Solve n+2/n-2+n-4/n+4=

$$\frac { n + 2 } { n - 2 } + \frac { n - 4 } { n + 4 } =$$

Evaluate

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

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of n-2 and n+4 is \left(n-2\right)\left(n+4\right). Multiply \frac{n+2}{n-2} times \frac{n+4}{n+4}. Multiply \frac{n-4}{n+4} times \frac{n-2}{n-2}.

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

Since \frac{\left(n+2\right)\left(n+4\right)}{\left(n-2\right)\left(n+4\right)} and \frac{\left(n-4\right)\left(n-2\right)}{\left(n-2\right)\left(n+4\right)} have the same denominator, add them by adding their numerators.

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

Do the multiplications in \left(n+2\right)\left(n+4\right)+\left(n-4\right)\left(n-2\right).

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

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

$$\frac{2n^{2}+16}{\left(n-2\right)\left(n+4\right)}$$

Expand \left(n-2\right)\left(n+4\right).

$$\frac{2n^{2}+16}{n^{2}+2n-8}$$

Expand

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

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of n-2 and n+4 is \left(n-2\right)\left(n+4\right). Multiply \frac{n+2}{n-2} times \frac{n+4}{n+4}. Multiply \frac{n-4}{n+4} times \frac{n-2}{n-2}.

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

Since \frac{\left(n+2\right)\left(n+4\right)}{\left(n-2\right)\left(n+4\right)} and \frac{\left(n-4\right)\left(n-2\right)}{\left(n-2\right)\left(n+4\right)} have the same denominator, add them by adding their numerators.

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

Do the multiplications in \left(n+2\right)\left(n+4\right)+\left(n-4\right)\left(n-2\right).

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

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

$$\frac{2n^{2}+16}{\left(n-2\right)\left(n+4\right)}$$

Expand \left(n-2\right)\left(n+4\right).

$$\frac{2n^{2}+16}{n^{2}+2n-8}$$