Question

Operations with Algebraic Expressions Perform the indicated operations and simplify.

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

Short answer

The simplified form of the given expression is $\frac{1}{2x+1}$.

Step-by-step solution

Step 1. Property of fractions.

To divide a fraction by another fraction, we invert the divisor and multiply.

$\frac{A}{B}÷\frac{C}{D}=\frac{A}{B}\times \frac{D}{C}$

Step 2. Given Information.

The given expression is:

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

Step 3. Simplify the given expression.

The given expression can be rewritten as:

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

Using the property of fractions the above expression can be rewritten as:

$\begin{array}{c}\frac{\frac{1}{x}-\frac{1}{x+1}}{\frac{1}{x}+\frac{1}{x+1}}=\frac{\frac{1}{x\left(x+1\right)}}{\frac{2x+1}{x\left(x+1\right)}}\\ =\frac{1}{x\left(x+1\right)}\times \frac{x\left(x+1\right)}{2x+1}\\ =\frac{1}{2x+1}\end{array}$

Thus, the simplified form of the given expression is $\frac{1}{2x+1}$.