Simplify (b+3)/(b^2-9)+1/(b-3)

$\frac{b + 3}{b^{2} - 9} + \frac{1}{b - 3}$

Step 1

Simplify each term.

Simplify the denominator.

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

$\frac{b + 3}{b^{2} - 3^{2}} + \frac{1}{b - 3}$

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

$\frac{b + 3}{(b + 3) (b - 3)} + \frac{1}{b - 3}$

Cancel the common factor of $b + 3$ .

Cancel the common factor.

$\frac{b + 3}{(b + 3) (b - 3)} + \frac{1}{b - 3}$

Rewrite the expression.

$\frac{1}{b - 3} + \frac{1}{b - 3}$

Step 2

Combine fractions.

Combine the numerators over the common denominator.

$\frac{1 + 1}{b - 3}$

Add $1$ and $1$ .

$\frac{2}{b - 3}$