Solve 10/x^2+x-6-8/x^2-4)

$$\frac { 10 } { x ^ { 2 } + x - 6 } - \frac { 8 } { x ^ { 2 } - 4 } )$$

Evaluate

$$\frac{2}{\left(x+2\right)\left(x+3\right)}$$

Short Solution Steps

Factor x^{2}+x-6. Factor x^{2}-4.

$$\frac{10}{\left(x-2\right)\left(x+3\right)}-\frac{8}{\left(x-2\right)\left(x+2\right)}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(x-2\right)\left(x+3\right) and \left(x-2\right)\left(x+2\right) is \left(x-2\right)\left(x+2\right)\left(x+3\right). Multiply \frac{10}{\left(x-2\right)\left(x+3\right)} times \frac{x+2}{x+2}. Multiply \frac{8}{\left(x-2\right)\left(x+2\right)} times \frac{x+3}{x+3}.

$$\frac{10\left(x+2\right)}{\left(x-2\right)\left(x+2\right)\left(x+3\right)}-\frac{8\left(x+3\right)}{\left(x-2\right)\left(x+2\right)\left(x+3\right)}$$

Since \frac{10\left(x+2\right)}{\left(x-2\right)\left(x+2\right)\left(x+3\right)} and \frac{8\left(x+3\right)}{\left(x-2\right)\left(x+2\right)\left(x+3\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{10\left(x+2\right)-8\left(x+3\right)}{\left(x-2\right)\left(x+2\right)\left(x+3\right)}$$

Do the multiplications in 10\left(x+2\right)-8\left(x+3\right).

$$\frac{10x+20-8x-24}{\left(x-2\right)\left(x+2\right)\left(x+3\right)}$$

Combine like terms in 10x+20-8x-24.

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

Factor the expressions that are not already factored in \frac{2x-4}{\left(x-2\right)\left(x+2\right)\left(x+3\right)}.

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

Cancel out x-2 in both numerator and denominator.

$$\frac{2}{\left(x+2\right)\left(x+3\right)}$$

Expand \left(x+2\right)\left(x+3\right).

$$\frac{2}{x^{2}+5x+6}$$

Factor

$$\frac{2}{\left(x+2\right)\left(x+3\right)}$$