Answer:
\[\frac{3t-2}{4}-\frac{2t+3}{3}=\frac{2}{3}-t\] or, \[\frac{3t-2}{4}-\frac{2t+3}{3}+t=\frac{2}{3}\] [Transposing t to LHS] or, \[\frac{3(3t-2)-4(2t+3)+12t}{12}=\frac{2}{3}\] or, \[9t-6-8t-12+12t=\frac{2}{3}\times 12\] or, \[13t-18=8\] or, \[~13t=8+18\] or, \[13t=26\] or, \[t=\frac{26}{13}\,\,or,\,\,t=2\] To check \[\frac{3t-2}{4}-\frac{2t+3}{3}=\frac{2}{3}-t\] or, \[\frac{3\times 2-2}{4}-\frac{2\times 2+3}{3}=\frac{2}{3}-2\] or, \[\frac{6-2}{4}-\frac{4+3}{3}=\frac{2}{3}-2\] or, \[1-\frac{7}{3}=\frac{2}{3}-2\] or, \[\frac{3-7}{3}=\frac{2-6}{3}\] or, \[\frac{-4}{3}=\frac{-4}{3}\] or, LHS = RHS
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