1. **State the problem:** We are given a system of linear equations: $$2x + 2y = -2$$ $$\frac{1}{3}x + y = 5$$ We want to solve for $x$ and $y$. 2. **Use substitution or elimination method:** Let's use substitution. From the second equation: $$\frac{1}{3}x + y = 5 \implies y = 5 - \frac{1}{3}x$$ 3. **Substitute $y$ into the first equation:** $$2x + 2\left(5 - \frac{1}{3}x\right) = -2$$ 4. **Simplify the equation:** $$2x + 10 - \frac{2}{3}x = -2$$ 5. **Combine like terms:** $$\left(2 - \frac{2}{3}\right)x + 10 = -2$$ 6. **Calculate the coefficient:** $$2 - \frac{2}{3} = \frac{6}{3} - \frac{2}{3} = \frac{4}{3}$$ So, $$\frac{4}{3}x + 10 = -2$$ 7. **Isolate $x$:** $$\frac{4}{3}x = -2 - 10$$ $$\frac{4}{3}x = -12$$ 8. **Divide both sides by $\frac{4}{3}$:** $$x = -12 \div \frac{4}{3} = -12 \times \frac{3}{4}$$ 9. **Simplify:** $$x = -12 \times \frac{3}{4} = -\cancel{12}^3 \times \frac{3}{\cancel{4}^1} = -9$$ 10. **Find $y$ using $y = 5 - \frac{1}{3}x$:** $$y = 5 - \frac{1}{3}(-9) = 5 + 3 = 8$$ **Final answer:** $$x = -9, \quad y = 8$$