Simplify (x/(x^2))÷((x^2+6x-27)/(x^2-9))

$\frac{x}{x^{2}} \div \frac{x^{2} + 6 x - 27}{x^{2} - 9}$

Step 1

To divide by a fraction, multiply by its reciprocal.

$\frac{x}{x^{2}} \cdot \frac{x^{2} - 9}{x^{2} + 6 x - 27}$

Step 2

Cancel the common factor of $x$ and $x^{2}$ .

Raise $x$ to the power of $1$ .

$\frac{x^{1}}{x^{2}} \cdot \frac{x^{2} - 9}{x^{2} + 6 x - 27}$

Factor $x$ out of $x^{1}$ .

$\frac{x \cdot 1}{x^{2}} \cdot \frac{x^{2} - 9}{x^{2} + 6 x - 27}$

Cancel the common factors.

Factor $x$ out of $x^{2}$ .

$\frac{x \cdot 1}{x \cdot x} \cdot \frac{x^{2} - 9}{x^{2} + 6 x - 27}$

Cancel the common factor.

$\frac{x \cdot 1}{x \cdot x} \cdot \frac{x^{2} - 9}{x^{2} + 6 x - 27}$

Rewrite the expression.

$\frac{1}{x} \cdot \frac{x^{2} - 9}{x^{2} + 6 x - 27}$

Step 3

Simplify the numerator.

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

$\frac{1}{x} \cdot \frac{x^{2} - 3^{2}}{x^{2} + 6 x - 27}$

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 $b = 3$ .

$\frac{1}{x} \cdot \frac{(x + 3) (x - 3)}{x^{2} + 6 x - 27}$

Step 4

Factor $x^{2} + 6 x - 27$ 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 $- 27$ and whose sum is $6$ .

$- 3, 9$

Write the factored form using these integers.

$\frac{1}{x} \cdot \frac{(x + 3) (x - 3)}{(x - 3) (x + 9)}$

Step 5

Simplify terms.

Cancel the common factor of $x - 3$ .

Cancel the common factor.

$\frac{1}{x} \cdot \frac{(x + 3) (x - 3)}{(x - 3) (x + 9)}$

Rewrite the expression.

$\frac{1}{x} \cdot \frac{x + 3}{x + 9}$

Multiply $\frac{1}{x}$ by $\frac{x + 3}{x + 9}$ .

$\frac{x + 3}{x (x + 9)}$

Simplify (x/(x^2))÷((x^2+6x-27)/(x^2-9))

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