ORCCA Solving Rational Equations

Section13.5Solving Rational Equations

Figure13.5.1Alternative Video Lesson

Subsection13.5.1Solving Rational Equations

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

Two towns are connected by a $12$ -mile-long river, which flows from Town A to Town B at the speed of $2$ miles per hour. A boat will travel at a constant speed of, $v$ miles per hour, from Town A to Town B, and then back to Town A. Due to the river's current, the actual speed of travel is $v+2$ miles per hour downstream from Town A to Town B, and $v-2$ miles per hour upstream back to Town A. If the boat driver plans to spend $8$ hours for the whole trip, how fast should they drive the boat?

The time it takes the boat to travel from Town A to Town B is $\frac{12}{v+2}\text{,}$ and $\frac{12}{v-2}$ from Town B to Town A. 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:

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

Instead of using the function's graph, we will solve this equation algebraically. Please review the technique of eliminating denominators discussed in Subsection 3.2.2. 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:

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

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

Remark13.5.2

It's possible that a false solution appears when we eliminate the denominator to solve a rational equation, and so it's critical that these proposed solutions are 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 are possible in the original equation.

In the original function, we can see the domain conditions are $v\ne2$ and $v\ne-2\text{,}$ because those two values would make the function undefined. Thus $v=4$ and $v=-1$ do not violate the domain conditions, so $4$ and $-1$ are solutions to the equation. In this context, the boat's speed shouldn't be negative, so we only take the solution $4\text{.}$

If the boat 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.

Example13.5.3

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

Solution

It takes Henry $3$ hours to paint the room. This implies Henry paints $\frac{1}{3}$ of the room each hour. Similarly, Dennis paints $\frac{1}{6}$ of the room each hour. If they work together, they paint $\frac{1}{3}+\frac{1}{6}$ of the room.

Assume it takes $x$ hours to paint the room if Henry and Dennis work together. This implies they paint $\frac{1}{x}$ of the room when working together. Now we can write this equation:

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

In this equation, the restriction is $x\ne0\text{.}$

To remove all fractions, 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{6x}\frac{1}{x}\\ 6x\cdot\frac{1}{3}+6x\cdot\frac{1}{6}\amp=6x\cdot\frac{1}{x}\\ 2x+x\amp=6\\ 3x\amp=6\\ x\amp=2 \end{align*}

The solution $x=2$ doesn't violate the restriction $x\ne0\text{,}$ so it is a solution.

If Henry and Dennis work together, it takes them $2$ hours to paint the room.

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

Example13.5.4

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

Solution

In this equation, we can see $y\ne-1$ and $y\ne0\text{.}$ 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(y+1)}\frac{2}{y+1}\amp=\multiplyleft{y(y+1)}\frac{3}{y}\\ 2y\amp=3(y+1)\\ 2y\amp=3y+3\\ -y\amp=3\\ y\amp=-3 \end{align*}

Since $y=-3$ doesn't violate the domain conditions, the solution to the equation is $-3\text{.}$ We write the solution set as $\{-3\}\text{.}$

Example13.5.5

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

Solution

In this equation, we can see $z\ne4\text{.}$ 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{(z-4)}\frac{z-3}{z-4}\\ (z-4)\cdot z+(z-4)\cdot\frac{1}{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*}

Since $z=4$ violates the domain condition $z\ne4\text{,}$ $4$ is not a solution. The only solution to the equation is $1\text{,}$ and thus we can write the solution set as $\{1\}\text{.}$

Example13.5.6

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

Solution

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 $(p+2)(p-2)\text{,}$ and the domain conditions are $p\ne2$ and $p\ne-2\text{.}$ 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)(p-2)\cdot\frac{3}{p-2}+(p+2)(p-2)\cdot\frac{5}{p+2}\amp=(p+2)(p-2)\cdot\frac{12}{(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*}

Since $p=2$ violates the domain condition $p\ne2\text{,}$ this equation has no solution.

