question_answer

                   Simplify and solve the following : \[\frac{3t-2}{4}-\frac{2t+3}{3}=\frac{2}{3}-t\]

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