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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 \begin{align*} \frac{3} {5} x + \frac{1} {2} = 4\end{align*}.

An example of the second kind of equation is \begin{align*} \frac{x} {x - 1} + 1 = \frac{4} {2x + 3}\end{align*}.

The first aim in solving a rational equation is to eliminate all denominators. In this 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 two terms is easily solved by the method of cross products or cross multiplication. Consider the following equation.

\begin{align*} \frac{x} {5} = \frac{x + 1} {2}\end{align*}

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

\begin{align*}\cancel{5} \cdot \frac{x}{\cancel{5}}=\frac{x+1}{2} \cdot 5\end{align*}

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

\begin{align*}2 \cdot x=\frac{5(x+1)}{\cancel{2}}\cancel{2}\end{align*}

The equation simplifies to \begin{align*}2x = 5 (x + 1)\end{align*}.

\begin{align*} 2x = 5x + \Rightarrow x = - \frac{5} {3} \ \text{Answer}\end{align*}

Notice that when we remove the denominators from the rational expressions we end up multiplying the numerator on one side of the equal sign with the denominator of the opposite fraction.

Once again, we obtain the simplified equation: \begin{align*}2x = 5(x + 1)\end{align*}, whose solution is \begin{align*} x = - \frac{5} {3}\end{align*}

We check the answer by plugging the answer back into the original equation.

Check

On the left-hand side, if \begin{align*}x= - \frac{5}{3}\end{align*}, then we have

\begin{align*} \frac{x} {5} = \frac{\frac{-5}{3}} {5} = - \frac{1}{3}\end{align*}

On the right hand side, we have

\begin{align*}\frac{x + 1} {2} = \frac{\frac{-5}{3 +1}}{2} = \frac{\frac{-2}{3}}{2} = -\frac{1}{3}\end{align*}

Since the two expressions are equal, the answer checks out.

Example 1

Solve the equation \begin{align*} \frac{2} {x - 2} = \frac{3} {x + 3}\end{align*}.

Solution

Use cross-multiplication to eliminate the denominators of both fractions.

The equation simplifies to

\begin{align*}2(x + 3) = 3(x - 2)\end{align*}

Simplify.

\begin{align*}2x + 6 & = 3x - 6 \\ x & = 12 \end{align*}

Check.

\begin{align*} \frac{2} {x - 2} & = \frac{2} {12 - 2} = \frac{2}{10} = \frac{1}{5} \\ \frac{3} {x + 3} & = \frac{3}{12+3} = \frac{3}{15} = \frac{1}{5}\end{align*}

Example 2

Solve the equation \begin{align*} \frac{23} {x + 4} = \frac{5} {x}\end{align*}.

Solution

Cross-multiply.

The equation simplifies to

\begin{align*} 2x^2 = 5(x + 4)\end{align*}

Simplify.

\begin{align*} 2x^2 = 5x + 20\end{align*}

Move all terms to one side of the equation.

\begin{align*} 2x^2 - 5x - 20 = 0\end{align*}

Notice that this equation has a degree of two, that is, it is a quadratic equation. We can solve it using the quadratic formula.

\begin{align*} x = \frac{5 \pm \sqrt{185}} {4} \Rightarrow x \approx - 2.15\end{align*} or \begin{align*} x \approx 4.65\end{align*}

It is important to check the answer in the original equation when the variable appears in any denominator of the equation because the answer might be an excluded value of any of the rational expression. If the answer obtained makes any denominator equal to zero, that value is not a solution to the equation.

Check:

First we check \begin{align*}x=-2.15\end{align*} by substituting it in the original equations. On the left hand side we get the following.

\begin{align*} \frac{2x} {x + 4} = \frac{2(-2.15)} {-2.15 + 4} \frac{-4.30} {1.85} = -2.3\end{align*}

Now, check on the right hand side.

\begin{align*}\frac{5} {x} =\frac{5} {-2.15}=-2.3\end{align*}

Thus, 2.15 checks out.

