Simplify ((x+3)/(x^2-9))÷((x+2)/(x^2-6x+9))

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

Step 1

To divide by a fraction, multiply by its reciprocal.

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

Step 2

Simplify the denominator.

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

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

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 $b = 3$ .

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

Step 3

Cancel the common factor of $x + 3$ .

Cancel the common factor.

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

Rewrite the expression.

$\frac{1}{x - 3} \cdot \frac{x^{2} - 6 x + 9}{x + 2}$

Step 4

Factor using the perfect square rule.

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

$\frac{1}{x - 3} \cdot \frac{x^{2} - 6 x + 3^{2}}{x + 2}$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$6 x = 2 \cdot x \cdot 3$

Rewrite the polynomial.

$\frac{1}{x - 3} \cdot \frac{x^{2} - 2 \cdot x \cdot 3 + 3^{2}}{x + 2}$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 3$ .

$\frac{1}{x - 3} \cdot \frac{{(x - 3)}^{2}}{x + 2}$

Step 5

Cancel the common factor of $x - 3$ .

Factor $x - 3$ out of ${(x - 3)}^{2}$ .

$\frac{1}{x - 3} \cdot \frac{(x - 3) (x - 3)}{x + 2}$

Cancel the common factor.

$\frac{1}{x - 3} \cdot \frac{(x - 3) (x - 3)}{x + 2}$

Rewrite the expression.

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