Chapter 1, Problem 62RE

To determine

To simplify $\frac{\frac{1}{x}-\frac{1}{x+1}}{\frac{1}{x}+\frac{1}{x+1}}$ by using appropriate operations.

Answer to Problem 62RE

The answer is $\frac{1}{2x+1}$ .

Explanation of Solution

Given information:

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

Formula used:

Properties of fractions:- $\frac{A}{C}-\frac{B}{C}=\frac{A-B}{C}$

Properties of fractions :- $\frac{A}{B}÷\frac{C}{D}=\frac{A}{B}.\frac{D}{C}$

Calculations:

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

Writing fraction in form of LCD (least common denominator)

  $\begin{array}{l}=\frac{x+1}{x(x+1)}-\frac{x}{x(x+1)}\\ =\frac{x+1-x}{x(x+1)}\\ =\frac{1}{x(x+1)}\\ \end{array}$ (i)

Denominator = $\frac{1}{x}+\frac{1}{x+1}$

Writing fraction in form of LCD (least common denominator)

  $\begin{array}{l}=\frac{x+1}{x(x+1)}+\frac{x}{x(x+1)}\\ =\frac{x+1+x}{x(x+1)}\\ =\frac{2x+1}{x(x+1)}\end{array}$  (ii)

Now,

  $\begin{array}{l}=\frac{1}{x(x+1)}÷\frac{2x+1}{x(x+1)}\\ =\frac{1}{x(x+1)}.\frac{x(x+1)}{2x+1}\end{array}$

  $=\frac{1}{2x+1}$ ( by cancelling $x(x+1)$ )

After simplifying we get $\frac{1}{2x+1}$ .