Solve =(1-5/x+3):(x+1/x+3+1/x-3-2/x^2-9)

$$= ( 1 - \frac { 5 } { x + 3 } ) : ( \frac { x + 1 } { x + 3 } + \frac { 1 } { x - 3 } - \frac { 2 } { x ^ { 2 } - 9 } )$$

Evaluate

$$\frac{x-3}{x+1}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{x+3}{x+3}.

$$\frac{\frac{x+3}{x+3}-\frac{5}{x+3}}{\frac{x+1}{x+3}+\frac{1}{x-3}-\frac{2}{x^{2}-9}}$$

Since \frac{x+3}{x+3} and \frac{5}{x+3} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{x+3-5}{x+3}}{\frac{x+1}{x+3}+\frac{1}{x-3}-\frac{2}{x^{2}-9}}$$

Combine like terms in x+3-5.

$$\frac{\frac{x-2}{x+3}}{\frac{x+1}{x+3}+\frac{1}{x-3}-\frac{2}{x^{2}-9}}$$

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

$$\frac{\frac{x-2}{x+3}}{\frac{\left(x+1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{x+3}{\left(x-3\right)\left(x+3\right)}-\frac{2}{x^{2}-9}}$$

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

$$\frac{\frac{x-2}{x+3}}{\frac{\left(x+1\right)\left(x-3\right)+x+3}{\left(x-3\right)\left(x+3\right)}-\frac{2}{x^{2}-9}}$$

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

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

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

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

Factor x^{2}-9.

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

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

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

Combine like terms in x^{2}-x-2.

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

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

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

Cancel out x+3 in both numerator and denominator.

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

Factor the expressions that are not already factored.

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

Cancel out x-2 in both numerator and denominator.

$$\frac{x-3}{x+1}$$

Expand

$$\frac{x-3}{x+1}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{x+3}{x+3}.

$$\frac{\frac{x+3}{x+3}-\frac{5}{x+3}}{\frac{x+1}{x+3}+\frac{1}{x-3}-\frac{2}{x^{2}-9}}$$

Since \frac{x+3}{x+3} and \frac{5}{x+3} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{x+3-5}{x+3}}{\frac{x+1}{x+3}+\frac{1}{x-3}-\frac{2}{x^{2}-9}}$$

Combine like terms in x+3-5.

$$\frac{\frac{x-2}{x+3}}{\frac{x+1}{x+3}+\frac{1}{x-3}-\frac{2}{x^{2}-9}}$$

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

$$\frac{\frac{x-2}{x+3}}{\frac{\left(x+1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{x+3}{\left(x-3\right)\left(x+3\right)}-\frac{2}{x^{2}-9}}$$

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

$$\frac{\frac{x-2}{x+3}}{\frac{\left(x+1\right)\left(x-3\right)+x+3}{\left(x-3\right)\left(x+3\right)}-\frac{2}{x^{2}-9}}$$

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

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

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

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

Factor x^{2}-9.

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

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

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

Combine like terms in x^{2}-x-2.

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

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

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

Cancel out x+3 in both numerator and denominator.

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

Factor the expressions that are not already factored.

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

Cancel out x-2 in both numerator and denominator.

$$\frac{x-3}{x+1}$$