Solve n^2-3n-18/n^2-6n/frac{n^2-7n-30{n^2-5n-50}}

$$\frac { \frac { n ^ { 2 } - 3 n - 18 } { n ^ { 2 } - 6 n } } { \frac { n ^ { 2 } - 7 n - 30 } { n ^ { 2 } - 5 n - 50 } }$$

Evaluate

$$\frac{n+5}{n}$$

Short Solution Steps

Divide \frac{n^{2}-3n-18}{n^{2}-6n} by \frac{n^{2}-7n-30}{n^{2}-5n-50} by multiplying \frac{n^{2}-3n-18}{n^{2}-6n} by the reciprocal of \frac{n^{2}-7n-30}{n^{2}-5n-50}.

$$\frac{\left(n^{2}-3n-18\right)\left(n^{2}-5n-50\right)}{\left(n^{2}-6n\right)\left(n^{2}-7n-30\right)}$$

Factor the expressions that are not already factored.

$$\frac{\left(n-10\right)\left(n-6\right)\left(n+3\right)\left(n+5\right)}{n\left(n-10\right)\left(n-6\right)\left(n+3\right)}$$

Cancel out \left(n-10\right)\left(n-6\right)\left(n+3\right) in both numerator and denominator.

$$\frac{n+5}{n}$$

Expand

$$\frac{n+5}{n}$$

Short Solution Steps

Divide \frac{n^{2}-3n-18}{n^{2}-6n} by \frac{n^{2}-7n-30}{n^{2}-5n-50} by multiplying \frac{n^{2}-3n-18}{n^{2}-6n} by the reciprocal of \frac{n^{2}-7n-30}{n^{2}-5n-50}.

$$\frac{\left(n^{2}-3n-18\right)\left(n^{2}-5n-50\right)}{\left(n^{2}-6n\right)\left(n^{2}-7n-30\right)}$$

Factor the expressions that are not already factored.

$$\frac{\left(n-10\right)\left(n-6\right)\left(n+3\right)\left(n+5\right)}{n\left(n-10\right)\left(n-6\right)\left(n+3\right)}$$

Cancel out \left(n-10\right)\left(n-6\right)\left(n+3\right) in both numerator and denominator.

$$\frac{n+5}{n}$$