ORCCA Solving Rational Equations

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Section 12.5 Solving Rational Equations

Objectives: PCC Course Content and Outcome Guide

Figure 12.5.1. Alternative Video Lesson

Subsection 12.5.1 Solving Rational Equations

We open this section looking back on Example 12.3.2. Julia is taking her family on a boat trip $12$ miles down the river and back. The river flows at a speed of $2$ miles per hour and she wants to drive the boat at a constant speed, $v$ miles per hour downstream and back upstream. Due to the current of the river, the actual speed of travel is $v+2$ miles per hour going downstream, and $v-2$ miles per hour going upstream. If Julia plans to spend $8$ hours for the whole trip, how fast should she drive the boat?

The time it takes Julia to drive the boat downstream is $\frac{12}{v+2}$ hours, and upstream is $\frac{12}{v-2}$ hours. The function to model the whole trip's time is

$\begin{equation*} t(v)=\frac{12}{v-2}+\frac{12}{v+2} \end{equation*}$

where $t$ stands for time in hours. The trip will take $8$ hours, so we want $t(v)$ to equal $8\text{,}$ and we have:

$\begin{equation*} \frac{12}{v-2}+\frac{12}{v+2}=8\text{.} \end{equation*}$

Instead of using the function's graph, we will solve this equation algebraically. You may wish to review the technique of eliminating denominators discussed in Subsection 2.3.2. We can use the same technique with variable expressions in the denominators. To remove the fractions in this equation, we will multiply both sides of the equation by the least common denominator $(v-2)(v+2)\text{,}$ and we have:

$\begin{align*} \frac{12}{v-2}+\frac{12}{v+2}\amp=8\\ \multiplyleft{(v+2)(v-2)}\left(\frac{12}{v-2}+\frac{12}{v+2}\right)\amp=\multiplyleft{(v+2)(v-2)}8\\ (v+2)\cancelhighlight{(v-2)}\cdot\frac{12}{\cancelhighlight{v-2}}+\secondcancelhighlight{(v+2)}(v-2)\cdot\frac{12}{\secondcancelhighlight{v+2}}\amp=(v+2)(v-2)\cdot8\\ 12(v+2)+12(v-2)\amp=8(v^2-4)\\ 12v+24+12v-24\amp=8v^2-32\\ 24v\amp=8v^2-32\\ 0\amp=8v^2-24v-32\\ 0\amp=8(v^2-3v-4)\\ 0\amp=8(v-4)(v+1) \end{align*}$

$\begin{align*} v-4\amp=0\amp\amp\text{or}\amp v+1\amp=0\\ v\amp=4\amp\amp\text{or}\amp v\amp=-1 \end{align*}$

Remark 12.5.2.

At this point, logically all that we know is that the only possible solutions are $-1$ and $4\text{.}$ Because of the step where factors were canceled, it's possible that these might not actually be solutions to the original equation. They each might be what is called an extraneous solution. An extraneous solution is a number that would appear to be a solution based on the solving process, but actually does not make the original equation true. Because of this, it is important that these proposed solutions be checked. Note that we're not checking to see if we made a calculation error, but are instead checking to see if the proposed solutions actually solve the original equation.

permalinkWe check these values.

$\begin{align*} \frac{12}{\substitute{-1}-2}+\frac{12}{\substitute{-1}+2}\amp\stackrel{?}{=}8\amp\frac{12}{\substitute{4}-2}+\frac{12}{\substitute{4}+2}\amp\stackrel{?}{=}8\\ \frac{12}{-3}+\frac{12}{1}\amp\stackrel{?}{=}8\amp\frac{12}{2}+\frac{12}{6}\amp\stackrel{?}{=}8\\ -4+12\amp\stackrel{\checkmark}{=}8\amp6+2\amp\stackrel{\checkmark}{=}8 \end{align*}$

Algebraically, both values do check out to be solutions. In the context of this scenario, the boat's speed can't be negative, so we only take the solution $4\text{.}$ If Julia drives at $4$ miles per hour, the whole trip would take $8$ hours. This result matches the solution in Example 12.3.2.

Definition 12.5.3. Rational Equation.

