Solve =(1/a-3+1/a+3)div1/a^2-9

$$= ( \frac { 1 } { a - 3 } + \frac { 1 } { a + 3 } ) \div \frac { 1 } { a ^ { 2 } - 9 }$$

Evaluate

$$2a$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of a-3 and a+3 is \left(a-3\right)\left(a+3\right). Multiply \frac{1}{a-3} times \frac{a+3}{a+3}. Multiply \frac{1}{a+3} times \frac{a-3}{a-3}.

$$\frac{\frac{a+3}{\left(a-3\right)\left(a+3\right)}+\frac{a-3}{\left(a-3\right)\left(a+3\right)}}{\frac{1}{a^{2}-9}}$$

Since \frac{a+3}{\left(a-3\right)\left(a+3\right)} and \frac{a-3}{\left(a-3\right)\left(a+3\right)} have the same denominator, add them by adding their numerators.

$$\frac{\frac{a+3+a-3}{\left(a-3\right)\left(a+3\right)}}{\frac{1}{a^{2}-9}}$$

Combine like terms in a+3+a-3.

$$\frac{\frac{2a}{\left(a-3\right)\left(a+3\right)}}{\frac{1}{a^{2}-9}}$$

Divide \frac{2a}{\left(a-3\right)\left(a+3\right)} by \frac{1}{a^{2}-9} by multiplying \frac{2a}{\left(a-3\right)\left(a+3\right)} by the reciprocal of \frac{1}{a^{2}-9}.

$$\frac{2a\left(a^{2}-9\right)}{\left(a-3\right)\left(a+3\right)}$$

Factor the expressions that are not already factored.

$$\frac{2a\left(a-3\right)\left(a+3\right)}{\left(a-3\right)\left(a+3\right)}$$

Cancel out \left(a-3\right)\left(a+3\right) in both numerator and denominator.

$$2a$$

Expand

$$2a$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of a-3 and a+3 is \left(a-3\right)\left(a+3\right). Multiply \frac{1}{a-3} times \frac{a+3}{a+3}. Multiply \frac{1}{a+3} times \frac{a-3}{a-3}.

$$\frac{\frac{a+3}{\left(a-3\right)\left(a+3\right)}+\frac{a-3}{\left(a-3\right)\left(a+3\right)}}{\frac{1}{a^{2}-9}}$$

Since \frac{a+3}{\left(a-3\right)\left(a+3\right)} and \frac{a-3}{\left(a-3\right)\left(a+3\right)} have the same denominator, add them by adding their numerators.

$$\frac{\frac{a+3+a-3}{\left(a-3\right)\left(a+3\right)}}{\frac{1}{a^{2}-9}}$$

Combine like terms in a+3+a-3.

$$\frac{\frac{2a}{\left(a-3\right)\left(a+3\right)}}{\frac{1}{a^{2}-9}}$$

Divide \frac{2a}{\left(a-3\right)\left(a+3\right)} by \frac{1}{a^{2}-9} by multiplying \frac{2a}{\left(a-3\right)\left(a+3\right)} by the reciprocal of \frac{1}{a^{2}-9}.

$$\frac{2a\left(a^{2}-9\right)}{\left(a-3\right)\left(a+3\right)}$$

Factor the expressions that are not already factored.

$$\frac{2a\left(a-3\right)\left(a+3\right)}{\left(a-3\right)\left(a+3\right)}$$

Cancel out \left(a-3\right)\left(a+3\right) in both numerator and denominator.

$$2a$$