Solve 7-6/x+1/x-7-x^2-4{x+frac{2/x+3}}=

$$\frac { 7 - \frac { 6 } { x + 1 } } { x - 7 - \frac { x ^ { 2 } - 4 } { x + \frac { 2 } { x + 3 } } } =$$

Evaluate

$$-1$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply 7 times \frac{x+1}{x+1}.

$$\frac{\frac{7\left(x+1\right)}{x+1}-\frac{6}{x+1}}{x-7-\frac{x^{2}-4}{x+\frac{2}{x+3}}}$$

Since \frac{7\left(x+1\right)}{x+1} and \frac{6}{x+1} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{7\left(x+1\right)-6}{x+1}}{x-7-\frac{x^{2}-4}{x+\frac{2}{x+3}}}$$

Do the multiplications in 7\left(x+1\right)-6.

$$\frac{\frac{7x+7-6}{x+1}}{x-7-\frac{x^{2}-4}{x+\frac{2}{x+3}}}$$

Combine like terms in 7x+7-6.

$$\frac{\frac{7x+1}{x+1}}{x-7-\frac{x^{2}-4}{x+\frac{2}{x+3}}}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x times \frac{x+3}{x+3}.

$$\frac{\frac{7x+1}{x+1}}{x-7-\frac{x^{2}-4}{\frac{x\left(x+3\right)}{x+3}+\frac{2}{x+3}}}$$

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

$$\frac{\frac{7x+1}{x+1}}{x-7-\frac{x^{2}-4}{\frac{x\left(x+3\right)+2}{x+3}}}$$

Do the multiplications in x\left(x+3\right)+2.

$$\frac{\frac{7x+1}{x+1}}{x-7-\frac{x^{2}-4}{\frac{x^{2}+3x+2}{x+3}}}$$

Divide x^{2}-4 by \frac{x^{2}+3x+2}{x+3} by multiplying x^{2}-4 by the reciprocal of \frac{x^{2}+3x+2}{x+3}.

$$\frac{\frac{7x+1}{x+1}}{x-7-\frac{\left(x^{2}-4\right)\left(x+3\right)}{x^{2}+3x+2}}$$

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

$$\frac{\frac{7x+1}{x+1}}{x-7-\frac{\left(x-2\right)\left(x+2\right)\left(x+3\right)}{\left(x+1\right)\left(x+2\right)}}$$

Cancel out x+2 in both numerator and denominator.

$$\frac{\frac{7x+1}{x+1}}{x-7-\frac{\left(x-2\right)\left(x+3\right)}{x+1}}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x-7 times \frac{x+1}{x+1}.

$$\frac{\frac{7x+1}{x+1}}{\frac{\left(x-7\right)\left(x+1\right)}{x+1}-\frac{\left(x-2\right)\left(x+3\right)}{x+1}}$$

Since \frac{\left(x-7\right)\left(x+1\right)}{x+1} and \frac{\left(x-2\right)\left(x+3\right)}{x+1} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{7x+1}{x+1}}{\frac{\left(x-7\right)\left(x+1\right)-\left(x-2\right)\left(x+3\right)}{x+1}}$$

Do the multiplications in \left(x-7\right)\left(x+1\right)-\left(x-2\right)\left(x+3\right).

$$\frac{\frac{7x+1}{x+1}}{\frac{x^{2}+x-7x-7-x^{2}-3x+2x+6}{x+1}}$$

Combine like terms in x^{2}+x-7x-7-x^{2}-3x+2x+6.

$$\frac{\frac{7x+1}{x+1}}{\frac{-7x-1}{x+1}}$$

Divide \frac{7x+1}{x+1} by \frac{-7x-1}{x+1} by multiplying \frac{7x+1}{x+1} by the reciprocal of \frac{-7x-1}{x+1}.

$$\frac{\left(7x+1\right)\left(x+1\right)}{\left(x+1\right)\left(-7x-1\right)}$$

Extract the negative sign in 7x+1.

$$\frac{-\left(-7x-1\right)\left(x+1\right)}{\left(-7x-1\right)\left(x+1\right)}$$

Cancel out \left(-7x-1\right)\left(x+1\right) in both numerator and denominator.

$$-1$$

Factor

$$-1$$