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# 12.7: Solutions of Rational Equations

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Solve rational equations using cross products.
• Solve rational equations using lowest common denominators.
• Solve real-world problems with rational equations.

Introduction

A rational equation is one that contains rational expressions. It can be an equation that contains rational coefficients or an equation that contains rational terms where the variable appears in the denominator.

An example of the first kind of equation is: $\frac{3}{5}x+\frac{1}{2}=4$.

An example of the second kind of equation is: $\frac{x}{x-1}+1=\frac{4}{2x+3}$.

The first aim in solving a rational equation is to eliminate all denominators. That way, we can change a rational equation to a polynomial equation which we can solve with the methods we have learned this far.

## Solve Rational Equations Using Cross Products

A rational equation that contains just one term on each side is easy to solve by cross multiplication. Consider the following equation:

$\frac{x}{5}=\frac{x+1}{2}$

Our first goal is to eliminate the denominators of both rational expressions. In order to remove the 5 from the denominator of the first fraction, we multiply both sides of the equation by 5:

$5 \cdot \frac{x}{5} &= 5 \cdot \frac{x+1}{2}\\x &= \frac{5(x+1)}{2}$

Now, we remove the 2 from the denominator of the second fraction by multiplying both sides of the equation by 2:

$2 \cdot x &= 2 \cdot \frac{5(x+1)}{2}\\2x &= 5(x+1)$

Then we can solve this equation for $x$.

Notice that this equation is what we would get if we simply multiplied each numerator in the original equation by the denominator from the opposite side of the equation. It turns out that we can always simplify a rational equation with just two terms by multiplying each numerator by the opposite denominator; this is called cross multiplication.

Example 1

Solve the equation $\frac{2x}{x+4}=\frac{5}{x}$.

Solution

$\text{Cross-multiply. The equation simplifies to:} \qquad \qquad \quad 2x^2=5(x+4)\!\\\\\text{Simplify:} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad 2x^2=5x+20\!\\\\\text{Move all terms to one side of the equation:} \qquad \qquad \quad \ 2x^2-5x-20=0\!\\\\\text{Solve using the quadratic formula:} \qquad \qquad \qquad \qquad \quad x=\frac{5 \pm \sqrt{185}}{4} \Rightarrow \underline{\underline{x=-2.15}} \ \text{or} \ \underline{\underline{x=4.65}}$

It’s important to plug the answer back into the original equation when the variable appears in any denominator of the equation, because the answer might be an excluded value of one of the rational expressions. If the answer obtained makes any denominator equal to zero, that value is not really a solution to the equation.

Check: $\frac{2x}{x+4} = \frac{5}{x} \Rightarrow \frac{2(-2.15)}{-2.15+4} {\overset{?}=} \frac{5}{-2.15} \Rightarrow \frac{-4.30}{1.85} {\overset{?}=} -2.3 \Rightarrow -2.3=-2.3$. The answer checks out.

$\frac{2x}{x+4} = \frac{5}{x} \Rightarrow \frac{2(4.65)}{4.65+4} {\overset{?}=} \frac{5}{4.65} \Rightarrow \frac{9.3}{8.65} {\overset{?}=} 1.08 \Rightarrow 1.08=1.08$. The answer checks out.

## Solve Rational Equations Using Lowest Common Denominators

Another way of eliminating the denominators in a rational equation is to multiply all the terms in the equation by the lowest common denominator. You can use this method even when there are more than two terms in the equation.

Example 2

Solve $\frac{3}{x+2}-\frac{4}{x-5}=\frac{2}{x^2-3x-10}$.

Solution

$\text{Factor all denominators:} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \frac{3}{x+2}-\frac{4}{x-5}=\frac{2}{(x+2)(x-5)}\!\\\\\text{Find the lowest common denominator:} \qquad \qquad \qquad \qquad \text{LCD} = (x+2)(x-5)\!\\\\\text{Multiply all terms in the equation by the LCD:}$

$(x+2)(x-5) \cdot \frac{3}{x+2}-(x+2)(x-5) \cdot \frac{4}{x-5}=(x+2)(x-5) \cdot \frac{2}{(x+2)(x-5)}$

$\text{The equation simplifies to:} \qquad \qquad \qquad \qquad \qquad \qquad \qquad 3(x-5)-4(x+2)=2\!\\\\\text{Simplify:} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad 3x-15-4x-8=2\!\\\\{\;} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \underline{\underline{x=-25}}$

Check: $\frac{3}{x+2}-\frac{4}{x-5}=\frac{2}{x^2-3x-10} \Rightarrow \frac{3}{-25+2}-\frac{4}{-25-5} {\overset{?}=} \frac{2}{(-25)^2-3(-25)-10} \Rightarrow .003=.003$. The answer checks out.

Example 3

Solve $\frac{2x}{2x+1}+\frac{x}{x+4}=1$.

