Pre-Algebra Examples

$\frac{4 x}{x^{2} - 49} \cdot \frac{x + 7}{16 x}$

Step 1

Simplify the denominator.

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Step 1.1

Rewrite $49$ as $7^{2}$ .

$\frac{4 x}{x^{2} - 7^{2}} \cdot \frac{x + 7}{16 x}$

Step 1.2

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 7$ .

$\frac{4 x}{(x + 7) (x - 7)} \cdot \frac{x + 7}{16 x}$

Step 2

Simplify terms.

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Step 2.1

Cancel the common factor of $4 x$ .

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Step 2.1.1

Factor $4 x$ out of $16 x$ .

$\frac{4 x}{(x + 7) (x - 7)} \cdot \frac{x + 7}{4 x (4)}$

Step 2.1.2

Cancel the common factor.

$\frac{4 x}{(x + 7) (x - 7)} \cdot \frac{x + 7}{4 x \cdot 4}$

Step 2.1.3

Rewrite the expression.

$\frac{1}{(x + 7) (x - 7)} \cdot \frac{x + 7}{4}$

Step 2.2

Cancel the common factor of $x + 7$ .

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Step 2.2.1

Cancel the common factor.

$\frac{1}{(x + 7) (x - 7)} \cdot \frac{x + 7}{4}$

Step 2.2.2

Rewrite the expression.

$\frac{1}{x - 7} \cdot \frac{1}{4}$

Step 2.3

Multiply $\frac{1}{x - 7}$ by $\frac{1}{4}$ .

$\frac{1}{(x - 7) \cdot 4}$

Step 2.4

Move $4$ to the left of $x - 7$ .

$\frac{1}{4 (x - 7)}$