Solve E(x)=(x+2/x-5-x^2-4/x^2+7x+10-27-5x/25-x^2):x+3/x^2-25

$$E ( x ) = ( \frac { x + 2 } { x - 5 } - \frac { x ^ { 2 } - 4 } { x ^ { 2 } + 7 x + 10 } - \frac { 27 - 5 x } { 25 - x ^ { 2 } } ) : \frac { x + 3 } { x ^ { 2 } - 25 }$$

Evaluate

$$9$$

Short Solution Steps

Factor the expressions that are not already factored in \frac{x^{2}-4}{x^{2}+7x+10}.

$$\frac{\frac{x+2}{x-5}-\frac{\left(x-2\right)\left(x+2\right)}{\left(x+2\right)\left(x+5\right)}-\frac{27-5x}{25-x^{2}}}{\frac{x+3}{x^{2}-25}}$$

Cancel out x+2 in both numerator and denominator.

$$\frac{\frac{x+2}{x-5}-\frac{x-2}{x+5}-\frac{27-5x}{25-x^{2}}}{\frac{x+3}{x^{2}-25}}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of x-5 and x+5 is \left(x-5\right)\left(x+5\right). Multiply \frac{x+2}{x-5} times \frac{x+5}{x+5}. Multiply \frac{x-2}{x+5} times \frac{x-5}{x-5}.

$$\frac{\frac{\left(x+2\right)\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}-\frac{\left(x-2\right)\left(x-5\right)}{\left(x-5\right)\left(x+5\right)}-\frac{27-5x}{25-x^{2}}}{\frac{x+3}{x^{2}-25}}$$

Since \frac{\left(x+2\right)\left(x+5\right)}{\left(x-5\right)\left(x+5\right)} and \frac{\left(x-2\right)\left(x-5\right)}{\left(x-5\right)\left(x+5\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{\left(x+2\right)\left(x+5\right)-\left(x-2\right)\left(x-5\right)}{\left(x-5\right)\left(x+5\right)}-\frac{27-5x}{25-x^{2}}}{\frac{x+3}{x^{2}-25}}$$

Do the multiplications in \left(x+2\right)\left(x+5\right)-\left(x-2\right)\left(x-5\right).

$$\frac{\frac{x^{2}+5x+2x+10-x^{2}+5x+2x-10}{\left(x-5\right)\left(x+5\right)}-\frac{27-5x}{25-x^{2}}}{\frac{x+3}{x^{2}-25}}$$

Combine like terms in x^{2}+5x+2x+10-x^{2}+5x+2x-10.

$$\frac{\frac{14x}{\left(x-5\right)\left(x+5\right)}-\frac{27-5x}{25-x^{2}}}{\frac{x+3}{x^{2}-25}}$$

Factor 25-x^{2}.

$$\frac{\frac{14x}{\left(x-5\right)\left(x+5\right)}-\frac{27-5x}{\left(x-5\right)\left(-x-5\right)}}{\frac{x+3}{x^{2}-25}}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(x-5\right)\left(x+5\right) and \left(x-5\right)\left(-x-5\right) is \left(x-5\right)\left(x+5\right). Multiply \frac{27-5x}{\left(x-5\right)\left(-x-5\right)} times \frac{-1}{-1}.

$$\frac{\frac{14x}{\left(x-5\right)\left(x+5\right)}-\frac{-\left(27-5x\right)}{\left(x-5\right)\left(x+5\right)}}{\frac{x+3}{x^{2}-25}}$$

Since \frac{14x}{\left(x-5\right)\left(x+5\right)} and \frac{-\left(27-5x\right)}{\left(x-5\right)\left(x+5\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{14x-\left(-\left(27-5x\right)\right)}{\left(x-5\right)\left(x+5\right)}}{\frac{x+3}{x^{2}-25}}$$

Do the multiplications in 14x-\left(-\left(27-5x\right)\right).

$$\frac{\frac{14x+27-5x}{\left(x-5\right)\left(x+5\right)}}{\frac{x+3}{x^{2}-25}}$$

Combine like terms in 14x+27-5x.

$$\frac{\frac{9x+27}{\left(x-5\right)\left(x+5\right)}}{\frac{x+3}{x^{2}-25}}$$

Divide \frac{9x+27}{\left(x-5\right)\left(x+5\right)} by \frac{x+3}{x^{2}-25} by multiplying \frac{9x+27}{\left(x-5\right)\left(x+5\right)} by the reciprocal of \frac{x+3}{x^{2}-25}.

$$\frac{\left(9x+27\right)\left(x^{2}-25\right)}{\left(x-5\right)\left(x+5\right)\left(x+3\right)}$$

Factor the expressions that are not already factored.

$$\frac{9\left(x-5\right)\left(x+3\right)\left(x+5\right)}{\left(x-5\right)\left(x+3\right)\left(x+5\right)}$$

Cancel out \left(x-5\right)\left(x+3\right)\left(x+5\right) in both numerator and denominator.

$$9$$

Factor

$$3^{2}$$