For \begin{align*}x =4.65\end{align*} we repeat the procedure.

\begin{align*}\frac{2x} {x + 4} & = \frac{2(4.65)} {4.65 + 4}=1.08. \\ \frac{5} {x} & =\frac{5} {4.65} =1.08.\end{align*}

4.65 also checks out.

## Solve Rational Equations Using the Lowest Common Denominators

An alternate way of eliminating the denominators in a rational equation is to multiply all terms in the equation by the lowest common denominator. This method is suitable even when there are more than two terms in the equation.

Example 3

Solve \begin{align*} \frac{3x} {35} = \frac{x^2} {5} - \frac{1} {21}\end{align*}.

Solution

Find the lowest common denominator:

LCM = 105

Multiply each term by the LCD.

\begin{align*} 105 \cdot \frac{3x} {35} = 105 \cdot \frac{x^2} {5} - 105 \cdot \frac{1} {21}\end{align*}

Cancel common factors.

\begin{align*}\cancel{105}^{\color{red}3}.\frac{3x}{\cancel{35}}=\cancel{105}^{\color{red}21}. \frac{{x}{^2{^{\ {\color{red}5}}}}}{\cancel{5}} -\cancel{105}. \frac{1}{\cancel{21}}\end{align*}

The equation simplifies to

\begin{align*}9x = 21x^{2} - 5\end{align*}

Move all terms to one side of the equation.

\begin{align*} 21x^2 - 9x - 5 = 0\end{align*}

\begin{align*} x = \frac{9 \pm \sqrt{501}} {42}\end{align*}

\begin{align*}x \approx -0.32\end{align*} or \begin{align*}x \approx 0.75 \ \text{Answer}\end{align*}

Check

We use the substitution \begin{align*}x=-0.32\end{align*}.

\begin{align*} \frac{3x} {35} & = \frac{3(-0.32)} {35} = -0.27 \\ \frac{x^2} {5} - \frac{1} {24} & = \frac{(-0.32)^2} {5} - \frac{1} {21} = -.027. \ \text{The answer checks out}.\end{align*}

Now we check the solution \begin{align*}x=0.75\end{align*}.

\begin{align*} \frac{3x} {35} & = \frac{3(0.75)} {35} = 0.64 \\ \frac{x^2} {5} - \frac{1} {21} \frac{(0.75)^2} {5} - \frac{1} {21} & = .064. \ \text{The answer checks out}.\end{align*}

Example 4

Solve \begin{align*} \frac{3} {x + 2} - \frac{4} {x = 5} = \frac{2} {x^2 - 3x - 10}\end{align*}.

Solution

Factor all denominators.

\begin{align*} \frac{3} {x + 2} - \frac{4} {x - 5} - \frac{2} {(x + 2)(x - 5)}\end{align*}

Find the lowest common denominator.

\begin{align*}\text{LCM} = (x + 2)(x - 5)\end{align*}

Multiply all terms in the equation by the LCM.

\begin{align*} (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)}\end{align*}

Cancel the common terms.

\begin{align*}\cancel{(x+2)}(x-5) \cdot \frac{3}{\cancel{x+2}}-(x+2)\cancel{(x-5)} \cdot \frac{4}{\cancel{x-5}}=\cancel{(x+2)}\cancel{(x-5)} \cdot \frac{2}{\cancel{(x+2)}\cancel{(x-5)}}\end{align*}

The equation simplifies to

\begin{align*}3(x - 5) - 4(x + 2) = 2\end{align*}

Simplify.

\begin{align*}3x - 15 - 4x - 8 & = 2 \\ x & = -25 \ \text{Answer}\end{align*}

Check.

\begin{align*} \frac{3} {x + 2} - \frac{4} {x - 5} & = \frac{3} {-25 + 2} - \frac{4} {-25-5} = 0.003 \\ \frac{2} {x^2 - 3x - 10} & = \frac{2} {(-25)^2 - 3 (-25) - 10} = 0.003 \end{align*}

Example 5

Solve \begin{align*} \frac{2x} {2x + 1} + \frac{x} {x + 4} = 1\end{align*}.

