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 $2$.

Updated on: 10-Oct-2022

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