrac{ay}{b}$$
22.
$$F=\dfrac{Gm_1m_2}{d^2},~~~~$$ for $$d$$
23.
$$C=\dfrac{rR}{r+R},~~~~$$ for $$R$$
$$R=\dfrac{Cr}{r-C}$$
24.
$$E=\dfrac{Ff}{(P-x)p},~~~~$$ for $$x$$

Applications

25.
Rani went hiking in the Santa Monica mountains last weekend. She drove 10 miles in her car and then walked 4 miles, and arrived at a small lake hours after she left home. If Rani drives 20 times faster than she walks, how fast does she walk?
2 mph
26.
Sam Scholarship and Reginald Privilege each travel the 360 miles to Fort Lauderdale on spring break, but Reginald drives his Porsche while Sam hitches a ride on a vegetable truck. Reginald travels 20 miles per hour faster than Sam and arrives in 3 hours less time. How fast did each travel?
27.
Walt and Irma use a tank of fuel oil for their furnace every 25 days during the winter. Last winter it was so cold that they also lit their space heater 10 days after they filled the fuel oil tank. If the space heater uses a tank of fuel oil every 40 days, how much longer will the fuel last with both heaters running?
24 days
28.
An underground spring fills a small pond in 12 days. Evaporation from the surface of the pond can empty the pond in 28 days. If the pond is completely dry, how long will it take to fill again?
29.
The cost, in thousands of dollars, for immunizing $$p$$ percent of the residents of Emporia against a dangerous new disease is given by the function
\begin{equation*} C(p) = \frac{72p}{100-p} \end{equation*}
Write and solve an equation to determine what percent of the population can be immunized for 168,000. Answer. $$168=\dfrac{72p}{100-p}\text{;}$$ $$p=70\%$$ 30. A school of bluefin tuna average 36 miles per hour on a 200-mile trip in still water, but this time they encounter a current. 1. Express the tuna’s travel time, $$t\text{,}$$ as a function of the current speed, $$v\text{,}$$ and graph the function in the window \begin{align*} {\text{Xmin}} \amp = 0 \amp\amp {\text{Xmax}} = 36\\ {\text{Ymin}} \amp = 0 \amp\amp {\text{Ymax}} =50 \end{align*} 2. Write and solve an equation to find the current speed if the school makes its trip in 8 hours. Label the corresponding point on your graph. 31. During the baseball season so far this year, Pete got hits 44 times out of 164 times at bat. 1. What is Pete’s batting average so far? (Batting average is the fraction of at-bats that resulted in hits.) 2. If Pete gets hits on every one of his next $$x$$ at-bats, write an expression for his new batting average. 3. How many consecutive hits does Pete need to raise his batting average to 0.350 ? Answer. 1. 0.268 2. $$\displaystyle \dfrac{44+x}{164+x}$$ 3. 21 32. The manager of Joe’s Burgers discovers that he will sell $$\dfrac{160}{x}$$ burgers per day if the price of a burger is $$x$$ dollars. On the other hand, he can afford to make $$6x+49$$ burgers if he charges $$x$$ dollars apiece for them. 1. Graph the demand function, $$D(x) = \dfrac{160}{x}$$ and the supply function, $$S(x) = 6x+49\text{,}$$ in the same window. At what price $$x$$ does the demand for burgers equal the number that Joe can afford to supply? This value for is called the equilibrium price. 2. Write and solve an equation to verify your equilibrium price. 33. A chartered sightseeing flight over the Grand Canyon is scheduled to return to its departure point in 3 hours. The pilot would like to cover a distance of 144 miles before turning around, and he hears on the Weather Service that there will be a headwind of 20 miles per hour on the outward journey. 1. Express the time it takes for the outward journey as a function of the airspeed of the plane. 2. Express the time it takes for the return journey as a function of the speed of the plane. 3. Graph the sum of the two functions and find the point on the graph with $$y$$-coordinate 3. Interpret the coordinates of the point in the context of the problem. 4. The pilot would like to know what airspeed to maintain in order to complete the tour in 3 hours. Write an equation to describe this situation. 5. Solve your equation to find the appropriate airspeed. Answer. 1. $$\displaystyle t=\dfrac{144}{v-20}$$ 2. $$\displaystyle t=\dfrac{144}{v+20}$$ 3. $$\displaystyle (100,3)$$ 4. $$\displaystyle t=\dfrac{144}{v-20} + t=\dfrac{144}{v+20} = 3$$ 5. 100 mph 34. The cost of wire fencing is7.50 per foot. A rancher wants to enclose a rectangular pasture of 1000 square feet with this fencing.
