Question

$4[8x-3(8x+4\left)\right]=8(-4-\frac{2x}{3})$. Find $x$.

• A. $\frac{11}{3}$
• B. $-\frac{11}{3}$
• C. $\frac{3}{11}$
• D. $-\frac{3}{11}$

Answer 1

The correct option is D $-\frac{3}{11}$

Given, $4[8x-3(8x+4\left)\right]=8(-4-\frac{2x}{3})$.

Opening the brackets on both sides, we get,
$4\times 8x-4\times 3(8x+4)=-8\times 4-8\times \frac{2}{3}x$ ...[By Distribution Law]

$⟹32x-(12\times 8)x-12\times 4=-32-\frac{16}{3}x$ ...[By Distribution Law]

$⟹32x-96x-48=-32-\frac{16}{3}x$.

Transposing $x$ terms to one side, we get,
$⟹32x-96x+\frac{16}{3}x=-32+48$

$⟹-64x+\frac{16}{3}x=16$ ...[Taking LCM]

$⟹(-64x\times 3)+16x=16\times 3$

$⟹-192x+16x=48$

$⟹-176x=48$

$⟹x=\frac{-48}{176}=-\frac{3}{11}$.

Hence, option $D$ is correct.