# 12.9: Clearing Denominators in Rational Equations

Difficulty Level: At Grade Created by: CK-12
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Practice Clearing Denominators in Rational Equations

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Suppose that you had 24 quarts of juice to equally distribute to each member of your class. However, 10 members of your class were absent, so the amount of juice distributed to each person increased by 15\begin{align*} \frac {1}{5}\end{align*} of a quart. How many people are in your class? What equation could you set up to find the answer to this question? In this Concept, you'll learn how to solve rational equations by clearing denominators so that you can solve problems like this one.

### Guidance

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 A

Solve 5x1x=4\begin{align*}5x-\frac{1}{x}=4\end{align*}.

Solution:

Start by clearing the denominators. In this case, there is only one denominator, x\begin{align*}x\end{align*}. This means that x\begin{align*}x\end{align*} is the LCM of all the denominators, since there is only one denominator. Simply multiply the whole equation by that denominator, . Then solve for x\begin{align*}x\end{align*} using methods learned previously.

Clear the denominator.This is quadratic, so make one side equal to zero.Factor.Use the Zero Product Property to solve for the variable.5x1x5x215x24x1(5x+1)(x1)x=15 or x=4=4x=0=0=1\begin{align*} && 5x-\frac{1}{x}&=4\\ \text{Clear the denominator.} && 5x^2-1&=4x\\ \text{This is quadratic, so make one side equal to zero.} && 5x^2-4x-1&=0\\ \text{Factor.} && (5x+1)(x-1)&=0\\ \text{Use the Zero Product Property to solve for the variable.} && x=-\frac{1}{5} \text{ or } x&=1 \end{align*}

#### Example B

Solve 3x+24x5=2x23x10\begin{align*}\frac{3}{x+2}-\frac{4}{x-5}=\frac{2}{x^2-3x-10}\end{align*}.

Solution:

Factor all denominators and find the least common multiple.

3x+2LCM4x52(x+2)(x5)=(x+2)(x5)\begin{align*}\frac{3}{x+2}- & \frac{4}{x-5}-\frac{2}{(x+2)(x-5)}\\ LCM &= (x+2)(x-5)\end{align*}

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

(x+2)(x5)3x+2(x+2)(x5)4x5(x+2)(x5)3x+2(x+2)(x5)4x5=(x+2)(x5)2(x+2)(x5)=(x+2)(x5)2(x+2)(x5)\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)}\\ \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)}}\end{align*}

Now solve and simplify.

\begin{align*}3(x-5)-4(x+2) &= 2\\ 3x-15-4x-8 &= 2\\ x &= -25 \ \text{Answer}\end{align*}

\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 C

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

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-12}\end{align*}

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