Solve 3a/a^2-9-3/3-a=

$$\frac { 3 a } { a ^ { 2 } - 9 } - \frac { 3 } { 3 - a } =$$

Evaluate

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

Short Solution Steps

Factor a^{2}-9.

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

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

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

Since \frac{3a}{\left(a-3\right)\left(a+3\right)} and \frac{3\left(-1\right)\left(a+3\right)}{\left(a-3\right)\left(a+3\right)} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in 3a-3\left(-1\right)\left(a+3\right).

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

Combine like terms in 3a+3a+9.

$$\frac{6a+9}{\left(a-3\right)\left(a+3\right)}$$

Expand \left(a-3\right)\left(a+3\right).

$$\frac{6a+9}{a^{2}-9}$$

Factor

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