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# Addition and Subtraction of Rational Expressions

## Add and subtract fractions with variables in the denominator

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Practice Addition and Subtraction of Rational Expressions
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Adding and Subtracting Rational Expressions with Like Denominators

In triangle ABC, side AB is x2+5x2+3x+2\begin{align*}\frac{x^2+5}{x^2 + 3x + 2}\end{align*} units long. Side AC is 3x23xx2+3x+2\begin{align*}\frac{3x^2 - 3x}{x^2 + 3x + 2}\end{align*} units long. How much longer is side AC than side AB?

### Guidance

Recall, that when you add or subtract fractions, the denominators must be the same. The same is true of adding and subtracting rational expressions. The denominators must be the same expression and then you can add or subtract the numerators.

#### Example A

Add xx6+7x6\begin{align*}\frac{x}{x-6} + \frac{7}{x-6}\end{align*}.

Solution: In this concept, the denominators will always be the same. Therefore, all you will need to do is add the numerators and simplify if needed.

xx6+7x6=x+7x6\begin{align*}\frac{x}{x-6} + \frac{7}{x-6} = \frac{x+7}{x-6}\end{align*}

#### Example B

Subtract x24x32x1x3\begin{align*}\frac{x^2-4}{x-3} - \frac{2x-1}{x-3}\end{align*}.

Solution: You need to be a little more careful with subtraction. The entire expression in the second numerator is being subtracted. Think of the minus sign like distributing -1 to that numerator.

x24x32x1x3=x24(2x+1)x3=x242x1x3=x22x3x3

At this point, factor the numerator if possible.

x22x3x3=(x3)(x+1)x3=x+1\begin{align*}\frac{x^2-2x-3}{x-3} = \frac{\cancel{(x-3)}(x+1)}{\cancel{x-3}} = x+1\end{align*}

#### Example C

Add x+72x2+14x+20+x+12x2+14x+20\begin{align*}\frac{x+7}{2x^2+14x+20} + \frac{x+1}{2x^2+14x+20}\end{align*}.

Solution: Add the numerators and simplify the denominator.

x+72x2+14x+20+x+12x2+14x+20=2x+82x2+14x+20=2(x+4)2(x+5)(x+2)=(x+4)(x+5)(x+2)

Intro Problem Revisit We need to subtract the length of side AB from the length of side AC.

3x23xx2+3x+2x2+5x2+3x+23x23x(x2+5)x2+3x+23x23xx25x2+3x+22x23x5x2+3x+2

Now we need to factor the numerator and the denominator.

2x23x5x2+3x+2=(2x5)(x+1)(x+2)(x+1)\begin{align*}\frac{2x^2 - 3x - 5}{x^2 + 3x + 2} = \frac{(2x-5)(x+1)}{(x+2)(x+1)}\end{align*}

The (x+1)\begin{align*}(x + 1)\end{align*} in the numerator and the denominator cancel out and we are left with 2x5x+2\begin{align*}\frac{2x-5}{x+2}\end{align*}. Therefore, side AC is 2x5x+2\begin{align*}\frac{2x-5}{x+2}\end{align*} units longer than side AB.

### Guided Practice

Add or subtract the following rational expressions.

1. 3x29x+7x29\begin{align*}\frac{3}{x^2-9} - \frac{x+7}{x^2-9}\end{align*}

2. 5x62x+3+x122x+3\begin{align*}\frac{5x-6}{2x+3} + \frac{x-12}{2x+3}\end{align*}

3. x2+24x24x3x22x+14x24x3\begin{align*}\frac{x^2+2}{4x^2-4x-3} - \frac{x^2-2x+1}{4x^2-4x-3}\end{align*}

1. 3x29x+7x29=3(x+7)x29=3x7x29=x4x29\begin{align*}\frac{3}{x^2-9} - \frac{x+7}{x^2-9} = \frac{3-(x+7)}{x^2-9} = \frac{3-x-7}{x^2-9} = \frac{-x-4}{x^2-9}\end{align*}

We did not bother to factor the denominator because we know that the factors of -9 are 3 and -3 and will not cancel with x4\begin{align*}-x-4\end{align*}.

2. 5x62x+3+x122x+3=6x182x+3=6(x3)2x+3\begin{align*}\frac{5x-6}{2x+3} + \frac{x-12}{2x+3} = \frac{6x-18}{2x+3} = \frac{6(x-3)}{2x+3}\end{align*}

3.

x2+24x24x3x22x+14x24x3=x2+2(x22x+1)4x24x3=x2+2x2+2x14x24x3=2x+14x24x3

At this point, we will factor the denominator to see if any factors cancel with the numerator.

2x+14x24x3=2x+1(2x+1)(2x3)=12x3\begin{align*}\frac{2x+1}{4x^2-4x-3} = \frac{\cancel{2x+1}}{\cancel{(2x+1)}(2x-3)} = \frac{1}{2x-3}\end{align*}

### Explore More

2. Explain why \begin{align*}\frac{2}{3} + \frac{4}{5} \neq \frac{3}{4}\end{align*}.

Add or subtract the following rational expressions.

1. \begin{align*}\frac{2}{x} + \frac{5}{x}\end{align*}
2. \begin{align*}\frac{5}{2x} + \frac{7}{2x}\end{align*}
3. \begin{align*}\frac{6}{5x} + \frac{3-2x}{5x}\end{align*}
4. \begin{align*}\frac{3}{x} + \frac{x+1}{x}\end{align*}
5. \begin{align*}\frac{5}{x+1} + \frac{x-4}{x+1}\end{align*}
6. \begin{align*}\frac{x+15}{x-2} - \frac{10}{x-2}\end{align*}
7. \begin{align*}\frac{4x-3}{x+3} + \frac{15}{x+3}\end{align*}
8. \begin{align*}\frac{3x+8}{x^2-4x-5} + \frac{2x+3}{x^2-4x-5}\end{align*}
9. \begin{align*}\frac{5x+3}{x^2-4} - \frac{2x+9}{x^2-4}\end{align*}
10. \begin{align*}\frac{3x^2+x}{x^3-8} + \frac{4}{x^3-8} - \frac{2x^2-x}{x^3-8}\end{align*}
11. \begin{align*}\frac{4x+3}{x^2+1} - \frac{x+2}{x^2+1} + \frac{1-x}{x^2+1}\end{align*}
12. \begin{align*}\frac{18x^2-7x+2}{8x^3+4x^2-18x-9} - \frac{3x^2+13x-4}{8x^3+4x^2-18x-9} + \frac{5x^2-13}{8x^3+4x^2-18x-9}\end{align*}
13. \begin{align*}\frac{2x^2+3x}{x^3+2x^2-16x-32} + \frac{5x^2-13}{x^3+2x^2-16x-32} - \frac{4x^2+9x+11}{x^3+2x^2-16x-32}\end{align*}

### Answers for Explore More Problems

To view the Explore More answers, open this PDF file and look for section 9.10.

### Vocabulary Language: English

Least Common Denominator

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.