A rational equation is an equation involving one or more rational expressions. Usually, we consider these to be equations that have the variable in the denominator of at least one term.

permalinkLet's look at another application problem.

Example 12.5.4.

It takes Ku $3$ hours to paint a room and it takes Jacob $6$ hours to paint the same room. If they work together, how long would it take them to paint the room?

Since it takes Ku \(3\) hours to paint the room, he paints \(\frac{1}{3}\) of the room each hour. Similarly, Jacob paints \(\frac{1}{6}\) of the room each hour. If they work together, they paint \(\frac{1}{3}+\frac{1}{6}\) of the room each hour.

Assume it takes \(x\) hours to paint the room if Ku and Jacob work together. This implies they paint \(\frac{1}{x}\) of the room together each hour. Now we can write this equation:

\begin{equation*} \frac{1}{3}+\frac{1}{6}=\frac{1}{x}\text{.} \end{equation*}

To clear away denominators, we multiply both sides of the equation by the common denominator of \(3\text{,}\) \(6\) and \(x\text{,}\) which is \(6x\text{:}\)

\begin{align*} \frac{1}{3}+\frac{1}{6}\amp=\frac{1}{x}\\ \multiplyleft{6x}\left(\frac{1}{3}+\frac{1}{6}\right)\amp=\multiplyleft{6\cancelhighlight{x}}\frac{1}{\cancelhighlight{x}}\\ 6x\cdot\frac{1}{3}+6x\cdot\frac{1}{6}\amp=6\\ 2x+x\amp=6\\ 3x\amp=6\\ x\amp=2 \end{align*}

Does the possible solution \(x=2\) check as an actual solution?

\begin{align*} \frac{1}{3}+\frac{1}{6}\amp\stackrel{?}{=}\frac{1}{\substitute{2}}\\ \frac{2}{6}+\frac{1}{6}\amp\stackrel{?}{=}\frac{1}{\substitute{2}}\\ \frac{3}{6}\amp\stackrel{\checkmark}{=}\frac{1}{\substitute{2}} \end{align*}

It does, so it is a solution. If Ku and Jacob work together, it would take them \(2\) hours to paint the room.

permalinkWe are ready to outline a general process for solving a rational equation.

Process 12.5.5. Solving Rational Equations.

To solve a rational equation,

Find the least common denominator for all terms in the equation.
Multiply every term in the equation by the least common denominator
Every denominator should cancel leaving a simpler kind of equation to solve. Use previous method to solve that equation.

permalinkLet's look at a few more examples of solving rational equations.

Example 12.5.6.

Solve for $y$ in $\frac{2}{y+1}=\frac{3}{y}\text{.}$

The common denominator is \(y(y+1)\text{.}\) We will multiply both sides of the equation by \(y(y+1)\text{:}\)

\begin{align*} \frac{2}{y+1}\amp=\frac{3}{y}\\ \multiplyleft{y\cancelhighlight{(y+1)}}\frac{2}{\cancelhighlight{y+1}}\amp=\multiplyleft{\secondcancelhighlight{y}(y+1)}\frac{3}{\secondcancelhighlight{y}}\\ 2y\amp=3(y+1)\\ 2y\amp=3y+3\\ -y\amp=3\\ y\amp=-3 \end{align*}

Does the possible solution \(y=-3\) check as an actual solution?

\begin{align*} \frac{2}{\substitute{-3}+1}\amp\stackrel{?}{=}\frac{3}{\substitute{-3}}\\ \frac{2}{-2}\amp\stackrel{\checkmark}{=}-1 \end{align*}

It checks, so \(-3\) is a solution. We write the solution set as \(\{-3\}\text{.}\)

Example 12.5.7.

