Solve x^2/x^2+x-5/x^2-3x=

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

Evaluate

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

Short Solution Steps

Factor the expressions that are not already factored in \frac{x^{2}}{x^{2}+x}.

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

Cancel out x in both numerator and denominator.

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

Factor x^{2}-3x.

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of x+1 and x\left(x-3\right) is x\left(x-3\right)\left(x+1\right). Multiply \frac{x}{x+1} times \frac{x\left(x-3\right)}{x\left(x-3\right)}. Multiply \frac{5}{x\left(x-3\right)} times \frac{x+1}{x+1}.

$$\frac{xx\left(x-3\right)}{x\left(x-3\right)\left(x+1\right)}-\frac{5\left(x+1\right)}{x\left(x-3\right)\left(x+1\right)}$$

Since \frac{xx\left(x-3\right)}{x\left(x-3\right)\left(x+1\right)} and \frac{5\left(x+1\right)}{x\left(x-3\right)\left(x+1\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{xx\left(x-3\right)-5\left(x+1\right)}{x\left(x-3\right)\left(x+1\right)}$$

Do the multiplications in xx\left(x-3\right)-5\left(x+1\right).

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

Expand x\left(x-3\right)\left(x+1\right).

$$\frac{x^{3}-3x^{2}-5x-5}{x^{3}-2x^{2}-3x}$$

Differentiate w.r.t. x

$$\frac{x^{4}+4x^{3}+14x^{2}-20x-15}{\left(x\left(x-3\right)\left(x+1\right)\right)^{2}}$$