Solve 1/x+1+1/x+2/frac{1{x+1}-1/x+2}

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

Evaluate

$$2x+3$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of x+1 and x+2 is \left(x+1\right)\left(x+2\right). Multiply \frac{1}{x+1} times \frac{x+2}{x+2}. Multiply \frac{1}{x+2} times \frac{x+1}{x+1}.

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

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

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

Combine like terms in x+2+x+1.

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of x+1 and x+2 is \left(x+1\right)\left(x+2\right). Multiply \frac{1}{x+1} times \frac{x+2}{x+2}. Multiply \frac{1}{x+2} times \frac{x+1}{x+1}.

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

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

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

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

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

Combine like terms in x+2-x-1.

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

Divide \frac{2x+3}{\left(x+1\right)\left(x+2\right)} by \frac{1}{\left(x+1\right)\left(x+2\right)} by multiplying \frac{2x+3}{\left(x+1\right)\left(x+2\right)} by the reciprocal of \frac{1}{\left(x+1\right)\left(x+2\right)}.

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

Cancel out \left(x+1\right)\left(x+2\right) in both numerator and denominator.

$$2x+3$$

Expand

$$2x+3$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of x+1 and x+2 is \left(x+1\right)\left(x+2\right). Multiply \frac{1}{x+1} times \frac{x+2}{x+2}. Multiply \frac{1}{x+2} times \frac{x+1}{x+1}.

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

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

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

Combine like terms in x+2+x+1.

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of x+1 and x+2 is \left(x+1\right)\left(x+2\right). Multiply \frac{1}{x+1} times \frac{x+2}{x+2}. Multiply \frac{1}{x+2} times \frac{x+1}{x+1}.

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

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

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

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

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

Combine like terms in x+2-x-1.

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

Divide \frac{2x+3}{\left(x+1\right)\left(x+2\right)} by \frac{1}{\left(x+1\right)\left(x+2\right)} by multiplying \frac{2x+3}{\left(x+1\right)\left(x+2\right)} by the reciprocal of \frac{1}{\left(x+1\right)\left(x+2\right)}.

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

Cancel out \left(x+1\right)\left(x+2\right) in both numerator and denominator.

$$2x+3$$