Solution

$\text{Find the lowest common denominator:} \qquad \qquad \quad \text{LCD} = (2x+1)(x+4)\!\\\\\text{Multiply all terms in the equation by the LCD:}$

$(2x+1)(x+4) \cdot \frac{2x}{2x+1}+(2x+1)(x+4) \cdot \frac{x}{x+4}=(2x+1)(x+4)$

$\text{Cancel all common terms.} \qquad \qquad \qquad \qquad \qquad \ 2x(x+4)+x(2x+1)=(2x+1)(x+4)\!\\\text{The simplified equation is:} \!\\\!\\\text{Eliminate parentheses:} \quad \ \qquad \qquad \qquad \qquad \ \qquad 2x^2+8x+2x^2+x=2x^2+9x+4\!\\\!\\\text{Collect like terms:} \qquad \quad \ \qquad \qquad \qquad \qquad \ \qquad 2x^2=4\!\\\!\\{\;} \qquad \qquad \qquad \quad \qquad \qquad \qquad \qquad \qquad \qquad \ \qquad x^2=2 \Rightarrow \underline{\underline{x=\pm\sqrt{2}}}$

Check: $\frac{2x}{2x+1}+\frac{x}{x+4} = \frac{2\sqrt{2}}{2\sqrt{2}+1}+\frac{\sqrt{2}}{\sqrt{2}+4}=0.739+0.261=1$. The answer checks out.

$\frac{2x}{2x+1}+\frac{x}{x+4} = \frac{2\left(-\sqrt{2}\right)}{2\left(-\sqrt{2}\right)+1}+\frac{-\sqrt{2}}{-\sqrt{2}+4}=1.547-0.547=1$. The answer checks out.

## Solve Real-World Problems Using Rational Equations

A motion problem with no acceleration is described by the formula $distance = speed \times time$. These problems can involve the addition and subtraction of rational expressions.

Example 4

Last weekend Nadia went canoeing on the Snake River. The current of the river is three miles per hour. It took Nadia the same amount of time to travel 12 miles downstream as it did to travel 3 miles upstream. Determine how fast Nadia’s canoe would travel in still water.

Solution

Define variables:

Let $s =$ speed of the canoe in still water

Then, $s + 3 =$ the speed of the canoe traveling downstream

$s - 3 =$ the speed of the canoe traveling upstream

Construct a table:

Direction Distance (miles) Rate Time
Downstream 12 $s + 3$ $t$
Upstream 3 $s - 3$ $t$

Write an equation:

Since $distance = rate \times time$, we can say that $time=\frac{distance}{rate}$.

$\text{The time to go downstream is:} \qquad \qquad \qquad \qquad \qquad \qquad \quad \qquad \qquad \qquad \qquad \qquad t=\frac{12}{s+3}\!\\\\\text{The time to go upstream is:} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \qquad \qquad \qquad \qquad \qquad t=\frac{3}{s-3}\!\\\\\text{Since the time it takes to go upstream and downstream are the same, we have:} \qquad \frac{3}{s-3}=\frac{12}{s+3}$

Solve the equation:

$\text{Cross-multiply:} \qquad \qquad \qquad \qquad 3(s+3)=12(s-3)\!\\\\\text{Simplify:} \qquad \qquad \qquad \qquad \qquad \ \ 3s+9=12s-36\!\\\\\text{Solve:} \qquad \qquad \qquad \qquad \qquad \qquad s=5 \ mi/h$

Check: Upstream: $t=\frac{12}{8}=1 \frac{1}{2} \ hour$; downstream: $t=\frac{3}{2}=1 \frac{1}{2} \ hour$. The answer checks out.

Example 5

Peter rides his bicycle. When he pedals uphill he averages a speed of eight miles per hour, when he pedals downhill he averages 14 miles per hour. If the total distance he travels is 40 miles and the total time he rides is four hours, how long did he ride at each speed?

Solution

Define variables:

Let $t =$ time Peter bikes at 8 miles per hour.

Construct a table:

Direction Distance (miles) Rate (mph) Time (hours)
Uphill $d$ 8 $t_1$
Downhill $40 - d$ 14 $t_2$

Write an equation:

We know that $time=\frac{distance}{rate}$.

$\text{The time to go uphill is:} \qquad \qquad \qquad \qquad \qquad \quad t_1=\frac{d}{8}\!\\\\\text{The time to go downhill is:} \qquad \qquad \qquad \qquad \qquad t_2=\frac{40-d}{14}\!\\\\\text{We also know that the total time is} \ 4 \ \text{hours:} \qquad \ \frac{d}{8}+\frac{40-d}{14}=4$

Solve the equation:

$\text{Find the lowest common denominator:} \qquad \qquad \qquad \qquad \ \text{LCD}=56\!\\\\\text{Multiply all terms by the common denominator:} \qquad \qquad 7d+160-4d=224\!\\\\\text{Solve:} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ \qquad \qquad \qquad d=21.3 \ mi$

Check: Uphill: $t=\frac{21.3}{8}=2.67 \ hours$; downhill: $t=\frac{40-21.3}{14}=1.33 \ hours$. The answer checks out.

