Question

Helpppppppp!!!!!! plz answer​

Answer 1

$\frac{\frac{1}{a}-\frac{1}{b}}{\frac{1}{a}+\frac{1}{b}}=\frac{b-a}{b+a}$

Solution:

Given expression:

$\frac{\frac{1}{a}-\frac{1}{b}}{\frac{1}{a}+\frac{1}{b}}$

To simplify the given expression:

$\frac{\frac{1}{a}-\frac{1}{b}}{\frac{1}{a}+\frac{1}{b}}$

Let us first solve the numerator of the expression $\frac{1}{a} -\frac{1}{b}$.

To add or subtract the fraction, the denominators of the fraction must be same.

To make it same, multiply $\frac{1}{a}$ by $\frac{b}{b}$ and $\frac{1}{b}$ by $\frac{a}{a}$.

$\frac{1}{a} -\frac{1}{b}=\frac{b}{ab} -\frac{a}{ab}$

          $=\frac{b-a}{ab}$ ------------------- (1)

Now, solve the denominator of the expression $\frac{1}{a} +\frac{1}{b}$.

To add or subtract the fraction, the denominators of the fraction must be same.

To make it same, multiply $\frac{1}{a}$ by $\frac{b}{b}$ and $\frac{1}{b}$ by $\frac{a}{a}$.

$\frac{1}{a} +\frac{1}{b}=\frac{b}{ab} +\frac{a}{ab}$

          $=\frac{b+a}{ab}$  ------------------- (2)

Substitute (1) and (2) in the given expression.

$\frac{\frac{1}{a}-\frac{1}{b}}{\frac{1}{a}+\frac{1}{b}}=\frac{\frac{b-a}{ab}}{\frac{b+a}{ab}}$

Using rational rule: $\frac{\frac{x}{y} }{\frac{w}{z} }=\frac{x}{y} \times\frac{z}{w}$

        $=\frac{b-a}{ab}}\times {\frac{ab}{b+a}}$

Common factor ab get canceled.

        $=\frac{b-a}{b+a}$    

$\frac{\frac{1}{a}-\frac{1}{b}}{\frac{1}{a}+\frac{1}{b}}=\frac{b-a}{b+a}$

Hence the simplified expression is $\frac{b-a}{b+a}$.