Solve for $z$ in $z+\frac{1}{z-4}=\frac{z-3}{z-4}\text{.}$

The common denominator is \(z-4\text{.}\) We will multiply both sides of the equation by \(z-4\text{:}\)

\begin{align*} z+\frac{1}{z-4}\amp=\frac{z-3}{z-4}\\ \multiplyleft{(z-4)}\left(z+\frac{1}{z-4}\right)\amp=\multiplyleft{\cancelhighlight{(z-4)}}\frac{z-3}{\cancelhighlight{z-4}}\\ (z-4)\cdot z+\cancelhighlight{(z-4)}\cdot\frac{1}{\cancelhighlight{z-4}}\amp=z-3\\ (z-4)\cdot z+1\amp=z-3\\ z^2-4z+1\amp=z-3\\ z^2-5z+4\amp=0\\ (z-1)(z-4)\amp=0 \end{align*}

\begin{align*} z-1\amp=0\amp\amp\text{or}\amp z-4\amp=0\\ z\amp=1\amp\amp\text{or}\amp z\amp=4 \end{align*}

Do the possible solutions \(z=1\) and \(z=4\) check as actual solutions?

\begin{align*} \substitute{1}+\frac{1}{\substitute{1}-4}\amp\stackrel{?}{=}\frac{\substitute{1}-3}{\substitute{1}-4}\amp\substitute{4}+\frac{1}{\substitute{4}-4}\amp\stackrel{?}{=}\frac{\substitute{4}-3}{\substitute{4}-4}\\ 1-\frac{1}{3}\amp\stackrel{\checkmark}{=}\frac{-2}{-3}\amp4+\frac{1}{0}\amp\stackrel{\text{no}}{=}\frac{1}{0} \end{align*}

The possible solution \(z=4\) does not actually work, since it leads to division by \(0\) in the equation. It is an extraneous solution. However, \(z=1\) is a valid solution. The only solution to the equation is \(1\text{,}\) and thus we can write the solution set as \(\{1\}\text{.}\)

Example 12.5.8.

Solve for $p$ in $\frac{3}{p-2}+\frac{5}{p+2}=\frac{12}{p^2-4}\text{.}$

To find the common denominator, we need to factor all denominators if possible:

\begin{equation*} \frac{3}{p-2}+\frac{5}{p+2}=\frac{12}{(p+2)(p-2)} \end{equation*}

Now we can see the common denominator is \((p+2)(p-2)\text{.}\) We will multiply both sides of the equation by \((p+2)(p-2)\text{:}\)

\begin{align*} \frac{3}{p-2}+\frac{5}{p+2}\amp=\frac{12}{p^2-4}\\ \frac{3}{p-2}+\frac{5}{p+2}\amp=\frac{12}{(p+2)(p-2)}\\ \multiplyleft{(p+2)(p-2)}\left(\frac{3}{p-2}+\frac{5}{p+2}\right)\amp=\multiplyleft{(p+2)(p-2)}\frac{12}{(p+2)(p-2)}\\ (p+2)\cancelhighlight{(p-2)}\cdot\frac{3}{\cancelhighlight{p-2}}+\secondcancelhighlight{(p+2)}(p-2)\cdot\frac{5}{\secondcancelhighlight{p+2}}\amp=\secondcancelhighlight{(p+2)}\cancelhighlight{(p-2)}\cdot\frac{12}{\secondcancelhighlight{(p+2)}\cancelhighlight{(p-2)}}\\ 3(p+2)+5(p-2)\amp=12\\ 3p+6+5p-10\amp=12\\ 8p-4\amp=12\\ 8p\amp=16\\ p\amp=2 \end{align*}

Does the possible solution \(p=2\) check as an actual solution?

\begin{align*} \frac{3}{\substitute{2}-2}+\frac{5}{\substitute{2}+2}\amp\stackrel{?}{=}\frac{12}{\substitute{2}^2-4}\\ \frac{3}{0}+\frac{5}{4}\amp\stackrel{\text{no}}{=}\frac{12}{0} \end{align*}

The possible solution \(p=2\) does not actually work, since it leads to division by \(0\) in the equation. So this is an extraneous solution, and the equation actually has no solution. We could say that its solution set is the empty set, \(\emptyset\text{.}\)

Example 12.5.9.

Solve $C(t)=0.35\text{,}$ where $C(t)=\frac{3t}{t^2+8}$ gives a drug's concentration in milligrams per liter $t$ hours since an injection. (This function was explored in the introduction of Section 12.1.)