Example13.5.7

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.)

Solution

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}{t^2+8}\multiplyright{(t^2+8)}\amp=0.35\multiplyright{(t^2+8)}\\ 3t\amp=0.35(t^2+8)\\ 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 to solve for $t\text{:}$

\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*}

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.

Remark13.5.8

In Example 13.5.7, we could have chosen to multiply each side by $100$ in order to eliminate the decimal coefficients before using the quadratic formula, like this:

$\begin{align*} 3t\amp=0.35t^2+2.8\\ 3t\multiplyright{100}\amp=0.35(t^2+8)\multiplyright{100}\\ 300t\amp=35(t^2+8)\\ 300t\amp=35t^2+280\\ 0\amp=35t^2-300t+280 \end{align*}$

Subsection13.5.2Solving 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. Constants are usually denoted with subscripts.

Example13.5.9

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

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

Solve for $R$ in this formula.

Solution

In this equation, the domain condition is $R\ne0\text{.}$ 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{R R_1 R_2}\frac{1}{R}\amp=\multiplyleft{R R_1 R_2}\left(\frac{1}{R_1}+\frac{1}{R_2}\right)\\ R_1 R_2\amp=R R_1 R_2\cdot\frac{1}{R_1}+R R_1 R_2\cdot\frac{1}{R_2}\\ R_1 R_2\amp=R R_2 + R R_1\\ R_1 R_2\amp=R(R_2+R_1)\\ \frac{R_1 R_2}{R_2+R_1}\amp=R\\ R\amp=\frac{R_1 R_2}{R_1+R_2} \end{align*}

The solution doesn't violate the domain condition.

Example13.5.10

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.

Solution

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{(x_2-x_1)}m\amp =\multiplyleft{(x_2-x_1)}\frac{y_2-y_1}{x_2-x_1}\\ m x_2-m x_1\amp=y_2-y_1\\ m x_2-m x_1\subtractright{m x_2}\amp=y_2-y_1\subtractright{m x_2}\\ -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*}

In this equation, we won't check the domain conditions because there are too many variables/constants.

Example13.5.11

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

Solution

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}{2y-3}\multiplyright{(2y-3)}\\ 2xy-3x\amp=4y-1\\ 2xy-3x\subtractright{4y}\addright{3x}\amp=4y-1\subtractright{4y}\addright{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*}

Subsection13.5.3Solving 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.

Example13.5.12

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

Solution

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.

Figure13.5.13Graph 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 $\{x\mid x\approx -1.524, 3.405\}\text{.}$

SubsectionExercises

Solving Rational Equations

1

Solve the equation.

$\displaystyle{ {\frac{15}{t}} = {5} }$

2

Solve the equation.

$\displaystyle{ {\frac{18}{t}} = {-2} }$

3

Solve the equation.

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

4

Solve the equation.

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

5

Solve the equation.

$\displaystyle{ {\frac{y-3}{5y-7}} = \frac{3}{19} }$

6

Solve the equation.

$\displaystyle{ {\frac{y-1}{3y-1}} = \frac{1}{2} }$

7

Solve the equation.

$\displaystyle{ {\frac{2r-3}{r+1}} = {\frac{2r}{r+4}} }$

8

Solve the equation.

$\displaystyle{ {\frac{-5r+6}{r-10}} = {-\frac{5r}{r-4}} }$

9

Solve the equation.

$\displaystyle{ {\frac{4}{r}} = {6+\frac{22}{r}} }$

10

Solve the equation.

$\displaystyle{ {\frac{4}{t}} = {-5+\frac{34}{t}} }$

11

Solve the equation.

$\displaystyle{ {\frac{4}{3q}+\frac{1}{2q}}={-2} }$

12

Solve the equation.

$\displaystyle{ {\frac{3}{2x}+\frac{1}{4x}}={-4} }$

13

Solve the equation.

