# 12.7: Addition and Subtraction of Rational Expressions

**Basic**Created by: CK-12

**Practice**Addition and Subtraction of Rational Expressions

Suppose that Jake can photocopy all the pages in a book in 3 hours, while Meredith can photocopy all the pages in the book in 2 hours. If Jake started at the beginning of the book and Meredith started at the end of the book and they worked together, how many hours would it take for them to photocopy all the pages? What equation could you set up to find this value?

### Adding and Subtracting Rational Expressions

Like numerical fractions, rational expressions represent a part of a whole quantity. Remember when adding or subtracting fractions, the denominators must be the same. Once the denominators are identical, the numerators are combined by adding or subtracting like terms.

#### Let's simplify \begin{align*}\frac{4x^2-3}{x+5}+\frac{2x^2-1}{x+5}\end{align*}:

The denominators are identical; therefore we can add the like terms of the numerator to simplify.

\begin{align*}\frac{4x^2-3}{x+5}+\frac{2x^2-1}{x+5}=\frac{6x^2-4}{x+5}\end{align*}

Not all denominators are the same however. In the case of **unlike denominators,** common denominators must be created through multiplication by finding the **least common multiple.**

The **least common multiple (LCM)** is the smallest number that is evenly divisible by every member of the set.

For example, what is the least common multiple of \begin{align*}2, 4x\end{align*}, and \begin{align*}6x^2\end{align*}?

The smallest number 2, 4, and 6 can divide into evenly is six. The largest exponent of \begin{align*}x\end{align*} is 2. Therefore, the LCM of \begin{align*}2, 4x\end{align*}, and \begin{align*}6x^2\end{align*} is \begin{align*}6x^2\end{align*}.

#### Now, let's find the least common multiple of the following two expressions:

\begin{align*}2x^2+8x+8\end{align*} and \begin{align*}x^3-4x^2-12x\end{align*}.

Factor the polynomials completely.

\begin{align*}2x^2+8x+8 &= 2(x^2+4x+4)=2(x+2)^2\\ x^3-4x^2-12x &= x(x^2-4x-12)=x(x-6)(x+2)\end{align*}

The LCM is found by taking each factor to the highest power that it appears in either expression. \begin{align*}LCM=2x(x+2)^2 (x-6)\end{align*}

#### Let's use this approach to subtract the following rational expressions with unlike denominators:

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

The denominators cannot be factored any further, so the LCM is just the product of the separate denominators.

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

The first fraction needs to be multiplied by the factor \begin{align*}(2x-5)\end{align*} and the second fraction needs to be multiplied by the factor \begin{align*}(x+2)\end{align*}.

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

We combine the numerators and simplify.

\begin{align*}\frac{2(2x-5)-3(x+2)}{(x+2)(2x-5)}=\frac{4x-10-3x-6}{(x+2)(2x-5)}\end{align*}

Combine like terms in the numerator.

\begin{align*}\frac{x-16}{(x+2)(2x-5)}\end{align*}

#### Work Problems

These are problems where two objects work together to complete a job. Work problems often contain rational expressions. Typically we set up such problems by looking at the part of the task completed by each person or machine. The completed task is the sum of the parts of the tasks completed by each individual or each machine.

Part of task completed by first person + Part of task completed by second person = One completed task

To determine the part of the task completed by each person or machine, we use the following fact.

Part of the task completed = rate of work \begin{align*}\times\end{align*} time spent on the task

In general, it is very useful to set up a table where we can list all the known and unknown variables for each person or machine and then combine the parts of the task completed by each person or machine at the end.

#### Let's solve the following real-world problem:

Mary can paint a house by herself in 12 hours. John can paint a house by himself in 16 hours. How long would it take them to paint the house if they worked together?

Let \begin{align*}t=\end{align*} the time it takes Mary and John to paint the house together.

Since Mary takes 12 hours to paint the house by herself, in one hour she paints \begin{align*}\frac{1}{12}\end{align*} of the house.

Since John takes 16 hours to pain the house by himself, in one hour he paints \begin{align*}\frac{1}{16}\end{align*} of the house.

Mary and John work together for \begin{align*}t\end{align*} hours to paint the house together. Using the formula:

Part of the task completed = rate of work \begin{align*}\times\end{align*} time spent on the task

we can write that Mary completed \begin{align*}\frac{t}{12}\end{align*} of the house and John completed \begin{align*}\frac{t}{16}\end{align*} of the house in this time and summarize the data in the following table.

