Simplify ((x+3)/(x^2-9))÷((x^2)/(x^2-9x+18))

$\frac{x + 3}{x^{2} - 9} \div \frac{x^{2}}{x^{2} - 9 x + 18}$

To divide by a fraction, multiply by its reciprocal.

$\frac{x + 3}{x^{2} - 9} \cdot \frac{x^{2} - 9 x + 18}{x^{2}}$

Simplify the denominator.

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

$\frac{x + 3}{x^{2} - 3^{2}} \cdot \frac{x^{2} - 9 x + 18}{x^{2}}$

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 $b = 3$ .

$\frac{x + 3}{(x + 3) (x - 3)} \cdot \frac{x^{2} - 9 x + 18}{x^{2}}$

Cancel the common factor of $x + 3$ .

Cancel the common factor.

$\frac{x + 3}{(x + 3) (x - 3)} \cdot \frac{x^{2} - 9 x + 18}{x^{2}}$

Rewrite the expression.

$\frac{1}{x - 3} \cdot \frac{x^{2} - 9 x + 18}{x^{2}}$

Factor $x^{2} - 9 x + 18$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $18$ and whose sum is $- 9$ .

$- 6, - 3$

Write the factored form using these integers.

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

Cancel the common factor of $x - 3$ .

Factor $x - 3$ out of $(x - 6) (x - 3)$ .

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

Cancel the common factor.

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

Rewrite the expression.

$\frac{x - 6}{x^{2}}$

Simplify ((x+3)/(x^2-9))÷((x^2)/(x^2-9x+18))

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