Solve the following



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


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

$\Rightarrow \frac{3 t-2}{4}-\frac{2 t+3}{3}=\frac{2-3 t}{3} \quad(3$ is the L.C.M. of 1 and 3$)$

$\Rightarrow 12\left(\frac{3 t-2}{4}\right)-12\left(\frac{2 t+3}{3}\right)=12\left(\frac{2-3 t}{3}\right) \quad($ multiplying throughout by 12, which is the L. C.M. of 4,3 and 3$)$

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

$\Rightarrow 9 t-6-8 t-12=8-12 t$

$\Rightarrow 9 t-8 t-6-12=8-12 t$

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

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

$\Rightarrow 13 t=26$

$\Rightarrow t=\frac{26}{13}=2$

$\therefore \mathrm{t}=2$

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