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Solve the following linear equation.
$ \frac{3 t-2}{4}-\frac{2 t+3}{3}=\frac{2}{3}-t $.
Given:
$\frac{3 t-2}{4}-\frac{2 t+3}{3}=\frac{2}{3}-t$.
To do:
We have to solve the given linear equation.
Solution:
$\frac{3 t-2}{4}-\frac{2 t+3}{3}=\frac{2}{3}-t$
This implies,
$\frac{3 t-2}{4}-\frac{2 t+3}{3}+t=\frac{2}{3}$
$\frac{3(3t-2) - 4( 2t+3) +12(t)}{12}=\frac{2}{3}$
$9t-6-8t-12+12t= \frac{2}{3}\times12$
$13t-18= 2(4)$
$13t=8+18$
$13t=26$
$t =\frac{26}{13}$
$t = 2$
The value of $t$ is $2$.
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