Solve (2/x+2-1)div(1/x+2+1)

$$( \frac { 2 } { x + 2 } - 1 ) \div ( \frac { 1 } { x + 2 } + 1 )$$

Evaluate

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

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{x+2}{x+2}.

$$\frac{\frac{2}{x+2}-\frac{x+2}{x+2}}{\frac{1}{x+2}+1}$$

Since \frac{2}{x+2} and \frac{x+2}{x+2} have the same denominator, subtract them by subtracting their numerators.

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

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

$$\frac{\frac{2-x-2}{x+2}}{\frac{1}{x+2}+1}$$

Combine like terms in 2-x-2.

$$\frac{\frac{-x}{x+2}}{\frac{1}{x+2}+1}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{x+2}{x+2}.

$$\frac{\frac{-x}{x+2}}{\frac{1}{x+2}+\frac{x+2}{x+2}}$$

Since \frac{1}{x+2} and \frac{x+2}{x+2} have the same denominator, add them by adding their numerators.

$$\frac{\frac{-x}{x+2}}{\frac{1+x+2}{x+2}}$$

Combine like terms in 1+x+2.

$$\frac{\frac{-x}{x+2}}{\frac{3+x}{x+2}}$$

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

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

Cancel out x+2 in both numerator and denominator.

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

Expand

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

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{x+2}{x+2}.

$$\frac{\frac{2}{x+2}-\frac{x+2}{x+2}}{\frac{1}{x+2}+1}$$

Since \frac{2}{x+2} and \frac{x+2}{x+2} have the same denominator, subtract them by subtracting their numerators.

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

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

$$\frac{\frac{2-x-2}{x+2}}{\frac{1}{x+2}+1}$$

Combine like terms in 2-x-2.

$$\frac{\frac{-x}{x+2}}{\frac{1}{x+2}+1}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{x+2}{x+2}.

$$\frac{\frac{-x}{x+2}}{\frac{1}{x+2}+\frac{x+2}{x+2}}$$

Since \frac{1}{x+2} and \frac{x+2}{x+2} have the same denominator, add them by adding their numerators.

$$\frac{\frac{-x}{x+2}}{\frac{1+x+2}{x+2}}$$

Combine like terms in 1+x+2.

$$\frac{\frac{-x}{x+2}}{\frac{3+x}{x+2}}$$

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

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

Cancel out x+2 in both numerator and denominator.

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