Question
Solve:
$\frac{\text{3t}-2}{4}-\frac{\text{2t}+3}{3}=\frac{2}{3}-\text{t}$

Answer

$\frac{\text{3t}-2}{4}-\frac{\text{2t}+3}{3}=\frac{2}{3}-\text{t}$
$\Rightarrow\frac{\text{3t}-2}{4}-\frac{\text{2t}+3}{3}=\frac{2-3\text{t}}{3}$
$(3$ is the $L.C.M.$ of $1$ and $3)$
$\Rightarrow12\Big(\frac{\text{3t}-2}{4}\Big)-12\Big(\frac{\text{2t}+3}{3}\Big)=12\Big(\frac{2-3\text{t}}{3}\Big)$
(multiplying throughout by $12,$ which is the $L.C.M.$ of $4, 3$ and $3):$
$\Rightarrow3(3\text{t}-2)-4(2\text{t}+3)=4(2-3\text{t})$
$\Rightarrow9\text{t} - 6 - 8\text{t} - 12 = 8 - 12\text{t}$
$\Rightarrow 9\text{t} - 8\text{t} - 6 - 12 = 8 - 12\text{t}$
$\Rightarrow\text{t} - 18 = 8 - 12\text{t}$
$\Rightarrow\text{t} + 12\text{t} = 18 + 8$
$\Rightarrow13\text{t} = 26$
$\Rightarrow\text{t} = \frac{26}{13} = 2$
$\therefore \text{t} = 2$

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