Question

Solve $\frac{x+3}{3}-\frac{x-2}{2}=1$ . Hence find $p$, if $\frac{1}{x}+p=1$

Answer 1

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

Multiplying above eq. by $6$, we get

$\Rightarrow 2(x+3)-3(x-2)=6$

$\Rightarrow 2x+6-3x+6=6$

$\Rightarrow 6-x=0$

$\Rightarrow x=6$

As $\frac{1}{x}+p=1$

$\Rightarrow p=1-\frac{1}{6}$

$\therefore$ $p=\frac{5}{6}$