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

Given that,\(\frac{3t-2}4-\frac{2t+3}3=\frac{2}3-t\)
LCM of denominators 3 and 4=12

So, \(\frac{3t-2}4×12-\frac{2t+3}3×12=\frac{2}3×12-t×12 \quad \)[Multiplying both the sides by 12]

\(\Rightarrow (3t – 2) × 3 – (2t + 3) × 4 = 2 × 4 – 12t\)

\(\Rightarrow 9t – 6 – 8t – 12 = 8 – 12t \quad \)[Solving the brackets]

\(\Rightarrow t – 18 = 8 – 12t \quad \)

\(\Rightarrow t + 12t = 8 + 18 \)[Transposing 12t to LHS and 18 to RHS]

\(\Rightarrow 13t = 26\)

\(\Rightarrow t = 2 \)[Transposing 13 to RHS]

Hence t = 2 is the required solution.