Solve 35n-7/5n^2-n-15n-3/5n^2-n=

$$\frac { 35 n - 7 } { 5 n ^ { 2 } - n } - \frac { 15 n - 3 } { 5 n ^ { 2 } - n } =$$

Evaluate

$$\frac{4}{n}$$

Short Solution Steps

Factor the expressions that are not already factored in \frac{35n-7}{5n^{2}-n}.

$$\frac{7\left(5n-1\right)}{n\left(5n-1\right)}-\frac{15n-3}{5n^{2}-n}$$

Cancel out 5n-1 in both numerator and denominator.

$$\frac{7}{n}-\frac{15n-3}{5n^{2}-n}$$

Factor the expressions that are not already factored in \frac{15n-3}{5n^{2}-n}.

$$\frac{7}{n}-\frac{3\left(5n-1\right)}{n\left(5n-1\right)}$$

Cancel out 5n-1 in both numerator and denominator.

$$\frac{7}{n}-\frac{3}{n}$$

Since \frac{7}{n} and \frac{3}{n} have the same denominator, subtract them by subtracting their numerators. Subtract 3 from 7 to get 4.

$$\frac{4}{n}$$

Expand

$$\frac{4}{n}$$

Short Solution Steps

Factor the expressions that are not already factored in \frac{35n-7}{5n^{2}-n}.

$$\frac{7\left(5n-1\right)}{n\left(5n-1\right)}-\frac{15n-3}{5n^{2}-n}$$

Cancel out 5n-1 in both numerator and denominator.

$$\frac{7}{n}-\frac{15n-3}{5n^{2}-n}$$

Factor the expressions that are not already factored in \frac{15n-3}{5n^{2}-n}.

$$\frac{7}{n}-\frac{3\left(5n-1\right)}{n\left(5n-1\right)}$$

Cancel out 5n-1 in both numerator and denominator.

$$\frac{7}{n}-\frac{3}{n}$$

Since \frac{7}{n} and \frac{3}{n} have the same denominator, subtract them by subtracting their numerators. Subtract 3 from 7 to get 4.

$$\frac{4}{n}$$