1. The problem is to solve the system of equations: $$y = -x - 2$$ $$y = \frac{2}{3}x + 3$$ 2. Since both expressions equal $y$, set them equal to each other: $$-x - 2 = \frac{2}{3}x + 3$$ 3. To isolate $x$, first subtract 3 from both sides: $$-x - 2 - 3 = \frac{2}{3}x + 3 - 3$$ $$-x - 5 = \frac{2}{3}x$$ 4. Add $x$ to both sides to get all $x$ terms on one side: $$-x - 5 + x = \frac{2}{3}x + x$$ $$-5 = \frac{2}{3}x + x$$ 5. Convert $x$ to a fraction with denominator 3: $$-5 = \frac{2}{3}x + \frac{3}{3}x = \frac{5}{3}x$$ 6. Solve for $x$ by dividing both sides by $\frac{5}{3}$: $$x = \frac{-5}{\frac{5}{3}}$$ Show cancellation: $$x = -5 \times \cancel{\frac{3}{5}} = -3$$ 7. Substitute $x = -3$ back into one of the original equations to find $y$. Using $y = -x - 2$: $$y = -(-3) - 2 = 3 - 2 = 1$$ 8. The solution to the system is: $$(x, y) = (-3, 1)$$