To solve \(C(t)=0.35\text{,}\) we'll begin by setting up \(\frac{3t}{t^2+8}=0.35\text{.}\) We'll begin by identifying that the LCD is \(t^2+8\text{,}\) and multiply each side of the equation by this:

\begin{align*} \frac{3t}{t^2+8}\amp=0.35\\ \frac{3t}{\cancelhighlight{t^2+8}}\multiplyright{\cancelhighlight{\left(t^2+8\right)}}\amp=0.35\multiplyright{\left(t^2+8\right)}\\ 3t\amp=0.35\left(t^2+8\right)\\ 3t\amp=0.35t^2+2.8 \end{align*}

This results in a quadratic equation so we will put it in standard form and use the quadratic formula:

\begin{align*} 0\amp=0.35t^2-3t+2.8\\ t\amp=\frac{-(-3)\pm \sqrt{(-3)^2-4(0.35)(2.8)}}{2(0.35)}\\ t\amp=\frac{3\pm \sqrt{5.08}}{0.7}\\ t\amp\approx 1.066 \text{ or } t\approx 7.506 \end{align*}

Each of these answers should be checked in the original equation; they both work. In context, this means that the drug concentration will reach \(0.35\) milligrams per liter about \(1.066\) hours after the injection was given, and again \(7.506\) hours after the injection was given.

Subsection 12.5.2 Solving Rational Equations for a Specific Variable

permalinkRational equations can contain many variables and constants and we can solve for any one of them. The process for solving still involves multiplying each side of the equation by the LCD. Instead of having a numerical answer though, our final result will contain other variables and constants.

Example 12.5.10.

In physics, when two resistances, $R_1$ and $R_2\text{,}$ are connected in a parallel circuit, the combined resistance, $R\text{,}$ can be calculated by the formula

$\begin{equation*} \frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}\text{.} \end{equation*}$

Solve for $R$ in this formula.

The common denominator is \(R R_1 R_2\text{.}\) We will multiply both sides of the equation by \(R R_1 R_2\text{:}\)

\begin{align*} \frac{1}{R}\amp=\frac{1}{R_1}+\frac{1}{R_2}\\ \multiplyleft{\cancelhighlight{R} R_1 R_2}\frac{1}{\cancelhighlight{R}}\amp=\multiplyleft{R R_1 R_2}\left(\frac{1}{R_1}+\frac{1}{R_2}\right)\\ R_1 R_2\amp=R \cancelhighlight{R_1} R_2\cdot\frac{1}{\cancelhighlight{R_1}}+R R_1 \secondcancelhighlight{R_2}\cdot\frac{1}{\secondcancelhighlight{R_2}}\\ R_1 R_2\amp=R R_2 + R R_1\\ R_1 R_2\amp=R\left(R_2+R_1\right)\\ \frac{R_1 R_2}{R_2+R_1}\amp=R\\ R\amp=\frac{R_1 R_2}{R_1+R_2} \end{align*}

Example 12.5.11.

Here is the slope formula

$\begin{equation*} m=\frac{y_2-y_1}{x_2-x_1}\text{.} \end{equation*}$

Solve for $x_1$ in this formula.

The common denominator is \(x_2-x_1\text{.}\) We will multiply both sides of the equation by \(x_2-x_1\text{:}\)

\begin{align*} m\amp =\frac{y_2-y_1}{x_2-x_1}\\ \multiplyleft{\left(x_2-x_1\right)}m\amp =\multiplyleft{\cancelhighlight{\left(x_2-x_1\right)}}\frac{y_2-y_1}{\cancelhighlight{x_2-x_1}}\\ m x_2-m x_1\amp=y_2-y_1\\ -m x_1\amp=y_2-y_1-m x_2\\ \frac{-m x_1}{-m}\amp=\frac{y_2-y_1-m x_2}{-m}\\ x_1\amp=-\frac{y_2-y_1-m x_2}{m} \end{align*}

Example 12.5.12.