Painter |
Rate of work (per hour) |
Time worked |
Part of Task |
---|---|---|---|

Mary |
\begin{align*}\frac{1}{12}\end{align*} | \begin{align*}t\end{align*} | \begin{align*}\frac{t}{12}\end{align*} |

John |
\begin{align*}\frac{1}{16}\end{align*} | \begin{align*}t\end{align*} | \begin{align*}\frac{t}{16}\end{align*} |

Using the formula:

Part of task completed by first person + Part of task completed by second person = One completed task

write an equation to model the situation,

\begin{align*}\frac{t}{12}+\frac{t}{16}=1.\end{align*}

Solve the equation by finding the least common multiple.

\begin{align*}LCM &= 48\\ 48 \cdot \frac{t}{12} + 48 \cdot \frac{t}{16} &= 48 \cdot 1\\ \cancel{48}^{4} \cdot \frac{t}{\cancel{12}}+\cancel{48}^3 \cdot \frac{t}{\cancel{16}} &= 48 \cdot 1\\ 4t+3t &= 48\\ 7t=48 \Rightarrow t &= \frac{48}{7}=6.86 \ hours \ {Answer}\end{align*}

### Examples

#### Example 1

Earlier, you were asked how many hours it would take for Jake and Meredith to photocopy all of the pages in a book if Jake started at the end and Meredith started at the beginning.

We know that Jake can photocopy all the pages in a book in 3 hours, while Meredith can photocopy all the pages in the book in 2 hours. Given this information, you can set up a work problem.

Let \begin{align*}t=\end{align*} the amount of time it takes to copy the book together.

Since it takes Meredith 2 hours to photocopy the book by herself, in one hour she photocopies \begin{align*}\frac{1}{2}\end{align*} of the book.

Since it takes Jake 3 hours to photocopy the book by himself, in one hour he photocopies \begin{align*}\frac{1}{3}\end{align*} of the book.

Photocopier | Rate of work (per hour) | Time worked | Part of task |

Meredith | \begin{align*}\frac{1}{2}\end{align*} | \begin{align*}t\end{align*} | \begin{align*}\frac{t}{2}\end{align*} |

Jake | \begin{align*}\frac{1}{3}\end{align*} | \begin{align*}t\end{align*} | \begin{align*}\frac{t}{3}\end{align*} |

Now, let's use the formula *Part of task complete by first person + Part of task completed by second person = One completed task *to set up an equation and solve for \begin{align*}t\end{align*}.

#### \begin{align*}LCM &= 6\\ 6 \cdot \frac{t}{2} + 6 \cdot \frac{t}{3} &= 6 \cdot 1\\ 3t+2t &= 6\\ 5t=6 \Rightarrow t &= \frac{6}{5}= 1.2 \ hours \ (1 \ hour \ and \ 12 \ minutes) \end{align*}

**Example 2**

One pipe fills a pool 2 times faster than another pipe. Together, they fill the pool in 5 hours. How long does it take for the faster pipe to fill the pool?

Let \begin{align*}f\end{align*} be the time it takes for the faster pipe to fill the pool. Since the pipe is two times faster, the slower pipe takes twice as long. So the time it take the slower pipe to fill the pool is \begin{align*} 2f\end{align*}. After 5 hours, the faster pipe will fill up a certain proportion of the pool. That proportion is found by dividing 5 hours by the time it takes the faster pipe to fill the pool.

Proportion of the pool filled by the faster pipe:

\begin{align*}\frac{5}{f}\end{align*}

Proportion of the pool filled by the slower pipe:

\begin{align*} \frac{5}{2f}\end{align*}

Since together they fill the whole pool in 5 hours, we add the proportions together:

\begin{align*} \text{Using the Work Problem formula.} && \frac{5}{f}+\frac{5}{2f}&=1\\ \text{Multiplying top and bottom of the first fraction by 2, to get common denominators.} && \frac{10}{2f} +\frac{5}{2f}&=1\\ \text{Adding.} && \frac{15}{2f}&=1\\ \text{Solving for the variable.} && 15=2f \Rightarrow 7.5&=f \end{align*}

It would take the faster pipe 7.5 hours to fill the whole pool by itself.

### Review

Perform the indicated operation and simplify. Leave the denominator in factored form.

