Find the value of $t$, if \( \frac{3 t-2}{4}-\frac{2 t+3}{3}=\frac{2}{3}-t \).
Given:
$\frac{3t-2}{4}$- $\frac{2t+3}{3}$ = $\frac{2}{3}-t$
To do:
We have to find the value of t.
Solution:
$\frac{3t-2}{4}$- $\frac{2t+3}{3}$ = $\frac{2}{3}-t$
The LCM of 3 and 4 is 12.
Therefore,
$\frac{3(3t - 2) -4(2t+3)}{12} = \frac{2-3t}{3}$
$\frac{9t - 6 -8t -12}{12} = \frac{2-3t}{3}$
$\frac{t-18}{4}= 2 - 3t$
$t - 18 = 4(2 - 3t)$
$t - 18 = 8 - 12t$
$t+12t=8+18$
$13t = 26$
$t = \frac{26}{13}=2$
Therefore, the value of $t$ is $2$.
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