Question

Solve the following linear equation. If $\frac{3t-2}{4}-\frac{2t+3}{3}=\frac{2}{3}-t$, then $t$ is equal to

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is B $2$

Given $\frac{3t-2}{4}-\frac{(2t+3)}{3}=\frac{2}{3}-t$

L.C.M. of $4,3,3$ is $12$

Multiplying with 12 on both sides of the equation, we get

$\Rightarrow 3(3t-2)-4(2t+3)=8-12t$

$\Rightarrow 9t-6-(8t+12)=8-12t$

$\Rightarrow t-18=8-12t$

$\Rightarrow 13t=26$

$\Rightarrow t=2$