$\displaystyle{ {\frac{x}{2x-6}+\frac{3}{x-3}}={2} }$

14

Solve the equation.

$\displaystyle{ {\frac{y}{3y-18}+\frac{5}{y-6}}={-2} }$

15

Solve the equation.

$\displaystyle{ {\frac{5}{y+3}}={\frac{1}{y-3}-\frac{2}{y^{2}-9}} }$

16

Solve the equation.

$\displaystyle{ {-\frac{2}{r+2}}={-\left(\frac{1}{r-2}+\frac{6}{r^{2}-4}\right)} }$

17

Solve the equation.

$\displaystyle{ {\frac{8}{r-1}+\frac{1}{r-6}}={-\frac{4}{r^{2}-7r+6}} }$

18

Solve the equation.

$\displaystyle{ {\frac{5}{r-7}-\frac{8}{r-3}}={-\frac{4}{r^{2}-10r+21}} }$

19

Solve the equation.

$\displaystyle{ {\frac{t-6}{t^{2}+5}} = 0 }$

20

Solve the equation.

$\displaystyle{ {\frac{t-7}{t^{2}+6}} = 0 }$

21

Solve the equation.

$\displaystyle{ {\frac{6}{x}} = 0 }$

22

Solve the equation.

$\displaystyle{ {-\frac{1}{x}} = 0 }$

23

Solve the equation.

$\displaystyle{ {\frac{y-8}{y^{2}-2y-48}} = 0 }$

24

Solve the equation.

$\displaystyle{ {\frac{y+5}{y^{2}-25}} = 0 }$

25

Solve the equation.

$\displaystyle{ {\frac{6}{y}+\frac{4}{y-4}} = 3 }$

26

Solve the equation.

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

27

Solve the equation.

$\displaystyle{ {\frac{1}{r+4}+\frac{4}{r^{2}+4r}} = \frac{1}{5} }$

28

Solve the equation.

$\displaystyle{ {\frac{1}{t-3}-\frac{3}{t^{2}-3t}} = -\frac{1}{9} }$

29

Solve the equation.

$\displaystyle{ {\frac{1}{t-1}+\frac{8}{t^{2}-t}} = {{\frac{1}{5}}} }$

30

Solve the equation.

$\displaystyle{ {\frac{1}{x-7}-\frac{2}{x^{2}-7x}} = {{\frac{1}{6}}} }$

31

Solve the equation.

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

32

Solve the equation.

$\displaystyle{ {-\frac{3}{y-3}+\frac{3y}{y-4}}={\frac{3}{y^{2}-7y+12}} }$

33

Solve the equation.

$\displaystyle{ {-\frac{4}{y+1}+\frac{6y}{y+6}}={-\frac{4}{y^{2}+7y+6}} }$

34

Solve the equation.

$\displaystyle{ {\frac{6}{y+3}+\frac{2y}{y-2}}={\frac{2}{y^{2}+y-6}} }$

35

Solve the equation.

$\displaystyle{ {\frac{r+6}{r+8}-\frac{6}{r+5}} = 2 }$

36

Solve the equation.

$\displaystyle{ {\frac{r+5}{r-7}+\frac{5}{r+3}} = 2 }$

Solving Rational Equations for a Specific Variable

37

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

$\displaystyle{ n = \frac{t}{C} }$

38

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

$\displaystyle{ q = \frac{p}{y} }$

39

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

$\displaystyle{ x = \frac{A}{c} }$

40

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

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

41

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

$\displaystyle{ \frac{1}{2t} = \frac{1}{n} }$

42

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

$\displaystyle{ \frac{1}{9b} = \frac{1}{c} }$

43

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

$\displaystyle{ \frac{1}{c} = \frac{8}{r+8} }$

44

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

$\displaystyle{ \frac{1}{B} = \frac{4}{n+8} }$

Solving Rational Equations Using Technology