ORCCA Solving Rational Equations

Section 13.5 Solving Rational Equations

Figure 13.5.1 Alternative Video Lesson

Subsection 13.5.1 Solving Rational Equations

To start this section, we will use a scenario we have seen before in Example 13.3.2:

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 substitute \(t(v)\) with \(8\text{,}\) 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 3.3.2. 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)\highlight{\cancel{(v-2)}}\cdot\frac{12}{\highlight{\cancel{v-2}}}+\lighthigh{\bcancel{(v+2)}}(v-2)\cdot\frac{12}{\lighthigh{\bcancel{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 13.5.2

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.

We 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 13.3.2.

Let's look at another application problem.

Example 13.5.3

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?

Explanation

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\highlight{\cancel{x}}}\frac{1}{\highlight{\cancel{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.

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

Example 13.5.4

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

Explanation

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\highlight{\cancel{(y+1)}}}\frac{2}{\highlight{\cancel{y+1}}}\amp=\multiplyleft{\lighthigh{\bcancel{y}}(y+1)}\frac{3}{\lighthigh{\bcancel{y}}}\\ 2y\amp=3(y+1)\\ 2y\amp=3y+3\\ 0\amp=y+3\\ -3\amp=y \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 13.5.5

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

Explanation

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{\highlight{\cancel{(z-4)}}}\frac{z-3}{\highlight{\cancel{z-4}}}\\ (z-4)\cdot z+\highlight{\cancel{(z-4)}}\cdot\frac{1}{\highlight{\cancel{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 13.5.6

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

Explanation

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)\highlight{\cancel{(p-2)}}\cdot\frac{3}{\highlight{\cancel{p-2}}}+\highlight{\bcancel{(p+2)}}(p-2)\cdot\frac{5}{\highlight{\bcancel{p+2}}}\amp=\highlight{\xcancel{(p+2)(p-2)}}\cdot\frac{12}{\highlight{\xcancel{(p+2)(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 13.5.7

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 13.1.)

Explanation

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}{\highlight{\cancel{t^2+8}}}\multiplyright{\highlight{\cancel{\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 13.5.2 Solving Rational Equations for a Specific Variable

Rational 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 13.5.8

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.

Explanation

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{\highlight{\cancel{R}} R_1 R_2}\frac{1}{\highlight{\cancel{R}}}\amp=\multiplyleft{R R_1 R_2}\left(\frac{1}{R_1}+\frac{1}{R_2}\right)\\ R_1 R_2\amp=R \highlight{\cancel{R_1}} R_2\cdot\frac{1}{\highlight{\cancel{R_1}}}+R R_1 \highlight{\bcancel{R_2}}\cdot\frac{1}{\highlight{\bcancel{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 13.5.9

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.

Explanation

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{\highlight{\cancel{\left(x_2-x_1\right)}}}\frac{y_2-y_1}{\highlight{\cancel{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 13.5.10

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

Explanation

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}{\highlight{\cancel{2y-3}}}\multiplyright{\highlight{\cancel{(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\highlight{\cancel{(2x-4)}}}{\highlight{\cancel{2x-4}}}\amp=\divideunder{3x-1}{2x-4}\\ y\amp=\frac{3x-1}{2x-4} \end{align*}

Subsection 13.5.3 Solving Rational Equations Using Technology

In 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 13.5.11

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

Explanation

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 13.5.12 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 commuincate that these solutions are approximations. Here we used \(\ldots\text{,}\) but you could also just say in words that the solutions are approximate.

Subsection 13.5.4 Exercises

Review and Warmup

1

Solve the equation.

\({10y+5} = {y+50}\)

2

Solve the equation.

\({9r+9} = {r+33}\)

3

Solve the equation.

\(\displaystyle{ {6}={3-3\!\left(a-9\right)} }\)

4

Solve the equation.

\(\displaystyle{ {52}={10-7\!\left(b-8\right)} }\)

5

Solve the equation.

\(\displaystyle{ {4\!\left(A+1\right)-7\!\left(A-7\right)}={35} }\)

6

Solve the equation.

\(\displaystyle{ {2\!\left(C+5\right)-6\!\left(C-2\right)}={-2} }\)

7

Solve the equation.

\(\left(x+5\right)^2 = 121\)

8

Solve the equation.

\(\left(x+8\right)^2 = 49\)

9

Solve the equation.

\({x^{2}+4x-96}= 0\)

10

Solve the equation.

\({x^{2}-x-90}= 0\)

11

Solve the equation.

\({x^{2}-6x+1}= 8\)

12

Solve the equation.

\({x^{2}-14x+64}= 19\)

Solving Rational Equations

13

Solve the equation.

\(\displaystyle{ {\frac{-12}{y}} = {6} }\)

14

Solve the equation.

\(\displaystyle{ {\frac{-6}{r}} = {-1} }\)

15

Solve the equation.

\(\displaystyle{ {\frac{r}{r+3}} = {4} }\)

16

Solve the equation.

\(\displaystyle{ {\frac{t}{t-4}} = {-3} }\)

17

Solve the equation.

\(\displaystyle{ {\frac{t+7}{5t+7}} = \frac{1}{19} }\)

18

Solve the equation.

\(\displaystyle{ {\frac{t-2}{4t-1}} = \frac{1}{3} }\)

19

Solve the equation.