Solve the rational equation $x=\frac{4y-1}{2y-3}$ for $y\text{.}$

Our first step will be to multiply each side by the LCD, which is simply \(2y-3\text{.}\) After that, we'll isolate all terms containing \(y\text{,}\) factor out \(y\text{,}\) and then finish solving for that variable.

\begin{align*} x\amp=\frac{4y-1}{2y-3}\\ x\multiplyright{(2y-3)}\amp=\frac{4y-1}{\cancelhighlight{2y-3}}\multiplyright{\cancelhighlight{(2y-3)}}\\ 2xy-3x\amp=4y-1\\ 2xy\amp=4y-1+3x\\ 2xy-4y\amp=-1+3x\\ y(2x-4)\amp=3x-1\\ \divideunder{y(2x-4)}{2x-4}\amp=\divideunder{3x-1}{2x-4}\\ y\amp=\frac{3x-1}{2x-4} \end{align*}

Subsection 12.5.3 Solving Rational Equations Using Technology

permalinkIn some instances, it may be difficult to solve rational equations algebraically. We can instead use graphing technology to obtain approximate solutions. Let's look at one such example.

Example 12.5.13.

Solve the equation $\frac{2}{x-3}=\frac{x^3}{8}$ using graphing technology.

We will define \(f(x)=\frac{2}{x-3}\) and \(g(x)=\frac{x^3}{8}\text{,}\) and then look for the points of intersection.

eps pdf png svg tex

Figure 12.5.14. Graph of \(f(x)=\frac{2}{x-3}\) and \(g(x)=\frac{x^3}{8}\)

Since the two functions intersect at approximately \((-1.524,-0.442)\) and \((3.405,4.936)\text{,}\) the solutions to \(\frac{2}{x-3}=\frac{x^3}{8}\) are approximately \(-1.524\) and \(3.405\text{.}\) We can write the solution set as \(\{-1.524\ldots, 3.405\ldots\}\) or in several other forms. It may be important to do something to communicate that these solutions are approximations. Here we used “\(\ldots\)”, but you could also just say in words that the solutions are approximate.

Reading Questions 12.5.4 Reading Questions

1.

Describe what an “extraneous solution” to a rational equation is.

2.

In general, when solving a rational equation, multiplying through by the will leave you with a simpler equation to solve.

3.

When you believe you have the solutions to a rational equation, what is more important than usual (compared to other kinds of equations) for you to do?

Exercises 12.5.5 Exercises

Review and Warmup

1.

Solve the equation.

${8A+4} = {A+60}$

2.

Solve the equation.

${6C+8} = {C+33}$

3.

Solve the equation.

$\displaystyle{ {50}={8-3\!\left(n-6\right)} }$

4.

Solve the equation.

$\displaystyle{ {-3}={4-7\!\left(p-6\right)} }$

5.

Solve the equation.

$\displaystyle{ {2\!\left(x+10\right)-8\!\left(x-2\right)}={36} }$

6.

Solve the equation.

$\displaystyle{ {4\!\left(y+5\right)-7\!\left(y-7\right)}={72} }$

7.

Solve the equation.

$\left(x - 6\right)^2 = 25$

8.

Solve the equation.

$\left(x - 3\right)^2 = 144$

9.

Solve the equation.

${x^{2}+2x-3}= 0$

10.

Solve the equation.

${x^{2}-3x-10}= 0$

11.

Solve the equation.

${x^{2}+15x+61}= 17$

12.

Solve the equation.

${x^{2}+9x+31}= 17$

13.

Recall the time that Filip traveled with his kids to kick a soccer ball on Mars? We should examine one more angle to our soccer kick question. The formula $H(t)=-6.07t^2+27.1t$ finds the height of the soccer ball in feet above the ground at a time $t$ seconds after being kicked.

Using technology, find out what the maximum height of the ball was and when it reached that height.
Using technology, solve for when $H(t)=20$ and interpret the meaning of this in a complete sentence.
Using technology, solve for when $H(t)=0$ and interpret the meaning of this in a complete sentence.

Solving Rational Equations

14.

Solve the equation.

$\displaystyle{ {\frac{20}{t}} = {-5} }$

15.

Solve the equation.

$\displaystyle{ {\frac{40}{x}} = {8} }$

16.

Solve the equation.

$\displaystyle{ {\frac{x}{x+3}} = {4} }$

17.

Solve the equation.

$\displaystyle{ {\frac{x}{x+4}} = {-3} }$

18.

Solve the equation.

$\displaystyle{ {\frac{y+10}{3y-8}} = \frac{9}{8} }$

19.

Solve the equation.

$\displaystyle{ {\frac{y-4}{3y-2}} = \frac{3}{4} }$

20.

