Solve x^2+8/(x+2)(x-2)+x/x+2-2x/x-2=

$$\frac { x ^ { 2 } + 8 } { ( x + 2 ) ( x - 2 ) } + \frac { x } { x + 2 } - \frac { 2 x } { x - 2 } =$$

Evaluate

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

Short Solution Steps

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

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

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

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

Do the multiplications in x^{2}+8+x\left(x-2\right).

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

Combine like terms in x^{2}+8+x^{2}-2x.

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

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

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

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

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

Do the multiplications in 2x^{2}+8-2x-2x\left(x+2\right).

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

Combine like terms in 2x^{2}+8-2x-2x^{2}-4x.

$$\frac{8-6x}{\left(x-2\right)\left(x+2\right)}$$

Expand \left(x-2\right)\left(x+2\right).

$$\frac{8-6x}{x^{2}-4}$$

Expand

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

Short Solution Steps

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

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

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

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

Do the multiplications in x^{2}+8+x\left(x-2\right).

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

Combine like terms in x^{2}+8+x^{2}-2x.

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

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

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

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

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

Do the multiplications in 2x^{2}+8-2x-2x\left(x+2\right).

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

Combine like terms in 2x^{2}+8-2x-2x^{2}-4x.

$$\frac{8-6x}{\left(x-2\right)\left(x+2\right)}$$

Expand \left(x-2\right)\left(x+2\right).

$$\frac{8-6x}{x^{2}-4}$$