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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?

### Adding and Subtracting Rational Expressions

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.

Let's add or subtract the following rational expressions.

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

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*}

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

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\begin{align*}\frac{x^2-4}{x-3} - \frac{2x-1}{x-3} &= \frac{x^2-4-(2x+1)}{x-3} \\ &= \frac{x^2-4-2x-1}{x-3} \\ &= \frac{x^2-2x-3}{x-3}\end{align*}

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*}

1. 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*}.

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)\begin{align*}\frac{x+7}{2x^2+14x+20} + \frac{x+1}{2x^2+14x+20} &= \frac{2x+8}{2x^2+14x+20} \\ &= \frac{\cancel{2}(x+4)}{\cancel{2}(x+5)(x+2)} \\ &= \frac{(x+4)}{(x+5)(x+2)}\end{align*}

### Examples

#### Example 1

Earlier, you were asked to find how much longer side AC is compared to side AB.

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

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.

#### Example 2

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

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*}.

#### Example 3

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

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*}

#### Example 4

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

x2+24x24x3x22x+14x24x3=x2+2(x22x+1)4x24x3=x2+2x2+2x14x24x3=2x+14x24x3\begin{align*}\frac{x^2+2}{4x^2-4x-3} - \frac{x^2-2x+1}{4x^2-4x-3} &= \frac{x^2+2-(x^2-2x+1)}{4x^2-4x-3} \\ &= \frac{x^2+2-x^2+2x-1}{4x^2-4x-3} \\ &= \frac{2x+1}{4x^2-4x-3}\end{align*}

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

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

### Review

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*}

To see the Review answers, open this PDF file and look for section 9.10.

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### Vocabulary Language: English

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.

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