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# 12.7: Solution of Rational Equations

Created by: CK-12

You are now ready to solve rational equations! There are two main methods you will learn in this lesson to solve rational equations:

• Cross products
• Lowest common denominators

## Solving a Rational Proportion

When two rational expressions are equal, a proportion is created and can be solved using its cross products. For example, to solve $\frac{x}{5}=\frac{(x+1)}{2}$, cross multiply and the products are equal.

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

Solve for $x$:

$2(x) &= 5(x+1) \rightarrow 2x=5x+5\\2x-5x &= 5x-5x+5\\-3x &= 5\\x &= -\frac{5}{3}$

Example 1: Solve $\frac{2x}{x+4}=\frac{5}{x}$.

Solution:

$\frac{2x}{x+4} &= \frac{5}{x} \rightarrow 2x^2=5(x+4)\\2x^2 &= 5(x+4) \rightarrow 2x^2=5x+20\\&2x^2-5x-20 = 0$

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

$x=\frac{5 \pm \sqrt{185}}{4} \Rightarrow x \approx -2.15 \ \text{or} \ x \approx 4.65$

## Solving by Clearing Denominators

When a rational equation has several terms, it may not be possible to use the method of cross products. A second method to solve rational equations is to clear the fractions by multiplying the entire equation by the least common multiple of the denominators.

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

Solution: Factor all denominators and find the least common multiple.

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

Multiply all terms in the equation by the LCM and cancel the common terms.

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

Now solve and simplify.

$3(x-5)-4(x+2) &= 2\\3x-15-4x-8 &= 2\\x &= -25 \ \text{Answer}$

$\frac{3}{x+2}-\frac{4}{x-5} &= \frac{3}{-25+2}-\frac{4}{-25-5}=0.003\\\frac{2}{x^2-3x-10} &= \frac{2}{(-25)^2-3(-25)-10}=0.003$

Example: A group of friends decided to pool their money 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: Let $x=$ the number of friends in the original group

Number of People Gift Price Share Amount
Original group $x$ 200 $\frac{200}{x}$
Later group $x-12$ 200 $\frac{200}{x-12}$

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 by clearing the fractions. Don’t forget to check! $LCM &= x(x-12)\\x(x-12) \cdot \frac{200}{x-12} &= x(x-12) \cdot \frac{200}{x} + x(x-12) \cdot 15\\x\cancel{(x-12)} \cdot \frac{200}{\cancel{x-12}} &= \cancel{x}(x-12) \cdot \frac{200}{\cancel{x}}+x(x-12) \cdot 15\\200x &= 200(x-12)+15x(x-12)\\200x &= 200x-2400+15x^2-180x\\0 &= 15x^2-180x-2400\\x &= 20, x=-8$ The answer is 20 people. We discard the negative solution since it does not make sense in the context of this problem. ## Practice Set Sample explanations for some of the practice exercises below are available by viewing the following videos. Note that there is not always a match between the number of the practice exercise in the videos and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both. 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. $\frac{7x}{x-5}=\frac{x+3}{x}$ 5. $\frac{2}{x+3}-\frac{1}{x+4}=0$ 6. $\frac{3x^2+2x-1}{x^2-1}=-2$ 7. $x+ \frac{1}{x}=2$ 8. $-3 + \frac{1}{x+1}=\frac{2}{x}$ 9. $\frac{1}{x}-\frac{x}{x-2}=2$ 10. $\frac{3}{2x-1}+\frac{2}{x+4}=2$ 11. $\frac{2x}{x-1}-\frac{x}{3x+4}=3$ 12. $\frac{x+1}{x-1}+\frac{x-4}{x+4}=3$ 13. $\frac{x}{x-2}+\frac{x}{x+3}=\frac{1}{x^2+x-6}$ 14. $\frac{2}{x^2+4x+3}=2+ \frac{x-2}{x+3}$ 15. $\frac{1}{x+5}-\frac{1}{x-5}=\frac{1-x}{x+5}$ 16. $\frac{x}{x^2-36}+\frac{1}{x-6}=\frac{1}{x+6}$ 17. $\frac{2x}{3x+3}-\frac{1}{4x+4}=\frac{2}{x+1}$ 18. $\frac{-x}{x-2}+\frac{3x-1}{x+4}=\frac{1}{x^2+2x-8}$ 19. Juan jogs a certain distance and then walks a certain distance. When he jogs he averages seven miles per hour. When he walks, he averages 3.5 miles per hour. If he walks and jogs a total of six miles in a total of seven 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$2,250 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?

Mixed Review

1. Divide $-2 \frac{9}{10} \div - \frac{15}{8}$.
2. Solve for $g: -1.5 \left(-3 \frac{4}{5}+g \right)=\frac{201}{20}$.
3. Find the discriminant of $6x^2+3x+4=0$ and determine the nature of the roots.
4. Simplify $\frac{6b}{2b+2}+3$.
5. Simplify $\frac{8}{2x-4}- \frac{5x}{x-5}$.
6. Divide $(7x^2+16x-10) \div (x+3)$.
7. Simplify $(n-1)*(3n+2)(n-4)$.

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Feb 22, 2012

Dec 11, 2014