Question

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Answer 1

Question :

$\sf Solve :$

• $\sf \dfrac{3t-2}{4} - \dfrac{2t + 3}{3} = \dfrac{2}{3} - t$

Solution :

$\sf : \implies \dfrac{3t-2}{4} - \dfrac{2t + 3}{3} = \dfrac{2}{3} - t$

$\sf : \implies \dfrac{3(3t-2)}{4 \times 3} - \dfrac{4(2t + 3)}{3\times 4} = \dfrac{2}{3} - \dfrac{3t}{3}$

$\sf : \implies \dfrac{9t-6}{12} - \dfrac{8t + 12}{12} = \dfrac{2- 3t}{3}$

$\sf : \implies \dfrac{(9t-6) - (8t + 12)}{12} = \dfrac{2- 3t}{3}$

$\sf : \implies \dfrac{9t-6 - 8t - 12}{\cancel{12}} \times \cancel{3} = 2- 3t$

$\sf : \implies \dfrac{t - 18}{4} = 2- 3t$

$\sf : \implies t - 18 = (2- 3t)4$

$\sf : \implies t - 18 = 8 - 12t$

$\sf : \implies t + 12t = 8 + 18$

$\sf : \implies 13t = 26$

$\sf : \implies t = \cancel{\dfrac{26}{13}}$

$\sf : \implies t = 2$

$\Large \underline{\boxed{\sf t = 2}}$

Hence, value of t is 2.