Solve 1/3x^2-3/frac{5{x+1}-x+4/x^2-3x-4}

$$\frac { \frac { 1 } { 3 x ^ { 2 } - 3 } } { \frac { 5 } { x + 1 } - \frac { x + 4 } { x ^ { 2 } - 3 x - 4 } }$$

Evaluate

$$\frac{x-4}{12\left(x-6\right)\left(x-1\right)}$$

Short Solution Steps

Factor x^{2}-3x-4.

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of x+1 and \left(x-4\right)\left(x+1\right) is \left(x-4\right)\left(x+1\right). Multiply \frac{5}{x+1} times \frac{x-4}{x-4}.

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

Since \frac{5\left(x-4\right)}{\left(x-4\right)\left(x+1\right)} and \frac{x+4}{\left(x-4\right)\left(x+1\right)} have the same denominator, subtract them by subtracting their numerators.

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

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

$$\frac{\frac{1}{3x^{2}-3}}{\frac{5x-20-x-4}{\left(x-4\right)\left(x+1\right)}}$$

Combine like terms in 5x-20-x-4.

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

Divide \frac{1}{3x^{2}-3} by \frac{4x-24}{\left(x-4\right)\left(x+1\right)} by multiplying \frac{1}{3x^{2}-3} by the reciprocal of \frac{4x-24}{\left(x-4\right)\left(x+1\right)}.

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

Factor the expressions that are not already factored.

$$\frac{\left(x-4\right)\left(x+1\right)}{3\times 4\left(x-6\right)\left(x-1\right)\left(x+1\right)}$$

Cancel out x+1 in both numerator and denominator.

$$\frac{x-4}{3\times 4\left(x-6\right)\left(x-1\right)}$$

Expand the expression.

$$\frac{x-4}{12x^{2}-84x+72}$$

Expand

$$\frac{x-4}{12\left(x-6\right)\left(x-1\right)}$$

Short Solution Steps

Factor x^{2}-3x-4.

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of x+1 and \left(x-4\right)\left(x+1\right) is \left(x-4\right)\left(x+1\right). Multiply \frac{5}{x+1} times \frac{x-4}{x-4}.

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

Since \frac{5\left(x-4\right)}{\left(x-4\right)\left(x+1\right)} and \frac{x+4}{\left(x-4\right)\left(x+1\right)} have the same denominator, subtract them by subtracting their numerators.

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

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

$$\frac{\frac{1}{3x^{2}-3}}{\frac{5x-20-x-4}{\left(x-4\right)\left(x+1\right)}}$$

Combine like terms in 5x-20-x-4.

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

Divide \frac{1}{3x^{2}-3} by \frac{4x-24}{\left(x-4\right)\left(x+1\right)} by multiplying \frac{1}{3x^{2}-3} by the reciprocal of \frac{4x-24}{\left(x-4\right)\left(x+1\right)}.

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

Factor the expressions that are not already factored.

$$\frac{\left(x-4\right)\left(x+1\right)}{3\times 4\left(x-6\right)\left(x-1\right)\left(x+1\right)}$$

Cancel out x+1 in both numerator and denominator.

$$\frac{x-4}{3\times 4\left(x-6\right)\left(x-1\right)}$$

Expand the expression.

$$\frac{x-4}{12x^{2}-84x+72}$$