Question:
$\frac{3 t-2}{3}+\frac{2 t+3}{2}=t+\frac{7}{6}$
Solution:
Given, $\frac{3 t-2}{3}+\frac{2 t+3}{2}=t+\frac{7}{6}$
$\Rightarrow$ $\frac{2(3 t-2)+3(2 t+3)}{6}=\frac{6 t+7}{6}$
$\Rightarrow$ $6 t-4+6 t+9=6 t+7$
$\Rightarrow$ $12 t+5=6 t+7$
$\Rightarrow$ $12 t-6 t=7-5$ [transposing $6 t$ to LHS and 5 to RHS]
$\Rightarrow$ $6 t=2$
$\Rightarrow$ $\frac{6 t}{6}=\frac{2}{6}$ [dividing both sides by 6 ]
$\therefore$ $t=\frac{1}{3}$