Solve x+2/x^2-x-12-x/x^2

$$\frac { x + 2 } { x ^ { 2 } - x - 12 } - \frac { x } { x ^ { 2 } }$$

Evaluate

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

Short Solution Steps

Rewrite x^{2} as xx. Cancel out x in both numerator and denominator.

$$\frac{x+2}{x^{2}-x-12}-\frac{1}{x}$$

Factor x^{2}-x-12.

$$\frac{x+2}{\left(x-4\right)\left(x+3\right)}-\frac{1}{x}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(x-4\right)\left(x+3\right) and x is x\left(x-4\right)\left(x+3\right). Multiply \frac{x+2}{\left(x-4\right)\left(x+3\right)} times \frac{x}{x}. Multiply \frac{1}{x} times \frac{\left(x-4\right)\left(x+3\right)}{\left(x-4\right)\left(x+3\right)}.

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

Since \frac{\left(x+2\right)x}{x\left(x-4\right)\left(x+3\right)} and \frac{\left(x-4\right)\left(x+3\right)}{x\left(x-4\right)\left(x+3\right)} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in \left(x+2\right)x-\left(x-4\right)\left(x+3\right).

$$\frac{x^{2}+2x-x^{2}-3x+4x+12}{x\left(x-4\right)\left(x+3\right)}$$

Combine like terms in x^{2}+2x-x^{2}-3x+4x+12.

$$\frac{3x+12}{x\left(x-4\right)\left(x+3\right)}$$

Expand x\left(x-4\right)\left(x+3\right).

$$\frac{3x+12}{x^{3}-x^{2}-12x}$$

Expand

$$\frac{3\left(x+4\right)}{x\left(x^{2}-x-12\right)}$$

Short Solution Steps

Rewrite x^{2} as xx. Cancel out x in both numerator and denominator.

$$\frac{x+2}{x^{2}-x-12}-\frac{1}{x}$$

Factor x^{2}-x-12.

$$\frac{x+2}{\left(x-4\right)\left(x+3\right)}-\frac{1}{x}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(x-4\right)\left(x+3\right) and x is x\left(x-4\right)\left(x+3\right). Multiply \frac{x+2}{\left(x-4\right)\left(x+3\right)} times \frac{x}{x}. Multiply \frac{1}{x} times \frac{\left(x-4\right)\left(x+3\right)}{\left(x-4\right)\left(x+3\right)}.

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

Since \frac{\left(x+2\right)x}{x\left(x-4\right)\left(x+3\right)} and \frac{\left(x-4\right)\left(x+3\right)}{x\left(x-4\right)\left(x+3\right)} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in \left(x+2\right)x-\left(x-4\right)\left(x+3\right).

$$\frac{x^{2}+2x-x^{2}-3x+4x+12}{x\left(x-4\right)\left(x+3\right)}$$

Combine like terms in x^{2}+2x-x^{2}-3x+4x+12.

$$\frac{3x+12}{x\left(x-4\right)\left(x+3\right)}$$

Expand x\left(x-4\right)\left(x+3\right).

$$\frac{3x+12}{x^{3}-x^{2}-12x}$$