Solution

Find the lowest common denominator.

\begin{align*}\text{LCM} = (2x + 1) (x + 4) \end{align*}

Multiply all terms in the equation by the LCM.

\begin{align*} (2x + 1) (x + 4) \cdot \frac{2x} {2x + 1} + (2x + 1) (x + 4) \cdot \frac{x} {x + 4} = (2x + 1) (x + 4)\end{align*}

Cancel all common terms.

\begin{align*}\cancel{(2x+1)}(x+4) \cdot \frac{2x}{\cancel{2x+1}}+(2x+1)\cancel{(x+4)} \cdot \frac{x}{\cancel{x+4}}=(2x+1)(x+4)\end{align*}

The simplified equation is

\begin{align*}2x(x + 4) + x (2x + 1) = (2x + 1) (x + 4) \end{align*}

Eliminate parentheses.

\begin{align*}2x^{2} + 8x + 2x^{2} + x = 2x^{2} + 9x + 4 \end{align*}

Collect like terms.

\begin{align*}2x^{2} & = 4 \\ x^2 & = 2 \Rightarrow x = \pm \sqrt{2} \ \text{Answer}\end{align*}

Check.

\begin{align*} \frac{2x} {2x + 1} + \frac{x} {x + 4} & = \frac{2\sqrt{2}} {2\sqrt{2} + 1} + \frac{\sqrt{2}} {\sqrt{2} + 4} \approx 0.739 + 0.261 = 1. \ \text{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} \approx 1.547 + 0.547 = 1. \ \text{The answer checks out}.\end{align*}

## Solve Real-World Problems Using Rational Equations

Motion Problems

A motion problem with no acceleration is described by the formula \begin{align*}\text{distance} = \text{speed} \times \text{time}\end{align*}.

These problems can involve the addition and subtraction of rational expressions.

Example 6

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 three miles upstream. Determine the speed at which Nadia’s canoe would travel in still water.

Solution

1. Define variables

Let \begin{align*}s =\end{align*} speed of the canoe in still water

Then, \begin{align*}s + 3 \ =\end{align*} the speed of the canoe traveling downstream

\begin{align*}s-3 \ =\end{align*} the speed of the canoe traveling upstream

2. Construct a table.

We make a table that displays the information we have in a clear manner:

Direction Distance (miles) Rate Time
Downstream \begin{align*}12\end{align*} \begin{align*}s + 3\end{align*} \begin{align*}t\end{align*}
Upstream \begin{align*}3\end{align*} \begin{align*}s - 3\end{align*} \begin{align*}t\end{align*}

3. Write an equation.

Since \begin{align*}\text{distance} = \text{rate} \times \text{time}\end{align*}, we can say that: \begin{align*}\text{time} = \frac{\text{distance}} {\text{rate}}\end{align*}.

The time to go downstream is

\begin{align*} t = \frac{12} {s + 3}\end{align*}

The time to go upstream is

\begin{align*} t = \frac{3} {s - 3}\end{align*}

Since the time it takes to go upstream and downstream are the same then: \begin{align*} \frac{3} {s - 3} = \frac{12} {s + 3}\end{align*}

4. Solve the equation

Cross-multiply.

\begin{align*}3(s + 3) = 12 (s - 3) \end{align*}

Simplify.

\begin{align*}3s + 9 = 12s - 36\end{align*}

Solve.

\begin{align*}s = 5 \ mi/hr \ \text{Answer}\end{align*}

5. Check

Upstream: \begin{align*} t = \frac{12} {8} = 1 \frac{1} {2} \ hour\end{align*}; Downstream: \begin{align*} t = \frac{3} {2} = 1\frac{1} {2} \ hour\end{align*}. The answer checks out.

