12.9: Clearing Denominators in Rational Equations
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 \begin{align*} \frac {1}{5}\end{align*}
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.
Let's solve the following equations by clearing the fractions:
- \begin{align*}5x-\frac{1}{x}=4\end{align*}
5x−1x=4
Start by clearing the denominators. In this case, there is only one denominator, \begin{align*}x\end{align*}
\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*}
- \begin{align*}\frac{3}{x+2}-\frac{4}{x-5}=\frac{2}{x^2-3x-10}\end{align*}
Factor all denominators and find the least common multiple.
\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.
\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*}
Check your answer.
\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*}
Now, let's solve the following real-world problem:
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?
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*} |
Since each person’s share went up by $15 after 12 people refused to pay, we write the equation:
\begin{align*}\frac{200}{x-12}=\frac{200}{x}+15\end{align*}
Solve by clearing the fractions. Don’t forget to check!
\begin{align*}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\end{align*}
The answer is 20 people. We discard the negative solution since it does not make sense in the context of this problem.
Examples
Example 1
Earlier, you were asked about distributing 24 quarts of juice equally among members of your class. Remember that 10 members of your class were absent, so the amount of juice distributed to each person increased by \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?
Let \begin{align*}x=\end{align*} the number of students in your class.
Class size | Total Amount of Juice | Amount of Juice Per Person | |
Total class size | x | 24 | 24/x |
Class size day of | x-10 | 24 | 25/x-10 |
Since each person's share went up by \begin{align*} \frac {1}{5}\end{align*} of a quart, we write the equation:
\begin{align*}\frac{24}{x-10}=\frac{24}{x}+\frac{1}{5}\end{align*}
Solve by clearing fractions.
\begin{align*}LCM &= x(x-10)\\ x(x-10) \cdot \frac{24}{x-10} &= x(x-10) \cdot \frac{24}{x} + x(x-10) \cdot \frac{1}{5}\\ x\cancel{(x-10)} \cdot \frac{24}{\cancel{x-10}} &= \cancel{x}(x-10) \cdot \frac{24}{\cancel{x}}+x(x-10) \cdot \frac{1}{5}\\ 24x &= 24(x-10)+\frac{1}{5}x(x-10)\\ 24x &= 24x-240+\frac{1}{5}x^2-2x\\ x &= 40, x=-30\end{align*}
We discard the answer of -30 because a negative number does not make sense in the context of this problem. Your class consists of 40 people.
Solve \begin{align*}\frac{1}{x-1} - \frac{1}{x+1} = \frac{1}{x}\end{align*}.
Since each denominator cannot be a multiple of another denominator, or in other words, the denominators do not share common factors, the \begin{align*}LCM\end{align*} is the product of the 3 denominators:
\begin{align*}LCM=(x-1)(x+1)(x).\end{align*}
\begin{align*} && \frac{1}{x-1} - \frac{1}{x+1} &= \frac{1}{x}\\ \text{Multiply by the }LCM. && (x-1)(x+1)(x)\cdot \frac{1}{x-1} - (x-1)(x+1)(x)\cdot \frac{1}{x+1} &= (x-1)(x+1)(x)\cdot\frac{1}{x}\\ \text{Simplify by cancelling.} && \cancel{(x-1)}(x+1)(x)\cdot \frac{1}{\cancel{x-1}} - (x-1)\cancel{(x+1)}(x)\cdot \frac{1}{\cancel{x+1}} &= (x-1)(x+1)\cancel{(x)}\cdot\frac{1}{\cancel{x}}\\ && (x+1)(x) - (x-1)(x) &= (x-1)(x+1)\\ \text{Simplify by distributing.} && x^2+x- x^2+x&=x^2-1\\ \text{Simplify by combining like terms.} && 2x&=x^2-1\\ \text{Arrange so that one side is equal to zero.} && 0&=x^2-2x-1 \end{align*}
This quadratic equation cannot be factored, so use the quadratic formula to get:
\begin{align*}x=1\pm \sqrt{2}.\end{align*}
Review
Solve the following equations.
- \begin{align*}x+ \frac{1}{x}=2\end{align*}
- \begin{align*}-3 + \frac{1}{x+1}=\frac{2}{x}\end{align*}
- \begin{align*}\frac{1}{x}-\frac{x}{x-2}=2\end{align*}
- \begin{align*}\frac{3}{2x-1}+\frac{2}{x+4}=2\end{align*}
- \begin{align*}\frac{2x}{x-1}-\frac{x}{3x+4}=3\end{align*}
- \begin{align*}\frac{x+1}{x-1}+\frac{x-4}{x+4}=3\end{align*}
- \begin{align*}\frac{x}{x-2}+\frac{x}{x+3}=\frac{1}{x^2+x-6}\end{align*}
- \begin{align*}\frac{2}{x^2+4x+3}=2+ \frac{x-2}{x+3}\end{align*}
- \begin{align*}\frac{1}{x+5}-\frac{1}{x-5}=\frac{1-x}{x+5}\end{align*}
- \begin{align*}\frac{x}{x^2-36}+\frac{1}{x-6}=\frac{1}{x+6}\end{align*}
- \begin{align*}\frac{2x}{3x+3}-\frac{1}{4x+4}=\frac{2}{x+1}\end{align*}
- \begin{align*}\frac{-x}{x-2}+\frac{3x-1}{x+4}=\frac{1}{x^2+2x-8}\end{align*}
- 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?
- 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.
- 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?
- 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?
- 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?
- 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?
Review (Answers)
To see the Review answers, open this PDF file and look for section 12.9.
Notes/Highlights Having trouble? Report an issue.
Color | Highlighted Text | Notes | |
---|---|---|---|
Please Sign In to create your own Highlights / Notes | |||
Show More |
Image Attributions
Here you'll learn how to clear denominators in order to solve rational equations.