- \begin{align*}\frac{5}{24}-\frac{7}{24}\end{align*}
- \begin{align*}\frac{10}{21}+\frac{9}{35}\end{align*}
- \begin{align*}\frac{5}{2x+3}+\frac{3}{2x+3}\end{align*}
- \begin{align*}\frac{3x-1}{x+9}-\frac{4x+3}{x+9}\end{align*}
- \begin{align*}\frac{4x+7}{2x^2}-\frac{3x-4}{2x^2}\end{align*}
- \begin{align*}\frac{x^2}{x+5}-\frac{25}{x+5}\end{align*}
- \begin{align*}\frac{2x}{x-4}+\frac{x}{4-x}\end{align*}
- \begin{align*}\frac{10}{3x-1}-\frac{7}{1-3x}\end{align*}
- \begin{align*}\frac{5}{2x+3}-3\end{align*}
- \begin{align*}\frac{5x+1}{x+4}+2\end{align*}
- \begin{align*}\frac{1}{x}+\frac{2}{3x}\end{align*}
- \begin{align*}\frac{4}{5x^2}-\frac{2}{7x^3}\end{align*}
- \begin{align*}\frac{4x}{x+1}-\frac{2}{2(x+1)}\end{align*}
- \begin{align*}\frac{10}{x+5}+\frac{2}{x+2}\end{align*}
- \begin{align*}\frac{2x}{x-3}-\frac{3x}{x+4}\end{align*}
- \begin{align*}\frac{4x-3}{2x+1}+\frac{x+2}{x-9}\end{align*}
- \begin{align*}\frac{x^2}{x+4}-\frac{3x^2}{4x-1}\end{align*}
- \begin{align*}\frac{2}{5x+2}-\frac{x+1}{x^2}\end{align*}
- \begin{align*}\frac{x+4}{2x}+\frac{2}{9x}\end{align*}
- \begin{align*}\frac{5x+3}{x^2+x}+\frac{2x+1}{x}\end{align*}
- \begin{align*}\frac{4}{(x+1)(x-1)}-\frac{5}{(x+1)(x+2)}\end{align*}
- \begin{align*}\frac{2x}{(x+2)(3x-4)}+\frac{7x}{(3x-4)^2}\end{align*}
- \begin{align*}\frac{3x+5}{x(x-1)}-\frac{9x-1}{(x-1)^2}\end{align*}
- \begin{align*}\frac{1}{(x-2)(x-3)}+\frac{4}{(2x+5)(x-6)}\end{align*}
- \begin{align*}\frac{3x-2}{x-2}+\frac{1}{x^2-4x+4}\end{align*}
- \begin{align*}\frac{-x^2}{x^2-7x+6}+x-4\end{align*}
- \begin{align*}\frac{2x}{x^2+10x+25}-\frac{3x}{2x^2+7x-15}\end{align*}
- \begin{align*}\frac{1}{x^2-9}+\frac{2}{x^2+5x+6}\end{align*}
- \begin{align*}\frac{-x+4}{2x^2-x-15}+\frac{x}{4x^2+8x-5}\end{align*}
- \begin{align*}\frac{4}{9x^2-49}-\frac{1}{3x^2+5x-28}\end{align*}
- One number is 5 less than another. The sum of their reciprocals is \begin{align*}\frac{13}{36}\end{align*}. Find the two numbers.
- One number is 8 times more than another. The difference in their reciprocals is \begin{align*}\frac{21}{20}\end{align*}. Find the two numbers.
- A pipe can fill a tank full of oil in 4 hours and another pipe can empty the tank in 8 hours. If the valves to both pipes are open, how long would it take to fill the tank?
- Stefan and Misha are washing cars. Stefan could wash the cars by himself in 6 hours and Misha could wash the cars by himself in 5 hours. Stefan starts washing the cars by himself, but he needs to go to his football game after 2.5 hours. Misha continues the task. How long did it take Misha to finish washing the cars?
- Amanda and her sister Chyna are shoveling snow to clear their driveway. Amanda can clear the snow by herself in three hours and Chyna can clear the snow by herself in four hours. After Amanda has been working by herself for one hour, Chyna joins her and they finish the job together. How long does it take to clear the snow from the driveway?
- At a soda bottling plant, one bottling machine can fulfill the daily quota in ten hours and a second machine can fill the daily quota in 14 hours. The two machines started working together but after four hours the slower machine broke and the faster machine had to complete the job by itself. How many hours does the fast machine work by itself?

**Mixed Review**

- Explain the difference between these two situations. Write an equation to model each situation. Assume the town started with 10,000 people. When will statement b become larger than statement a?
- For the past seven years, the population grew by 500 people every year.
- For the past seven years, the population grew by 5% every year.

- Simplify. Your answer should have only positive exponents. \begin{align*}\frac{16x^2 y^7}{-2x^8 y} \cdot \frac{1}{2} x^{-10}\end{align*}
- Solve for \begin{align*}j: -12=j^2-8j\end{align*}. Which method did you use? Why did you choose this method?
- The distance you travel varies directly as the speed at which you drive. If you can drive 245 miles in five hours, how long will it take you to drive 90 miles?
- Two cities are 3.78 centimeters apart on an atlas. The atlas has a scale of \begin{align*}\frac{1}{2} cm=14 \ miles\end{align*}. What is the true distance between these cities?

### Review (Answers)

To see the Review answers, open this PDF file and look for section 12.7.

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work problems

A**is one where two or more people are involved in working to complete a task. The basic formula for two people is: Part of task completed by first person + Part of task completed by second person = One completed task**

*work problem*Least Common Denominator

The least common denominator or lowest common denominator of two fractions is the smallest number that is a multiple of both of the original denominators.### Image Attributions

Here you'll learn how to determine the sum and the difference of two rational expressions and how to find LCMs.

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