1. **State the problem:** We need to determine which of the given equations does NOT have $\frac{3}{2}$ as a solution. 2. **Recall the method:** To check if $x=\frac{3}{2}$ is a solution, substitute $x=\frac{3}{2}$ into each equation and verify if the equation holds true. 3. **Check equation A: $2x + 3 = 6$** Substitute $x=\frac{3}{2}$: $$2 \times \frac{3}{2} + 3 = 3 + 3 = 6$$ This is true, so $x=\frac{3}{2}$ is a solution for A. 4. **Check equation B: $x + \frac{5}{2} = 4$** Substitute $x=\frac{3}{2}$: $$\frac{3}{2} + \frac{5}{2} = \frac{8}{2} = 4$$ This is true, so $x=\frac{3}{2}$ is a solution for B. 5. **Check equation C: $x + \frac{2}{3} = 1$** Substitute $x=\frac{3}{2}$: $$\frac{3}{2} + \frac{2}{3} = \frac{9}{6} + \frac{4}{6} = \frac{13}{6} \neq 1$$ Since $\frac{13}{6} \neq 1$, $x=\frac{3}{2}$ is NOT a solution for C. 6. **Check equation D: $4x + 4 = 10$** Substitute $x=\frac{3}{2}$: $$4 \times \frac{3}{2} + 4 = 6 + 4 = 10$$ This is true, so $x=\frac{3}{2}$ is a solution for D. **Final answer:** Equation C does NOT have $\frac{3}{2}$ as a solution.