1. **Stating the problem:** We have the equation $$5(3x - 2) = ax + b$$ where $a$ and $b$ are constants. We want to find: - Part A: Values of $a$ and $b$ for which the equation has exactly one solution. - Part B: Values of $a$ and $b$ for which the equation has no solution. 2. **Rewrite the equation:** $$5(3x - 2) = ax + b$$ Distribute the 5: $$15x - 10 = ax + b$$ 3. **Bring all terms to one side:** $$15x - 10 = ax + b$$ Subtract $ax$ and $b$ from both sides: $$15x - ax - 10 - b = 0$$ Which simplifies to: $$(15 - a)x - (10 + b) = 0$$ 4. **Analyze the equation:** This is a linear equation in $x$ of the form: $$m x + c = 0$$ where $m = 15 - a$ and $c = -(10 + b)$. 5. **Part A: Exactly one solution** A linear equation has exactly one solution if the coefficient of $x$ is not zero: $$m \neq 0 \implies 15 - a \neq 0 \implies a \neq 15$$ Then the solution is: $$x = -\frac{c}{m} = -\frac{-(10 + b)}{15 - a} = \frac{10 + b}{15 - a}$$ 6. **Part B: No solution** A linear equation has no solution if it reduces to a contradiction, which happens when the coefficient of $x$ is zero but the constant term is not zero: $$m = 0 \text{ and } c \neq 0$$ That is: $$15 - a = 0 \implies a = 15$$ and $$-(10 + b) \neq 0 \implies 10 + b \neq 0 \implies b \neq -10$$ 7. **Summary:** - Exactly one solution if $$a \neq 15$$ (any $b$). - No solution if $$a = 15$$ and $$b \neq -10$$. 8. **Additional note:** If $$a = 15$$ and $$b = -10$$, then the equation becomes: $$15x - 10 = 15x - 10$$ which is true for all $x$, so there are infinitely many solutions. **Final answers:** - Part A: $$a \neq 15$$ for exactly one solution. - Part B: $$a = 15$$ and $$b \neq -10$$ for no solution.