Example 6

A group of friends decided to pool together and buy a birthday gift that cost $200. Later 12 of the friends decided not to participate any more. This meant that each person paid$15 more than their original share. How many people were in the group to begin with?

Solution

Define variables:

Let $x =$ the number of friends in the original group.

Make a table:

Number of people Gift price Share amount
Original group $x$ 200 $\frac{200}{x}$
Later group $x - 12$ 200 $\frac{200}{x-12}$

Write an equation:

Since each person’s share went up by $15 after 2 people refused to pay, we write the equation $\frac{200}{x-12}=\frac{200}{x}+15$ Solve the equation: $\text{Find the lowest common denominator:} \qquad \quad \text{LCD} =x(x-12)\!\\\\\text{Multiply all terms by the LCD:} \qquad \qquad \qquad x(x-12) \cdot \frac{200}{x-12}=x(x-12) \cdot \frac{200}{x}+x(x-12) \cdot 15\!\\\\\text{Cancel common factors and simplify:} \qquad \qquad \ 200x=200(x-12)+15x(x-12)\!\\\\\text{Eliminate parentheses:} \qquad \qquad \qquad \qquad \qquad \ 200x=200x-2400+15x^2-180x\!\\\\\text{Get all terms on one side of the equation:} \qquad 0=15x^2=180x-2400\!\\\\\text{Divide all terms by} \ 15: \qquad \qquad \qquad \qquad \quad \ \ 0=x^2-12x-160\!\\\\\text{Factor:} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad 0=(x-20)(x+8)\!\\\\\text{Solve:} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \quad \ x=20, x=-8$ The answer that makes sense is $x = \mathbf{20}$ people. Check: Originally$200 shared among 20 people is $10 each. After 12 people leave,$200 shared among 8 people is $25 each. So each person pays$15 more. The answer checks out.

## Review Questions

Solve the following equations.

1. $\frac{2x+1}{4}=\frac{x-3}{10}$
2. $\frac{4x}{x+2}=\frac{5}{9}$
3. $\frac{5}{3x-4}=\frac{2}{x+1}$
4. $\frac{7}{x+3}=\frac{x+1}{2x-3}$
5. $\frac{7x}{x-5}=\frac{x+3}{x}$
6. $\frac{2}{x+3}-\frac{1}{x+4}=0$
7. $\frac{3x^2+2x-1}{x^2-1}=-2$
8. $x+\frac{1}{x}=2$
9. $-3+\frac{1}{x+1}=\frac{2}{x}$
10. $\frac{1}{x}-\frac{x}{x-2}=2$
11. $\frac{3}{2x-1}+\frac{2}{x+4}=2$
12. $\frac{2x}{x-1}-\frac{x}{3x+4}=3$
13. $\frac{x+1}{x-1}+\frac{x-4}{x+4}=3$
14. $\frac{x}{x-2}+\frac{x}{x+3}=\frac{1}{x^2+x-6}$
15. $\frac{2}{x^2+4x+3}=2+\frac{x-2}{x+3}$
16. $\frac{1}{x+5}-\frac{1}{x-5}=\frac{1-x}{x+5}$
17. $\frac{x}{x^2-36}+\frac{1}{x-6}=\frac{1}{x+6}$
18. $\frac{2x}{3x+3}-\frac{1}{4x+4}=\frac{2}{x+1}$
19. $\frac{-x}{x-2}+\frac{3x-1}{x+4}=\frac{1}{x^2+2x-8}$
20. Juan jogs a certain distance and then walks a certain distance. When he jogs he averages 7 miles/hour and when he walks he averages 3.5 miles per hour. If he walks and jogs a total of 6 miles in a total of 1.2 hours, how far does he jog and how far does he walk?
21. A boat travels 60 miles downstream in the same time as it takes it to travel 40 miles upstream. The boat’s speed in still water is 20 miles per hour. Find the speed of the current.
22. Paul leaves San Diego driving at 50 miles per hour. Two hours later, his mother realizes that he forgot something and drives in the same direction at 70 miles per hour. How long does it take her to catch up to Paul?
23. On a trip, an airplane flies at a steady speed against the wind and on the return trip the airplane flies with the wind. The airplane takes the same amount of time to fly 300 miles against the wind as it takes to fly 420 miles with the wind. The wind is blowing at 30 miles per hour. What is the speed of the airplane when there is no wind?
24. A debt of $420 is shared equally by a group of friends. When five of the friends decide not to pay, the share of the other friends goes up by$25. How many friends were in the group originally?
25. A non-profit organization collected $2250 in equal donations from their members to share the cost of improving a park. If there were thirty more members, then each member could contribute$20 less. How many members does this organization have?

Feb 23, 2012

Dec 23, 2014