Solve 4x-5/x^2+x-12+9/18-3x-x^2+2/x^2+10x+24

$$\frac { 4 x - 5 } { x ^ { 2 } + x - 12 } + \frac { 9 } { 18 - 3 x - x ^ { 2 } } + \frac { 2 } { x ^ { 2 } + 10 x + 24 }$$

Evaluate

$$\frac{4}{x+4}$$

Short Solution Steps

Factor x^{2}+x-12. Factor 18-3x-x^{2}.

$$\frac{4x-5}{\left(x-3\right)\left(x+4\right)}+\frac{9}{\left(x+6\right)\left(-x+3\right)}+\frac{2}{x^{2}+10x+24}$$

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-3\right)\left(x+4\right)\left(x+6\right). Multiply \frac{4x-5}{\left(x-3\right)\left(x+4\right)} times \frac{x+6}{x+6}. Multiply \frac{9}{\left(x+6\right)\left(-x+3\right)} times \frac{-\left(x+4\right)}{-\left(x+4\right)}.

$$\frac{\left(4x-5\right)\left(x+6\right)}{\left(x-3\right)\left(x+4\right)\left(x+6\right)}+\frac{9\left(-1\right)\left(x+4\right)}{\left(x-3\right)\left(x+4\right)\left(x+6\right)}+\frac{2}{x^{2}+10x+24}$$

Since \frac{\left(4x-5\right)\left(x+6\right)}{\left(x-3\right)\left(x+4\right)\left(x+6\right)} and \frac{9\left(-1\right)\left(x+4\right)}{\left(x-3\right)\left(x+4\right)\left(x+6\right)} have the same denominator, add them by adding their numerators.

$$\frac{\left(4x-5\right)\left(x+6\right)+9\left(-1\right)\left(x+4\right)}{\left(x-3\right)\left(x+4\right)\left(x+6\right)}+\frac{2}{x^{2}+10x+24}$$

Do the multiplications in \left(4x-5\right)\left(x+6\right)+9\left(-1\right)\left(x+4\right).

$$\frac{4x^{2}+24x-5x-30-9x-36}{\left(x-3\right)\left(x+4\right)\left(x+6\right)}+\frac{2}{x^{2}+10x+24}$$

Combine like terms in 4x^{2}+24x-5x-30-9x-36.

$$\frac{4x^{2}+10x-66}{\left(x-3\right)\left(x+4\right)\left(x+6\right)}+\frac{2}{x^{2}+10x+24}$$

Factor the expressions that are not already factored in \frac{4x^{2}+10x-66}{\left(x-3\right)\left(x+4\right)\left(x+6\right)}.

$$\frac{2\left(x-3\right)\left(2x+11\right)}{\left(x-3\right)\left(x+4\right)\left(x+6\right)}+\frac{2}{x^{2}+10x+24}$$

Cancel out x-3 in both numerator and denominator.

$$\frac{2\left(2x+11\right)}{\left(x+4\right)\left(x+6\right)}+\frac{2}{x^{2}+10x+24}$$

Factor x^{2}+10x+24.

$$\frac{2\left(2x+11\right)}{\left(x+4\right)\left(x+6\right)}+\frac{2}{\left(x+4\right)\left(x+6\right)}$$

Since \frac{2\left(2x+11\right)}{\left(x+4\right)\left(x+6\right)} and \frac{2}{\left(x+4\right)\left(x+6\right)} have the same denominator, add them by adding their numerators.

$$\frac{2\left(2x+11\right)+2}{\left(x+4\right)\left(x+6\right)}$$

Do the multiplications in 2\left(2x+11\right)+2.

$$\frac{4x+22+2}{\left(x+4\right)\left(x+6\right)}$$

Combine like terms in 4x+22+2.

$$\frac{4x+24}{\left(x+4\right)\left(x+6\right)}$$

Factor the expressions that are not already factored in \frac{4x+24}{\left(x+4\right)\left(x+6\right)}.

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

Cancel out x+6 in both numerator and denominator.

$$\frac{4}{x+4}$$

Expand

$$\frac{4}{x+4}$$

Short Solution Steps

Factor x^{2}+x-12. Factor 18-3x-x^{2}.

$$\frac{4x-5}{\left(x-3\right)\left(x+4\right)}+\frac{9}{\left(x+6\right)\left(-x+3\right)}+\frac{2}{x^{2}+10x+24}$$

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-3\right)\left(x+4\right)\left(x+6\right). Multiply \frac{4x-5}{\left(x-3\right)\left(x+4\right)} times \frac{x+6}{x+6}. Multiply \frac{9}{\left(x+6\right)\left(-x+3\right)} times \frac{-\left(x+4\right)}{-\left(x+4\right)}.

$$\frac{\left(4x-5\right)\left(x+6\right)}{\left(x-3\right)\left(x+4\right)\left(x+6\right)}+\frac{9\left(-1\right)\left(x+4\right)}{\left(x-3\right)\left(x+4\right)\left(x+6\right)}+\frac{2}{x^{2}+10x+24}$$

Since \frac{\left(4x-5\right)\left(x+6\right)}{\left(x-3\right)\left(x+4\right)\left(x+6\right)} and \frac{9\left(-1\right)\left(x+4\right)}{\left(x-3\right)\left(x+4\right)\left(x+6\right)} have the same denominator, add them by adding their numerators.

$$\frac{\left(4x-5\right)\left(x+6\right)+9\left(-1\right)\left(x+4\right)}{\left(x-3\right)\left(x+4\right)\left(x+6\right)}+\frac{2}{x^{2}+10x+24}$$

Do the multiplications in \left(4x-5\right)\left(x+6\right)+9\left(-1\right)\left(x+4\right).

$$\frac{4x^{2}+24x-5x-30-9x-36}{\left(x-3\right)\left(x+4\right)\left(x+6\right)}+\frac{2}{x^{2}+10x+24}$$

Combine like terms in 4x^{2}+24x-5x-30-9x-36.

$$\frac{4x^{2}+10x-66}{\left(x-3\right)\left(x+4\right)\left(x+6\right)}+\frac{2}{x^{2}+10x+24}$$

Factor the expressions that are not already factored in \frac{4x^{2}+10x-66}{\left(x-3\right)\left(x+4\right)\left(x+6\right)}.

$$\frac{2\left(x-3\right)\left(2x+11\right)}{\left(x-3\right)\left(x+4\right)\left(x+6\right)}+\frac{2}{x^{2}+10x+24}$$

Cancel out x-3 in both numerator and denominator.

$$\frac{2\left(2x+11\right)}{\left(x+4\right)\left(x+6\right)}+\frac{2}{x^{2}+10x+24}$$

Factor x^{2}+10x+24.

$$\frac{2\left(2x+11\right)}{\left(x+4\right)\left(x+6\right)}+\frac{2}{\left(x+4\right)\left(x+6\right)}$$

Since \frac{2\left(2x+11\right)}{\left(x+4\right)\left(x+6\right)} and \frac{2}{\left(x+4\right)\left(x+6\right)} have the same denominator, add them by adding their numerators.

$$\frac{2\left(2x+11\right)+2}{\left(x+4\right)\left(x+6\right)}$$

Do the multiplications in 2\left(2x+11\right)+2.

$$\frac{4x+22+2}{\left(x+4\right)\left(x+6\right)}$$

Combine like terms in 4x+22+2.

$$\frac{4x+24}{\left(x+4\right)\left(x+6\right)}$$

Factor the expressions that are not already factored in \frac{4x+24}{\left(x+4\right)\left(x+6\right)}.

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

Cancel out x+6 in both numerator and denominator.

$$\frac{4}{x+4}$$