1. **State the problem:** We are given the equation $$2(3x + 2) + 2y = 66$$ and the relation $$y = 3x + 1$$. We need to find: a) The value of $$x$$ when $$y = 3x + 1$$. b) The corresponding value of $$y$$. 2. **Use substitution:** Since $$y = 3x + 1$$, substitute this into the first equation: $$2(3x + 2) + 2(3x + 1) = 66$$ 3. **Expand and simplify:** $$2 \times 3x + 2 \times 2 + 2 \times 3x + 2 \times 1 = 66$$ $$6x + 4 + 6x + 2 = 66$$ Combine like terms: $$6x + 6x + 4 + 2 = 66$$ $$12x + 6 = 66$$ 4. **Isolate $$x$$:** Subtract 6 from both sides: $$12x + \cancel{6} - \cancel{6} = 66 - 6$$ $$12x = 60$$ Divide both sides by 12: $$\frac{12x}{\cancel{12}} = \frac{60}{\cancel{12}}$$ $$x = 5$$ 5. **Find $$y$$:** Use $$y = 3x + 1$$: $$y = 3(5) + 1 = 15 + 1 = 16$$ **Final answers:** $$x = 5$$ $$y = 16$$