Solve the equation.

$\displaystyle{ {\frac{-7r+4}{r-4}} = {-\frac{7r}{r-8}} }$

21.

Solve the equation.

$\displaystyle{ {\frac{2r+9}{r+8}} = {\frac{2r}{r+7}} }$

22.

Solve the equation.

$\displaystyle{ {\frac{3}{t}} = {4+\frac{23}{t}} }$

23.

Solve the equation.

$\displaystyle{ {\frac{3}{t}} = {-2+\frac{13}{t}} }$

24.

Solve the equation.

$\displaystyle{ {\frac{5}{4x}+\frac{1}{3x}}={-5} }$

25.

Solve the equation.

$\displaystyle{ {\frac{3}{5y}+\frac{4}{3y}}={-5} }$

26.

Solve the equation.

$\displaystyle{ {\frac{x}{5x-30}-\frac{2}{x-6}}={1} }$

27.

Solve the equation.

$\displaystyle{ {\frac{y}{2y+12}+\frac{6}{y+6}}={2} }$

28.

Solve the equation.

$\displaystyle{ {\frac{y-2}{y^{2}+2}} = 0 }$

29.

Solve the equation.

$\displaystyle{ {\frac{r-9}{r^{2}+6}} = 0 }$

30.

Solve the equation.

$\displaystyle{ {\frac{3}{r}} = 0 }$

31.

Solve the equation.

$\displaystyle{ {-\frac{3}{t}} = 0 }$

32.

Solve the equation.

$\displaystyle{ {\frac{t+9}{t^{2}+15t+54}} = 0 }$

33.

Solve the equation.

$\displaystyle{ {\frac{t+3}{t^{2}-2t-15}} = 0 }$

34.

Solve the equation.

$\displaystyle{ {-\frac{2}{x}+\frac{8}{x+7}} = -1 }$

35.

Solve the equation.

$\displaystyle{ {-\frac{8}{x}-\frac{5}{x+9}} = 1 }$

36.

Solve the equation.

$\displaystyle{ {\frac{1}{y+2}+\frac{2}{y^{2}+2y}} = \frac{1}{5} }$

37.

Solve the equation.

$\displaystyle{ {\frac{1}{y-5}-\frac{5}{y^{2}-5y}} = -\frac{1}{9} }$

38.

Solve the equation.

$\displaystyle{ {\frac{1}{r+9}-\frac{5}{r^{2}+9r}} = {{\frac{1}{2}}} }$

39.

Solve the equation.

$\displaystyle{ {\frac{1}{r-7}-\frac{4}{r^{2}-7r}} = {{\frac{1}{2}}} }$

40.

Solve the equation.

$\displaystyle{ {\frac{t-4}{t+8}+\frac{2}{t+3}} = -4 }$

41.

Solve the equation.

$\displaystyle{ {\frac{t+9}{t-3}+\frac{4}{t-8}} = 3 }$

42.

Solve the equation.

$\displaystyle{ {\frac{2}{t+1}}={\frac{3}{t-1}-\frac{2}{t^{2}-1}} }$

43.

Solve the equation.

$\displaystyle{ {-\frac{4}{x+3}}={-\left(\frac{2}{x-3}+\frac{6}{x^{2}-9}\right)} }$

44.

Solve the equation.

$\displaystyle{ {\frac{8}{x+6}-\frac{9}{x+9}}={-\frac{9}{x^{2}+15x+54}} }$

45.

Solve the equation.

$\displaystyle{ {\frac{2}{y+5}-\frac{5}{y+1}}={-\frac{2}{y^{2}+6y+5}} }$

46.

Solve the equation.

$\displaystyle{ {-\frac{8}{y-6}+\frac{4y}{y-5}}={-\frac{8}{y^{2}-11y+30}} }$

47.

Solve the equation.

$\displaystyle{ {\frac{4}{r-7}+\frac{2r}{r-5}}={\frac{8}{r^{2}-12r+35}} }$

48.

Solve the equation.

$\displaystyle{ {-\frac{6}{r-3}+\frac{8r}{r+9}}={-\frac{4}{r^{2}+6r-27}} }$

49.

Solve the equation.

