Question

What is the value of $x$ in the given equation?
$\frac{(3x+1)}{16}+\frac{(2x-3)}{7}=\frac{(x+3)}{8}+\frac{(3x-1)}{14}$

• A. $4$
• B. $5$
• C. $2$
• D. $3$

Answer 1

Answer: (B)

Given,

$\frac{3x+1}{16}+\frac{2x-3}{7}=\frac{x+3}{8}+\frac{3x-1}{14}$.

Taking LCM on both sides, we get,

LCM of $16,7=$ LCM of $8,14=112$.

Then,

$⟹$ $(\frac{3x+1}{16}\times \frac{112}{112})+(\frac{2x-3}{7}\times \frac{112}{112})=(\frac{x+3}{8}\times \frac{112}{112})+(\frac{3x-1}{14}\times \frac{112}{112})$

$⟹$$\frac{7(3x+1)}{112}+\frac{16(2x-3)}{112}=\frac{14(x+3)}{112}+\frac{8(3x-1)}{112}$

$⟹$ $\frac{7(3x+1)+16(2x-3)}{112}=\frac{14(x+3)+8(3x-1)}{112}$.

Now, cancelling $112$ from both sides, we get,

$⟹$ $7(3x+1)+16(2x-3)=14(x+3)+8(3x-1)$

$⟹$ $21x+7+32x-48=14x+42+24x-8$

$⟹$ $21x+32x-14x-24x=42-8-7+48$

$⟹$ $15x=75$

$⟹$ $x=5$.

Therefore, option $D$ is correct.