Solve 2x/x^2+x-12+4x/x^2-9x+18=

$$\frac { 2 x } { x ^ { 2 } + x - 12 } + \frac { 4 x } { x ^ { 2 } - 9 x + 18 } =$$

Evaluate

$$\frac{2x\left(3x+2\right)}{\left(x-6\right)\left(x-3\right)\left(x+4\right)}$$

Short Solution Steps

Factor x^{2}+x-12. Factor x^{2}-9x+18.

$$\frac{2x}{\left(x-3\right)\left(x+4\right)}+\frac{4x}{\left(x-6\right)\left(x-3\right)}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(x-3\right)\left(x+4\right) and \left(x-6\right)\left(x-3\right) is \left(x-6\right)\left(x-3\right)\left(x+4\right). Multiply \frac{2x}{\left(x-3\right)\left(x+4\right)} times \frac{x-6}{x-6}. Multiply \frac{4x}{\left(x-6\right)\left(x-3\right)} times \frac{x+4}{x+4}.

$$\frac{2x\left(x-6\right)}{\left(x-6\right)\left(x-3\right)\left(x+4\right)}+\frac{4x\left(x+4\right)}{\left(x-6\right)\left(x-3\right)\left(x+4\right)}$$

Since \frac{2x\left(x-6\right)}{\left(x-6\right)\left(x-3\right)\left(x+4\right)} and \frac{4x\left(x+4\right)}{\left(x-6\right)\left(x-3\right)\left(x+4\right)} have the same denominator, add them by adding their numerators.

$$\frac{2x\left(x-6\right)+4x\left(x+4\right)}{\left(x-6\right)\left(x-3\right)\left(x+4\right)}$$

Do the multiplications in 2x\left(x-6\right)+4x\left(x+4\right).

$$\frac{2x^{2}-12x+4x^{2}+16x}{\left(x-6\right)\left(x-3\right)\left(x+4\right)}$$

Combine like terms in 2x^{2}-12x+4x^{2}+16x.

$$\frac{6x^{2}+4x}{\left(x-6\right)\left(x-3\right)\left(x+4\right)}$$

Expand \left(x-6\right)\left(x-3\right)\left(x+4\right).

$$\frac{6x^{2}+4x}{x^{3}-5x^{2}-18x+72}$$

Factor

$$\frac{2x\left(3x+2\right)}{\left(x-6\right)\left(x-3\right)\left(x+4\right)}$$