Example 8

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

1. Define variables.

Let \begin{align*}t_1 \ =\end{align*} time Peter bikes uphill, \begin{align*}t_2 \ =\end{align*} time Peter bikes downhill, and \begin{align*}d \ =\end{align*} distance he rides uphill.

2. Construct a table

We make a table that displays the information we have in a clear manner:

Direction Distance (miles) Rate (mph) Time (hours)
Uphill \begin{align*}d\end{align*} 8 \begin{align*}t_1\end{align*}
Downhill \begin{align*}40 - d\end{align*} 14 \begin{align*}t_2\end{align*}

3. Write an equation

We know that

\begin{align*} \text{time} = \frac{\text{distance}} {\text{rate}}\end{align*}

The time to go uphill is

\begin{align*} t_{1} = \frac{d} {8}\end{align*}

The time to go downhill is

\begin{align*} t_{2} = \frac{40 - d} {14}\end{align*}

We also know that the total time is 4 hours.

\begin{align*} \frac{d} {8} + \frac{40 - d} {14} = 4\end{align*}

4. Solve the equation.

Find the lowest common denominator:

LCM = 56

Multiply all terms by the common denominator:

\begin{align*}56\cdot \frac{d}{8} + 56 \cdot \frac{40-d}{14} & = 4 \cdot 56 \\ 7d + 160 - 4d & = 224 \\ 3d & = 64\end{align*}

Solve.

\begin{align*}d = 21.3 \ miles \ \text{Answer}\end{align*}

5. Check.

Uphill: \begin{align*} t = \frac{21.3} {8} = 2.67 \ hours\end{align*}; Downhill: \begin{align*} t = \frac{40 -21.3} {14} = 1.33 \ hours\end{align*}. The answer checks out.

Shares

Example 8

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 the original share. How many people were in the group to start?

Solution

1. Define variables.

Let \begin{align*}x \ =\end{align*} the number of friends in the original group

2. Make a table.

We make a table that displays the information we have in a clear manner:

Number of People Gift Price Share Amount
Original group \begin{align*}x\end{align*} 200 \begin{align*} \frac{200} {x}\end{align*}
Later group \begin{align*}x - 12\end{align*} 200 \begin{align*} \frac{200} {x}\end{align*}

3. Write an equation.

Since each person’s share went up by 15 after 2 people refused to pay, we write the equation: \begin{align*} \frac{200} {x - 12} = \frac{200} {x} + 15\end{align*} 4. Solve the equation. Find the lowest common denominator. \begin{align*}\text{LCM} = x(x - 12)\end{align*} Multiply all terms by the LCM. \begin{align*} x(x - 12) \cdot \frac{200} {x - 12} = x(x - 12) \cdot \frac{200} {x} + x (x - 12) \cdot 15\end{align*} Cancel common factors in each term: \begin{align*}x \cancel{(x-12)} \cdot \frac{200}{\cancel{x-12}}=\cancel{x}(x-12) \cdot \frac{200}{\cancel{x}}+x(x-12)\cdot 15\end{align*} Simplify. \begin{align*} 200x = 200(x - 12) + 15x (x - 12)\end{align*} Eliminate parentheses. \begin{align*} 200x = 200x - 2400 + 15x^2 - 180x\end{align*} Collect all terms on one side of the equation. \begin{align*} 0 = 15x^2 - 180x - 2400\end{align*} Divide all terms by 15. \begin{align*} 0 = x^2 - 12x - 160\end{align*} Factor. \begin{align*} 0 = (x - 20) (x + 8)\end{align*} Solve. \begin{align*} x = 20, x = -8\end{align*} The answer is \begin{align*}x = 20\end{align*} people. We discard the negative solution since it does not make sense in the context of this problem. 5. Check. Originally200 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.

## Review Questions

Solve the following equations.

