Simplify 4/(x+3)-(3x)/(x^2-9)

$\frac{4}{x + 3} - \frac{3 x}{x^{2} - 9}$

Step 1

Simplify the denominator.

Rewrite $9$ as $3^{2}$ .

$\frac{4}{x + 3} - \frac{3 x}{x^{2} - 3^{2}}$

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 $b = 3$ .

$\frac{4}{x + 3} - \frac{3 x}{(x + 3) (x - 3)}$

Step 2

To write $\frac{4}{x + 3}$ as a fraction with a common denominator, multiply by $\frac{x - 3}{x - 3}$ .

$\frac{4}{x + 3} \cdot \frac{x - 3}{x - 3} - \frac{3 x}{(x + 3) (x - 3)}$

Step 3

Simplify terms.

Multiply $\frac{4}{x + 3}$ by $\frac{x - 3}{x - 3}$ .

$\frac{4 (x - 3)}{(x + 3) (x - 3)} - \frac{3 x}{(x + 3) (x - 3)}$

Combine the numerators over the common denominator.

$\frac{4 (x - 3) - 3 x}{(x + 3) (x - 3)}$

Step 4

Simplify the numerator.

Apply the distributive property.

$\frac{4 x + 4 \cdot - 3 - 3 x}{(x + 3) (x - 3)}$

Multiply $4$ by $- 3$ .

$\frac{4 x - 12 - 3 x}{(x + 3) (x - 3)}$

Subtract $3 x$ from $4 x$ .

$\frac{x - 12}{(x + 3) (x - 3)}$