Question

Solve each of the following equations and also check your results in each case:
$\frac{1-2x}{7}-\frac{2-3x}{8}=\frac{3}{2}+\frac{x}{4}$

Answer 1

Given: $\frac{1-2x}{7}-\frac{2-3x}{8}=\frac{3}{2}+\frac{x}{4}$

$\Rightarrow \frac{1-2x}{7}-\frac{2-3x}{8}-\frac{x}{4}=\frac{3}{2}$

$\Rightarrow \frac{(1-2x)8-(2-3x)7-14x}{56}=\frac{3}{2}$

$\Rightarrow \frac{8-16x-14+21x-14x}{56}=\frac{3}{2}$

$\Rightarrow \frac{-9x-6}{56}=\frac{3}{2}$

$\Rightarrow 2(-9x-6)=3\left(56\right)$

$\Rightarrow -18x-12=168$

$\Rightarrow -18x=168+12$

$\Rightarrow -18x=180$

$\Rightarrow x=\frac{180}{-18}$

$\Rightarrow x=-10$

To check $\frac{1-2x}{7}-\frac{2-3x}{8}=\frac{3}{2}+\frac{x}{4}$ for $x=-10$

$L.H.S=\frac{1-2x}{7}-\frac{2-3x}{8}$

$=\frac{1-2(-10)}{7}-\frac{2-3(-10)}{8}$

$=\frac{21}{7}-\frac{32}{8}$

$=3-4=-1$

$R.H.S=\frac{3}{2}+\frac{x}{4}$

$=\frac{3}{2}+\frac{-10}{4}$

$=\frac{3}{2}-\frac{5}{2}$

$=\frac{3-5}{2}$

$=-\frac{2}{2}$

$=-1$

$\therefore L.H.S=R.H.S$

Hence, the given equation is verified for $x=-10$.