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Encyclopedia>Algebra>Basic Algebra>Fractions and Factorization>How to Multiply Fractions with Variables

When multiplying fractions, I am a fan of factorizing and canceling common factors instead of multiplying and then simplifying.

The reason for this is that it makes the calculations much easier, and I like it when things are easy. By cross-canceling when possible instead of expanding the parentheses, the calculations become simpler and less messy.

Here you will learn to multiply fractions with variables by factorizing and cross-canceling. Beware that you can only cancel factors that are on both sides of the fraction bar.

Rule

Multiplication Through Canceling

1.: Factorize the numerators.
2.: Factorize the denominators.
3.: Cancel common factors.
4.: Find the answer.

Example 1

Evaluate $\frac{x + 1}{x + 2} \cdot \frac{2 x + 4}{3 x + 3}$

You factorize and cross-cancel in this way:

\begin{array}{l} \frac{x + 1}{x + 2} \cdot \frac{2 x + 4}{3 x + 3} & = \frac{(x + 1)}{(x + 2)} \cdot \frac{2 (x + 2)}{3 (x + 1)} \\ = \frac{(x+1)}{(x+2)} \cdot \frac{2 (x+2)}{3 (x+1)} \\ = \frac{2}{3} \end{array}

Example 2

Evaluate $\frac{2 x + 6}{x + 3} \cdot \frac{x^{2} - 9}{x + 3}$

You factorize using the third algebraic identity of quadratic expressions, and then you cancel and multiply like this:

\begin{matrix} \frac{2 x + 6}{x + 3} \cdot \frac{x^{2} - 9}{x + 3} \\ = \\ \frac{2 (x + 3)}{(x + 3)} \cdot \frac{(x + 3) (x - 3)}{(x + 3)} \\ = \\ \frac{2 (x+3)}{(x+3)} \cdot \frac{(x+3) (x - 3)}{(x+3)} \\ = \\ 2 (x - 3) \\ = \\ 2 x - 6 \end{matrix}

\begin{array}{l} \frac{2 x + 6}{x + 3} \cdot \frac{x^{2} - 9}{x + 3} & = \frac{2 (x + 3)}{(x + 3)} \cdot \frac{(x + 3) (x - 3)}{(x + 3)} \\ = \frac{2 (x+3)}{(x+3)} \cdot \frac{(x+3) (x - 3)}{(x+3)} \\ = 2 (x - 3) \\ = 2 x - 6 \end{array}

Example 3

Evaluate $\frac{x^{2} - 1}{x + 5} \cdot \frac{x^{2} - 25}{x - 1}$

Use the third algebraic identity of quadratic expressions to factorize, and then cross-cancel and multiply in this manner:

\begin{matrix} \frac{x^{2} - 1}{x + 5} \cdot \frac{x^{2} - 25}{x - 1} \\ = \\ \frac{(x - 1) (x + 1)}{(x + 5)} \cdot \frac{(x + 5) (x - 5)}{(x - 1)} \\ = \\ \frac{(x-1) (x + 1)}{(x+5)} \cdot \frac{(x+5) (x - 5)}{(x-1)} \\ = \\ (x + 1) (x - 5) \\ = \\ x^{2} - 5 x + x - 5 \\ = \\ x^{2} - 4 x - 5 \end{matrix}

\begin{array}{l} \frac{x^{2} - 1}{x + 5} \cdot \frac{x^{2} - 25}{x - 1} & = \frac{(x - 1) (x + 1)}{(x + 5)} \cdot \frac{(x + 5) (x - 5)}{(x - 1)} \\ = \frac{(x-1) (x + 1)}{(x+5)} \cdot \frac{(x+5) (x - 5)}{(x-1)} \\ = (x + 1) (x - 5) \\ = x^{2} - 5 x + x - 5 \\ = x^{2} - 4 x - 5 \end{array}

Example 4

Evaluate $\frac{2 x^{2} - 32}{x - 4} \cdot \frac{x + 5}{2 (x^{2} - 25)}$

You factorize with the help of the third algebraic identity of quadratic expressions, and then you cancel and multiply the fractions in this way:

\begin{array}{l} \frac{2 x^{2} - 32}{x - 4} \cdot \frac{x + 5}{2 (x^{2} - 25)} \\ = \frac{2 (x^{2} - 16)}{(x - 4)} \cdot \frac{(x + 5)}{2 (x + 5) (x - 5)} \\ = \frac{2 (x + 4) (x-4)}{(x-4)} \cdot \frac{(x+5)}{2 (x+5) (x - 5)} \\ = (x + 4) \cdot \frac{1}{(x - 5)} \\ = \frac{x + 4}{x - 5} \end{array}

\begin{array}{l} \frac{2 x^{2} - 32}{x - 4} \cdot \frac{x + 5}{2 (x^{2} - 25)} & = \frac{2 (x^{2} - 16)}{(x - 4)} \cdot \frac{(x + 5)}{2 (x + 5) (x - 5)} \\ = \frac{2 (x + 4) (x-4)}{(x-4)} \cdot \frac{(x+5)}{2 (x+5) (x - 5)} \\ = (x + 4) \cdot \frac{1}{(x - 5)} \\ = \frac{x + 4}{x - 5} \end{array}

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How to Simplify Fractions with Variables

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