Question

Find the value of $t$, $\frac{3t-2}{4}-\frac{2t+3}{3}=\frac{2}{3}-t$.

Answer 1

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

$\Rightarrow \frac{3(3t-2)-4(2t+3)}{4\times 3}=\frac{2}{3}-t$ …[Cross-multiplying the denominators on the LHS]

$\Rightarrow \frac{9t-6-8t-12}{12}=\frac{2}{3}-t$ ...[By Distribution Law]

$\Rightarrow \frac{t-18}{12}=\frac{2}{3}-t$

$\Rightarrow \frac{t}{12}-\frac{18}{12}=\frac{2}{3}-t$.

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

$\Rightarrow \frac{t}{12}+t=\frac{2}{3}+\frac{18}{12}$

$\Rightarrow \frac{t+12t}{12}=\frac{2}{3}+\frac{3}{2}$ ...[On simplifying]

$\Rightarrow \frac{13t}{12}=\frac{4+9}{6}$

$\Rightarrow \frac{13t}{12}=\frac{13}{6}$

$\Rightarrow t=2$.