Simplify (6t-18)/(t^2-9)

$\frac{6 t - 18}{t^{2} - 9}$

Step 1

Factor $6$ out of $6 t - 18$ .

Factor $6$ out of $6 t$ .

$\frac{6 (t) - 18}{t^{2} - 9}$

Factor $6$ out of $- 18$ .

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

Factor $6$ out of $6 t + 6 \cdot - 3$ .

$\frac{6 (t - 3)}{t^{2} - 9}$

Step 2

Simplify the denominator.

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

$\frac{6 (t - 3)}{t^{2} - 3^{2}}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = t$ and $b = 3$ .

$\frac{6 (t - 3)}{(t + 3) (t - 3)}$

Step 3

Cancel the common factor of $t - 3$ .

Cancel the common factor.

$\frac{6 (t - 3)}{(t + 3) (t - 3)}$

Rewrite the expression.

$\frac{6}{t + 3}$