1. **State the problem:** Solve the system of linear equations: $$\frac{2}{3}x + y = -2$$ $$x + 3y = -3$$ Find the values of $x$ and $y$. 2. **Use substitution or elimination method:** Here, we use substitution. From the first equation, solve for $y$: $$y = -2 - \frac{2}{3}x$$ 3. **Substitute $y$ into the second equation:** $$x + 3\left(-2 - \frac{2}{3}x\right) = -3$$ 4. **Simplify the equation:** $$x - 6 - 2x = -3$$ $$\cancel{x} - 6 - \cancel{2x} = -3$$ $$-x - 6 = -3$$ 5. **Solve for $x$:** Add 6 to both sides: $$-x = -3 + 6$$ $$-x = 3$$ Multiply both sides by $-1$: $$x = -3$$ 6. **Substitute $x = -3$ back into the expression for $y$:** $$y = -2 - \frac{2}{3}(-3)$$ $$y = -2 + 2 = 0$$ 7. **Final answer:** $$\boxed{(-3, 0)}$$ This is the solution to the system of equations, representing the point where the two lines intersect on the coordinate plane.