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Addition and Subtraction of Rational Expressions

Add and subtract fractions with variables in the denominator

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Adding and Subtracting Rational Expressions with Like Denominators

In triangle ABC, side AB is \begin{align*}\frac{x^2+5}{x^2 + 3x + 2}\end{align*} units long. Side AC is \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 \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.

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

  1. Subtract \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.

\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.

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

  1. Add \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.

\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.

\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.

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

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

Example 2

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

\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 \begin{align*}-x-4\end{align*}.

Example 3

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

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

\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

  1. Explain how you add fractions. Assume your audience knows nothing about math.
  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 Review Problems

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

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Vocabulary

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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