Solve 3-9/n/n-frac{9{n}}

$$\frac { 3 - \frac { 9 } { n } } { n - \frac { 9 } { n } }$$

Evaluate

$$\frac{3}{n+3}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply 3 times \frac{n}{n}.

$$\frac{\frac{3n}{n}-\frac{9}{n}}{n-\frac{9}{n}}$$

Since \frac{3n}{n} and \frac{9}{n} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{3n-9}{n}}{n-\frac{9}{n}}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply n times \frac{n}{n}.

$$\frac{\frac{3n-9}{n}}{\frac{nn}{n}-\frac{9}{n}}$$

Since \frac{nn}{n} and \frac{9}{n} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{3n-9}{n}}{\frac{nn-9}{n}}$$

Do the multiplications in nn-9.

$$\frac{\frac{3n-9}{n}}{\frac{n^{2}-9}{n}}$$

Divide \frac{3n-9}{n} by \frac{n^{2}-9}{n} by multiplying \frac{3n-9}{n} by the reciprocal of \frac{n^{2}-9}{n}.

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

Cancel out n in both numerator and denominator.

$$\frac{3n-9}{n^{2}-9}$$

Factor the expressions that are not already factored.

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

Cancel out n-3 in both numerator and denominator.

$$\frac{3}{n+3}$$

Expand

$$\frac{3}{n+3}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply 3 times \frac{n}{n}.

$$\frac{\frac{3n}{n}-\frac{9}{n}}{n-\frac{9}{n}}$$

Since \frac{3n}{n} and \frac{9}{n} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{3n-9}{n}}{n-\frac{9}{n}}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply n times \frac{n}{n}.

$$\frac{\frac{3n-9}{n}}{\frac{nn}{n}-\frac{9}{n}}$$

Since \frac{nn}{n} and \frac{9}{n} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{3n-9}{n}}{\frac{nn-9}{n}}$$

Do the multiplications in nn-9.

$$\frac{\frac{3n-9}{n}}{\frac{n^{2}-9}{n}}$$

Divide \frac{3n-9}{n} by \frac{n^{2}-9}{n} by multiplying \frac{3n-9}{n} by the reciprocal of \frac{n^{2}-9}{n}.

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

Cancel out n in both numerator and denominator.

$$\frac{3n-9}{n^{2}-9}$$

Factor the expressions that are not already factored.

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

Cancel out n-3 in both numerator and denominator.

$$\frac{3}{n+3}$$