Question

Verify that $\frac{a}{(b+c)}\ne \frac{a}{b}+\frac{a}{c}$ for each of the following values of $a$, $\mathit{b}\mathbf{}$and $c$.

$a=12,b=-4,c=2$

Answer 1

Given,$a=12,b=–4,c=2$

On considering $LHS=\frac{a}{b+c}$

As per the division rule, When we divide a positive integer by a negative integer, we first divide them as whole numbers and then put minus sign (-) before the quotient.

$=\frac{12}{-4+2}$

$$\begin{gathered} =\frac{12}{-2} \\ =-6 \end{gathered}$$

Then, On solving for $RHS=\frac{a}{b}+\frac{a}{c}$

$=\frac{12}{-4}+\frac{12}{2}$

$=(-3)+\left(6\right)=3$

Comparing both $LHS$and $RHS$ , we get;

$=-6\ne 3$

$=LHS\ne RHS$

Hence, $\frac{a}{(b+c)}\ne \frac{a}{b}+\frac{a}{c}$is verified.