Solve 6n/n^2-9-3/n+3=

$$\frac { 6 n } { n ^ { 2 } - 9 } - \frac { 3 } { n + 3 } =$$

Evaluate

$$\frac{3}{n-3}$$

Short Solution Steps

Factor n^{2}-9.

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

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

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

Since \frac{6n}{\left(n-3\right)\left(n+3\right)} and \frac{3\left(n-3\right)}{\left(n-3\right)\left(n+3\right)} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in 6n-3\left(n-3\right).

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

Combine like terms in 6n-3n+9.

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

Factor the expressions that are not already factored in \frac{3n+9}{\left(n-3\right)\left(n+3\right)}.

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

Cancel out n+3 in both numerator and denominator.

$$\frac{3}{n-3}$$

Factor

$$\frac{3}{n-3}$$