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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Updated on: 10-Oct-2022

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