Question

Solve the following linear equation:

$\frac{(3t-2)}{4}-\frac{(2t+3)}{3}=\frac{2}{3}-t$

Answer 1

Given, $\frac{(3t-2)}{4}-\frac{(2t+3)}{3}=\frac{2}{3}-t$
$$\begin{gathered} \Rightarrow \frac{3(3t-2)-4(2t+3)}{12}=\frac{2-3t}{3} \\ \Rightarrow \frac{9t-6-8t-12}{12}=\frac{2-3t}{3} \\ \Rightarrow \frac{t-18}{12}=\frac{2-3t}{3} \\ \Rightarrow 3(t-18)=12(2-3t) \\ \Rightarrow 3t-54=24-36t \\ \Rightarrow 3t+36t=24+54 \\ \Rightarrow 39t=78 \\ \Rightarrow t=\frac{78}{39} \\ \Rightarrow t=2 \end{gathered}$$

Hence, $\mathit{t}\mathbf{}\mathbf{=}\mathbf{}\mathbf{2}$ is the solution of given equation.