Question

Solve for $x$: $\frac{2}{3}(4x-1)-[4x-\frac{1-3x}{2}]=\frac{x-7}{2}$.

• A. $0$
• B. $1$
• C. $2$
• D. $5$

Answer 1

The correct option is B $1$

Given, $\frac{2}{3}(4x-1)-[4x-\frac{1-3x}{2}]=\frac{x-7}{2}$

$\Rightarrow \frac{8x}{3}-\frac{2}{3}-\left[\frac{8x-1+3x}{2}\right]=\frac{x-7}{2}$

$\Rightarrow \frac{8x}{3}-\frac{2}{3}-\left[\frac{11x-1}{2}\right]=\frac{x-7}{2}$.

Transposing $x$ terms to one side, we get

$\Rightarrow \frac{8x}{3}-\frac{11x}{2}-\frac{x}{2}=\frac{2}{3}-\frac{7}{2}-\frac{1}{2}$

$\Rightarrow \frac{16x-33x-3x}{6}=\frac{4-21-3}{6}$

$\Rightarrow \frac{-20x}{6}=\frac{-20}{6}$

$\Rightarrow x=1$.

Hence, $B$ is the correct option.