$\displaystyle{ {\frac{6}{t-7}+\frac{2t}{t-3}}={-\frac{8}{t^{2}-10t+21}} }$

Solving Rational Equations for a Specific Variable

50.

Solve this equation for $a\text{:}$

$\displaystyle{ p = \frac{m}{a} }$

51.

Solve this equation for $m\text{:}$

$\displaystyle{ q = \frac{b}{m} }$

52.

Solve this equation for $x\text{:}$

$\displaystyle{ y = \frac{x}{r} }$

53.

Solve this equation for $C\text{:}$

$\displaystyle{ r = \frac{C}{B} }$

54.

Solve this equation for $a\text{:}$

$\displaystyle{ \frac{1}{2a} = \frac{1}{x} }$

55.

Solve this equation for $c\text{:}$

$\displaystyle{ \frac{1}{8c} = \frac{1}{q} }$

56.

Solve this equation for $B\text{:}$

$\displaystyle{ \frac{1}{A} = \frac{8}{B+2} }$

57.

Solve this equation for $a\text{:}$

$\displaystyle{ \frac{1}{C} = \frac{8}{a+6} }$

Solving Rational Equations Using Technology

58.

Use technology to solve the equation.

$\begin{equation*} \frac{10}{x^2+3}=\frac{x+1}{x+5} \text{.} \end{equation*}$

59.

Use technology to solve the equation.

$\begin{equation*} \frac{x-9}{x^5+1}=-3x-7 \text{.} \end{equation*}$

60.

Use technology to solve the equation.

$\begin{equation*} \frac{1}{x}+\frac{1}{x^2}=\frac{1}{x^3} \text{.} \end{equation*}$

61.

Use technology to solve the equation.

$\begin{equation*} \frac{12x}{x-5}+\frac{3}{x+1}=\frac{x-5}{x^2} \text{.} \end{equation*}$

62.

Use technology to solve the equation.

$\begin{equation*} 2x-\frac{1}{x+4}=\frac{3}{x+6} \text{.} \end{equation*}$

63.

Use technology to solve the equation.

$\begin{equation*} \frac{1}{x^2-1}-\frac{2}{x-4}=\frac{3}{x-2} \text{.} \end{equation*}$

Application Problems

64.

Scot and Jay are working together to paint a room. If Scot paints the room alone, it would take him $18$ hours to complete the job. If Jay paints the room alone, it would take him $12$ hours to complete the job. Answer the following question:

If they work together, it would take them hours to complete the job. Use a decimal in your answer if needed.

65.

There are three pipes at a tank. To fill the tank, it would take Pipe A $3$ hours, Pipe B $12$ hours, and Pipe C $4$ hours. Answer the following question:

If all three pipes are turned on, it would take hours to fill the tank.

66.

Casandra and Tien are working together to paint a room. Casandra works $1.5$ times as fast as Tien does. If they work together, it took them $9$ hours to complete the job. Answer the following questions:

If Casandra paints the room alone, it would take her hours to complete the job.

If Tien paints the room alone, it would take him hours to complete the job.

67.

Two pipes are being used to fill a tank. Pipe A can fill the tank $4.5$ times as fast as Pipe B does. When both pipes are turned on, it takes $18$ hours to fill the tank. Answer the following questions:

If only Pipe A is turned on, it would take hours to fill the tank.

If only Pipe B is turned on, it would take hours to fill the tank.

68.

Kandace and Jenny worked together to paint a room, and it took them $2$ hours to complete the job. If they work alone, it would take Jenny $3$ more hours than Kandace to complete the job. Answer the following questions:

If Kandace paints the room alone, it would take her hours to complete the job.

If Jenny paints the room alone, it would take her hours to complete the job.

69.

If both Pipe A and Pipe B are turned on, it would take $2$ hours to fill a tank. If each pipe is turned on alone, it takes Pipe B $3$ fewer hours than Pipe A to fill the tank. Answer the following questions:

If only Pipe A is turned on, it would take hours to fill the tank.

If only Pipe B is turned on, it would take hours to fill the tank.

70.