1. Express the length of the pasture as a function of its width.
2. Express the cost of the fence as a function of its width.
3. Graph your function for the cost and find the coordinates of the lowest point on the graph. Interpret those coordinates in the context of the problem.
4. The rancher has \$1050 to spend on the fence, and she would like to know what the width of the pasture should be. Write an equation to describe this situation.
5. Solve your equation and find the dimensions of the pasture.
35.
Distances on a map vary directly with actual distances. The scale on a map of Michigan uses $$\dfrac{3}{8}$$ inch to represent 10 miles. If Isle Royale is $$1\dfrac{11}{16}$$ inches long on the map, what is the actual length of the island?
$$28 \dfrac{1}{3}$$ miles
36.
The highest point on Earth is Mount Everest in Tibet, with an elevation of 8848 meters. The deepest part of the ocean is the Challenger Deep in the Mariana Trench, near Indonesia, 11,034 meters below sea level.
1. What is the total height variation in the surface of the Earth?
2. What percentage of the Earth’s radius, 6400 kilometers, is this variation?
3. If the Earth were shrunk to the size of a basketball, with a radius of 4.75 inches, what would be the corresponding height of Mount Everest?
37.
The rectangle $$ABCD$$ is divided into a square and a smaller rectangle, $$CDEF\text{.}$$ The two rectangles $$ABCD$$ and $$CDEF$$ are similar (their corresponding sides are proportional.) A rectangle $$ABCD$$ with this property is called a golden rectangle, and the ratio of its length to its width is called the golden ratio.
The golden ratio appears frequently in art and nature, and it is considered to give the most pleasing proportions to many figures. We will compute the golden ratio as follows.
1. Let $$AB = 1$$ and $$AD = x\text{.}$$ What are the lengths of $$AE\text{,}$$ $$DE\text{,}$$ and $$CD\text{?}$$
2. Write a proportion in terms of $$x$$ for the similarity of rectangles $$ABCD$$ and $$CDEF\text{.}$$ Be careful to match up the corresponding sides.
3. Solve your proportion for $$x\text{.}$$ Find the golden ratio, $$\dfrac{AD}{AB}=\dfrac{x}{1} \text{.}$$
1. $$\displaystyle AE = 1,~ DE = x - 1,~ CD = 1$$
2. $$\displaystyle \dfrac{1}{x}=\dfrac{x-1}{x}$$
3. $$\displaystyle \dfrac{1+\sqrt{5}}{2}$$
38.
The figure shows the graphs of two equations,
\begin{equation*} y = x~~~~\text{and}~~~~y=\dfrac{1}{x}+1 \text{.} \end{equation*}
1. Find the $$x$$-coordinate of the intersection point of the two graphs.
2. Compare your answer to the golden ratio you computed in Problem 37.
39.
Find the $$\alert{\text{error}}$$ in the following "proof" that $$1=0\text{:}$$ Start by letting $$x=1\text{.}$$
\begin{align*} x \amp = 1 \amp \amp \blert{\text{Multiply both sides by}~x.}\\ x^2 \amp = x \amp \amp \blert{\text{Subtract 1 from both sides.}}\\ x^2-1 \amp = x-1 \amp \amp \blert{\text{Factor the left side.}}\\ (x-1)(x+1) \amp = x-1 \amp \amp \blert{\text{Divide both sides by}~x-1.}\\ x+1 \amp = 1 \amp \amp \blert{\text{Subtract 1 from both sides.}}\\ x \amp = 0 \end{align*}
Because $$x=1$$ and $$x=0\text{,}$$ we have "proved" that $$1=0\text{.}$$
Because $$x=1\text{,}$$ dividing by $$x-1$$ in the fourth step is dividing by $$0\text{.}$$
Exercise Group.
For Problems 40–42,
1. Solve the equation graphically by graphing two functions, one for each side of the equation.
2. Solve the equation algebraically.
40.
$$\dfrac{2x}{x+1}=\dfrac{x+1}{2}$$
41.
$$\dfrac{2}{x+1}=\dfrac{x}{x+1}+1$$
1. $$\displaystyle x=\dfrac{1}{2}$$
$$\dfrac{15x}{1+x^2}=6$$