\(\displaystyle{ {\frac{-x-2}{x-1}} = {-\frac{x}{x-6}} }\)

20

Solve the equation.

\(\displaystyle{ {\frac{-x+10}{x-6}} = {-\frac{x}{x-1}} }\)

21

Solve the equation.

\(\displaystyle{ {\frac{6}{y}} = {2-\frac{4}{y}} }\)

22

Solve the equation.

\(\displaystyle{ {\frac{6}{y}} = {-5+\frac{16}{y}} }\)

23

Solve the equation.

\(\displaystyle{ {\frac{4}{5A}-\frac{1}{4A}}={-2} }\)

24

Solve the equation.

\(\displaystyle{ {\frac{1}{5B}+\frac{5}{6B}}={4} }\)

25

Solve the equation.

\(\displaystyle{ {\frac{t}{2t-10}+\frac{1}{t-5}}={4} }\)

26

Solve the equation.

\(\displaystyle{ {\frac{t}{6t+24}-\frac{3}{t+4}}={2} }\)

27

Solve the equation.

\(\displaystyle{ {\frac{t+8}{t^{2}+3}} = 0 }\)

28

Solve the equation.

\(\displaystyle{ {\frac{x+2}{x^{2}+6}} = 0 }\)

29

Solve the equation.

\(\displaystyle{ {-\frac{5}{x}} = 0 }\)

30

Solve the equation.

\(\displaystyle{ {\frac{7}{y}} = 0 }\)

31

Solve the equation.

\(\displaystyle{ {\frac{y+1}{y^{2}+8y+7}} = 0 }\)

32

Solve the equation.

\(\displaystyle{ {\frac{r-6}{r^{2}-10r+24}} = 0 }\)

33

Solve the equation.

\(\displaystyle{ {-\frac{9}{r}-\frac{6}{r+6}} = 1 }\)

34

Solve the equation.

\(\displaystyle{ {\frac{6}{r}+\frac{6}{r-9}} = 1 }\)

35

Solve the equation.

\(\displaystyle{ {\frac{1}{t-7}-\frac{7}{t^{2}-7t}} = \frac{1}{6} }\)

36

Solve the equation.

\(\displaystyle{ {\frac{1}{t+6}+\frac{6}{t^{2}+6t}} = -\frac{1}{9} }\)

37

Solve the equation.

\(\displaystyle{ {\frac{1}{x-8}-\frac{6}{x^{2}-8x}} = {{\frac{1}{8}}} }\)

38

Solve the equation.

\(\displaystyle{ {\frac{1}{x-5}-\frac{4}{x^{2}-5x}} = {{\frac{1}{6}}} }\)

39

Solve the equation.

\(\displaystyle{ {\frac{y+3}{y+7}-\frac{2}{y+5}} = -1 }\)

40

Solve the equation.

\(\displaystyle{ {\frac{y+1}{y+3}-\frac{8}{y+6}} = -1 }\)

41

Solve the equation.

\(\displaystyle{ {-\frac{3}{r+5}}={-\left(\frac{6}{r-5}+\frac{3}{r^{2}-25}\right)} }\)

42

Solve the equation.

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

43

Solve the equation.

\(\displaystyle{ {\frac{7}{r+4}-\frac{8}{r+1}}={-\frac{9}{r^{2}+5r+4}} }\)

44

Solve the equation.

\(\displaystyle{ {\frac{9}{t+7}-\frac{7}{t+3}}={\frac{4}{t^{2}+10t+21}} }\)

45

Solve the equation.

\(\displaystyle{ {-\frac{2}{t-8}+\frac{2t}{t-6}}={-\frac{4}{t^{2}-14t+48}} }\)

46

Solve the equation.

\(\displaystyle{ {\frac{2}{x-7}+\frac{2x}{x-5}}={\frac{4}{x^{2}-12x+35}} }\)

47

Solve the equation.

\(\displaystyle{ {\frac{4}{x-4}+\frac{8x}{x+1}}={-\frac{8}{x^{2}-3x-4}} }\)

48

Solve the equation.

\(\displaystyle{ {\frac{2}{y+5}+\frac{2y}{y+7}}={-\frac{2}{y^{2}+12y+35}} }\)

Solving Rational Equations for a Specific Variable

49

Solve this equation for \(y\text{:}\)

\(\displaystyle{ b = \frac{a}{y} }\)

50

Solve this equation for \(n\text{:}\)

\(\displaystyle{ A = \frac{q}{n} }\)

51

Solve this equation for \(t\text{:}\)

\(\displaystyle{ B = \frac{t}{r} }\)

52

Solve this equation for \(t\text{:}\)

\(\displaystyle{ m = \frac{t}{C} }\)

53

Solve this equation for \(n\text{:}\)

\(\displaystyle{ \frac{1}{5n} = \frac{1}{p} }\)

54

Solve this equation for \(q\text{:}\)

\(\displaystyle{ \frac{1}{3q} = \frac{1}{A} }\)

55

Solve this equation for \(r\text{:}\)

\(\displaystyle{ \frac{1}{y} = \frac{3}{r+5} }\)

56

Solve this equation for \(n\text{:}\)

\(\displaystyle{ \frac{1}{r} = \frac{9}{n+6} }\)