Town A and Town B are $570$ miles apart. A boat traveled from Town A to Town B, and then back to Town A. Since the river flows from Town B to Town A, the boat’s speed was $25$ miles per hour faster when it traveled from Town B to Town A. The whole trip took $19$ hours. Answer the following questions:

The boat traveled from Town A to Town B at the speed of miles per hour.

The boat traveled from Town B back to Town A at the speed of miles per hour.

71.

A river flows at $7$ miles per hour. A boat traveled with the current from Town A to Town B, which are $260$ miles apart. Then, the boat turned around, and traveled against the current to reach Town C, which is $160$ miles away from Town B. The second leg of the trip (Town B to Town C) took the same time as the first leg (Town A to Town B). During this whole trip, the boat was driving at a constant still-water speed. Answer the following question:

During this trip, the boat’s speed on still water was miles per hour.

72.

A river flows at $5$ miles per hour. A boat traveled with the current from Town A to Town B, which are $100$ miles apart. The boat stayed overnight at Town B. The next day, the water’s current stopped, and boat traveled on still water to reach Town C, which is $190$ miles away from Town B. The second leg of the trip (Town B to Town C) took $9$ hours longer than the first leg (Town A to Town B). During this whole trip, the boat was driving at a constant still-water speed. Find this speed.

Note that you should not consider the unreasonable answer.

During this trip, the boat’s speed on still water was miles per hour.

73.

Town A and Town B are $600$ miles apart. With a constant still-water speed, a boat traveled from Town A to Town B, and then back to Town A. During this whole trip, the river flew from Town A to Town B at $20$ miles per hour. The whole trip took $16$ hours. Answer the following question:

During this trip, the boat’s speed on still water was miles per hour.

74.

Town A and Town B are $350$ miles apart. With a constant still-water speed of $24$ miles per hour, a boat traveled from Town A to Town B, and then back to Town A. During this whole trip, the river flew from Town B to Town A at a constant speed. The whole trip took $30$ hours. Answer the following question:

During this trip, the river’s speed was miles per hour.

75.

Suppose that a large pump can empty a swimming pool in $43\ {\rm hr}$ and that a small pump can empty the same pool in $53\ {\rm hr}\text{.}$ If both pumps are used at the same time, how long will it take to empty the pool?

If both pumps are used at the same time, it will take to empty the pool.

76.

The winner of a $9\ {\rm mi}$ race finishes $14.73\ {\rm min}$ ahead of the second-place runner. On average, the winner ran $0.6\ {\textstyle\frac{\rm\mathstrut mi}{\rm\mathstrut hr}}$ faster than the second place runner. Find the average running speed for each runner.

The winner's average speed was and the second-place runner's average speed was .

77.

In still water a tugboat can travel $15\ {\textstyle\frac{\rm\mathstrut mi}{\rm\mathstrut hr}}\text{.}$ It travels $42\ {\rm mi}$ upstream and then $42\ {\rm mi}$ downstream in a total time of $5.96\ {\rm hr}\text{.}$ Find the speed of the current.

The current's speed is .

78.

Without any wind an airplane flies at $300\ {\textstyle\frac{\rm\mathstrut mi}{\rm\mathstrut hr}}\text{.}$ The plane travels $600\ {\rm mi}$ into the wind and then returns with the wind in a total time of $4.04\ {\rm hr}\text{.}$ Find the average speed of the wind.

The wind's speed is .

79.

When there is a $11.8\ {\textstyle\frac{\rm\mathstrut mi}{\rm\mathstrut hr}}$ wind, an airplane can fly $770\ {\rm mi}$ with the wind in the same time that it can fly $702\ {\rm mi}$ against the wind. Find the speed of the plane when there is no wind.

The plane's airspeed is .

80.

It takes one employee $2.5\ {\rm hr}$ longer to mow a football field than it does a more experienced employee. Together they can mow the grass in $1.9\ {\rm hr}\text{.}$ How long does it take each person to mow the football field working alone?

The more experienced worker takes to mow the field alone, and the less experienced worker takes .

81.

It takes one painter $13\ {\rm hr}$ longer to paint a house than it does a more experienced painter. Together they can paint the house in $30\ {\rm hr}\text{.}$ How long does it take for each painter to paint the house working alone?

The more experienced painter takes to paint the house alone, and the less experienced painter takes .