1. \begin{align*} \frac{2x + 1} {4} = \frac{x - 3} {10}\end{align*}
2. \begin{align*} \frac{4x} {x + 2} = \frac{5} {9}\end{align*}
3. \begin{align*} \frac{5} {3x - 4} = \frac{2} {x + 1}\end{align*}
4. \begin{align*} \frac{7x} {x - 5} = \frac{x + 3} {x}\end{align*}
5. \begin{align*} \frac{2} {x + 3} - \frac{1} {x + 4} = 0\end{align*}
6. \begin{align*} \frac{3x^2 + 2x - 1} {x^2 - 1} = -2\end{align*}
7. \begin{align*} x + \frac{1} {x} = 2\end{align*}
8. \begin{align*} -3 + \frac{1} {x + 1} = \frac{2} {x}\end{align*}
9. \begin{align*} \frac{1} {x} - \frac{x} {x - 2} = 2\end{align*}
10. \begin{align*} \frac{3} {2x - 1} + \frac{2} {x + 4} = 2\end{align*}
11. \begin{align*} \frac{2x} {x - 1} - \frac{x} {3x + 4} = 3\end{align*}
12. \begin{align*} \frac{x + 1} {x - 1} + \frac{x - 4} {x + 4} = 3\end{align*}
13. \begin{align*} \frac{x} {x - 2} + \frac{x} {x + 3} = \frac{1} {x^2 + x - 6}\end{align*}
14. \begin{align*} \frac{2} {x^2 + 4x + 3} = 2 + \frac{x - 2} {x + 3}\end{align*}
15. \begin{align*} \frac{1} {x + 5} - \frac{1} {x - 5} = \frac{1 - x} {x + 5}\end{align*}
16. \begin{align*} \frac{x} {x^2 - 36} + \frac{1} {x -6} = \frac{1} {x + 6}\end{align*}
17. \begin{align*} \frac{2x} {3x+3} - \frac{1} {4x + 4} = \frac{2} {x + 1}\end{align*}
18. \begin{align*} \frac{-x} {x - 2} + \frac{3x - 1} {x + 4} = \frac{1} {x^2 + 2x - 8}\end{align*}
19. Juan jogs a certain distance and then walks a certain distance. When he jogs he averages 7 miles/hour. When he walks, he averages 3.5 miles/hour. If he walks and jogs a total of 6 miles in a total of 7 hours, how far does he jog and how far does he walk?
20. 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/hour. Find the speed of the current.
21. Paul leaves San Diego driving at 50 miles/hour. Two hours later, his mother realizes that he forgot something and drives in the same direction at 70 miles/hour. How long does it take her to catch up to Paul?
22. On a trip, an airplane flies at a steady speed against the wind. 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/hour. What is the speed of the airplane when there is no wind?
23. 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?
24. 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?

1. \begin{align*} x = - \frac{11} {8}\end{align*}
2. \begin{align*} x = \frac{10} {31}\end{align*}
3. \begin{align*}x = 13\end{align*}
4. no real solution
5. \begin{align*}x = -5\end{align*}
6. \begin{align*}x = \frac{3}{5}\end{align*}
7. \begin{align*}x = 1\end{align*}
8. \begin{align*} x = \frac{1} {3}\end{align*}
9. \begin{align*}x = 1, x = \frac{2}{3}\end{align*}
10. \begin{align*}x = -3.17, x = 1.42\end{align*}
11. \begin{align*}x = -1.14, x = 2.64\end{align*}
12. \begin{align*}x = -10.84, x = 1.84\end{align*}
13. \begin{align*} x = -1, x = \frac{1} {2}\end{align*}
14. \begin{align*} x = -2, x = - \frac{1} {3}\end{align*}
15. \begin{align*}x = -.74, x = 6.74\end{align*}
16. \begin{align*}x = -12\end{align*}
17. \begin{align*} x = \frac{27} {8}\end{align*}
18. \begin{align*}x = .092, x = 5.41\end{align*}
19. jogs 3.6 miles and walks 2.4 miles
20. 4 miles/hour
21. 5 hours
22. 180 miles/hour
23. 12 friends
24. 45 members

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