Solve x-2-8/x/x-3-frac{4{x}}

$$\frac { x - 2 - \frac { 8 } { x } } { x - 3 - \frac { 4 } { x } }$$

Evaluate

$$\frac{x+2}{x+1}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply x-2 times \frac{x}{x}.

$$\frac{\frac{\left(x-2\right)x}{x}-\frac{8}{x}}{x-3-\frac{4}{x}}$$

Since \frac{\left(x-2\right)x}{x} and \frac{8}{x} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{\left(x-2\right)x-8}{x}}{x-3-\frac{4}{x}}$$

Do the multiplications in \left(x-2\right)x-8.

$$\frac{\frac{x^{2}-2x-8}{x}}{x-3-\frac{4}{x}}$$

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

$$\frac{\frac{x^{2}-2x-8}{x}}{\frac{\left(x-3\right)x}{x}-\frac{4}{x}}$$

Since \frac{\left(x-3\right)x}{x} and \frac{4}{x} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{x^{2}-2x-8}{x}}{\frac{\left(x-3\right)x-4}{x}}$$

Do the multiplications in \left(x-3\right)x-4.

$$\frac{\frac{x^{2}-2x-8}{x}}{\frac{x^{2}-3x-4}{x}}$$

Divide \frac{x^{2}-2x-8}{x} by \frac{x^{2}-3x-4}{x} by multiplying \frac{x^{2}-2x-8}{x} by the reciprocal of \frac{x^{2}-3x-4}{x}.

$$\frac{\left(x^{2}-2x-8\right)x}{x\left(x^{2}-3x-4\right)}$$

Cancel out x in both numerator and denominator.

$$\frac{x^{2}-2x-8}{x^{2}-3x-4}$$

Factor the expressions that are not already factored.

$$\frac{\left(x-4\right)\left(x+2\right)}{\left(x-4\right)\left(x+1\right)}$$

Cancel out x-4 in both numerator and denominator.

$$\frac{x+2}{x+1}$$

Expand

$$\frac{x+2}{x+1}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply x-2 times \frac{x}{x}.

$$\frac{\frac{\left(x-2\right)x}{x}-\frac{8}{x}}{x-3-\frac{4}{x}}$$

Since \frac{\left(x-2\right)x}{x} and \frac{8}{x} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{\left(x-2\right)x-8}{x}}{x-3-\frac{4}{x}}$$

Do the multiplications in \left(x-2\right)x-8.

$$\frac{\frac{x^{2}-2x-8}{x}}{x-3-\frac{4}{x}}$$

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

$$\frac{\frac{x^{2}-2x-8}{x}}{\frac{\left(x-3\right)x}{x}-\frac{4}{x}}$$

Since \frac{\left(x-3\right)x}{x} and \frac{4}{x} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{x^{2}-2x-8}{x}}{\frac{\left(x-3\right)x-4}{x}}$$

Do the multiplications in \left(x-3\right)x-4.

$$\frac{\frac{x^{2}-2x-8}{x}}{\frac{x^{2}-3x-4}{x}}$$

Divide \frac{x^{2}-2x-8}{x} by \frac{x^{2}-3x-4}{x} by multiplying \frac{x^{2}-2x-8}{x} by the reciprocal of \frac{x^{2}-3x-4}{x}.

$$\frac{\left(x^{2}-2x-8\right)x}{x\left(x^{2}-3x-4\right)}$$

Cancel out x in both numerator and denominator.

$$\frac{x^{2}-2x-8}{x^{2}-3x-4}$$

Factor the expressions that are not already factored.

$$\frac{\left(x-4\right)\left(x+2\right)}{\left(x-4\right)\left(x+1\right)}$$

Cancel out x-4 in both numerator and denominator.

$$\frac{x+2}{x+1}$$