Question

Solve the linear equation:

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

Answer 1

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

${\displaystyle \frac{3t}{4}-\frac{2}{4}-[\frac{2t}{3}+\frac{2}{3}]=\frac{2}{3}-t}$

${\displaystyle \frac{3t}{4}-\frac{1}{2}-\frac{2t}{3}-1=\frac{2}{3}-t}$

${\displaystyle \frac{3t}{4}-\frac{2t}{3}+t=\frac{2}{3}+\frac{1}{2}+1}$

${\displaystyle \frac{9t-8t+12t}{12}=\frac{4+3+6}{6}}$

${\displaystyle \frac{13t}{12}=\frac{13}{6}}$

${\displaystyle t=\frac{12}{6}}$

t = 2