Solve (1/x^2+x-2-1/2x^2+4x):1/4x^2+8x

$$( \frac { 1 } { x ^ { 2 } + x - 2 } - \frac { 1 } { 2 x ^ { 2 } + 4 x } ) : \frac { 1 } { 4 x ^ { 2 } + 8 x }$$

Evaluate

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

Short Solution Steps

Factor x^{2}+x-2. Factor 2x^{2}+4x.

$$\frac{\frac{1}{\left(x-1\right)\left(x+2\right)}-\frac{1}{2x\left(x+2\right)}}{\frac{1}{4x^{2}+8x}}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(x-1\right)\left(x+2\right) and 2x\left(x+2\right) is 2x\left(x-1\right)\left(x+2\right). Multiply \frac{1}{\left(x-1\right)\left(x+2\right)} times \frac{2x}{2x}. Multiply \frac{1}{2x\left(x+2\right)} times \frac{x-1}{x-1}.

$$\frac{\frac{2x}{2x\left(x-1\right)\left(x+2\right)}-\frac{x-1}{2x\left(x-1\right)\left(x+2\right)}}{\frac{1}{4x^{2}+8x}}$$

Since \frac{2x}{2x\left(x-1\right)\left(x+2\right)} and \frac{x-1}{2x\left(x-1\right)\left(x+2\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{2x-\left(x-1\right)}{2x\left(x-1\right)\left(x+2\right)}}{\frac{1}{4x^{2}+8x}}$$

Do the multiplications in 2x-\left(x-1\right).

$$\frac{\frac{2x-x+1}{2x\left(x-1\right)\left(x+2\right)}}{\frac{1}{4x^{2}+8x}}$$

Combine like terms in 2x-x+1.

$$\frac{\frac{x+1}{2x\left(x-1\right)\left(x+2\right)}}{\frac{1}{4x^{2}+8x}}$$

Divide \frac{x+1}{2x\left(x-1\right)\left(x+2\right)} by \frac{1}{4x^{2}+8x} by multiplying \frac{x+1}{2x\left(x-1\right)\left(x+2\right)} by the reciprocal of \frac{1}{4x^{2}+8x}.

$$\frac{\left(x+1\right)\left(4x^{2}+8x\right)}{2x\left(x-1\right)\left(x+2\right)}$$

Factor the expressions that are not already factored.

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

Cancel out 2x\left(x+2\right) in both numerator and denominator.

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

Expand the expression.

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

Expand

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

Short Solution Steps

Factor x^{2}+x-2. Factor 2x^{2}+4x.

$$\frac{\frac{1}{\left(x-1\right)\left(x+2\right)}-\frac{1}{2x\left(x+2\right)}}{\frac{1}{4x^{2}+8x}}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(x-1\right)\left(x+2\right) and 2x\left(x+2\right) is 2x\left(x-1\right)\left(x+2\right). Multiply \frac{1}{\left(x-1\right)\left(x+2\right)} times \frac{2x}{2x}. Multiply \frac{1}{2x\left(x+2\right)} times \frac{x-1}{x-1}.

$$\frac{\frac{2x}{2x\left(x-1\right)\left(x+2\right)}-\frac{x-1}{2x\left(x-1\right)\left(x+2\right)}}{\frac{1}{4x^{2}+8x}}$$

Since \frac{2x}{2x\left(x-1\right)\left(x+2\right)} and \frac{x-1}{2x\left(x-1\right)\left(x+2\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{2x-\left(x-1\right)}{2x\left(x-1\right)\left(x+2\right)}}{\frac{1}{4x^{2}+8x}}$$

Do the multiplications in 2x-\left(x-1\right).

$$\frac{\frac{2x-x+1}{2x\left(x-1\right)\left(x+2\right)}}{\frac{1}{4x^{2}+8x}}$$

Combine like terms in 2x-x+1.

$$\frac{\frac{x+1}{2x\left(x-1\right)\left(x+2\right)}}{\frac{1}{4x^{2}+8x}}$$

Divide \frac{x+1}{2x\left(x-1\right)\left(x+2\right)} by \frac{1}{4x^{2}+8x} by multiplying \frac{x+1}{2x\left(x-1\right)\left(x+2\right)} by the reciprocal of \frac{1}{4x^{2}+8x}.

$$\frac{\left(x+1\right)\left(4x^{2}+8x\right)}{2x\left(x-1\right)\left(x+2\right)}$$

Factor the expressions that are not already factored.

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

Cancel out 2x\left(x+2\right) in both numerator and denominator.

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

Expand the expression.

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