Question

The value of $x$ in $\frac{3}{4}(7x-1)-\left(2x-\frac{1-x}{2}\right)=x+\frac{3}{2}$ is ___________.

• A. $2$
• B. $3$
• C. $1$
• D. $0$

Answer 1

The correct option is C $1$

Given, $\frac{3}{4}(7x-1)-\left(2x-\frac{1-x}{2}\right)=x+\frac{3}{2}$

$\Rightarrow$ $\frac{21x-3}{4}-\left(\frac{4x-1+x}{2}\right)=\frac{2x+3}{2}$

$\Rightarrow$ $\frac{21x-3}{4}-\left(\frac{5x-1}{2}\right)=\frac{2x+3}{2}$

$\Rightarrow$ $\frac{42x-6-20x+4}{8}=\frac{2x+3}{2}$

$\Rightarrow$ $\frac{22x-2}{8}=\frac{2x+3}{2}$

$\Rightarrow$ $\frac{11x}{4}-\frac{1}{4}=\frac{4x}{4}+\frac{6}{4}$.

Transposing $x$ terms to one side, we get,

$\Rightarrow$ $\frac{11x}{4}-\frac{4x}{4}=\frac{1}{4}+\frac{6}{4}$

$\Rightarrow$ $7x=7$

$\Rightarrow$ $x=1$.

Hence, option $C$ is correct.