1. **State the problem:** Solve the system of equations algebraically: $$3x - 2y = 2$$ $$5x - 5y = 10$$ 2. **Choose a method:** We will use the elimination method to solve for $x$ and $y$. 3. **Make coefficients of one variable equal:** Multiply the first equation by 5 and the second equation by 2 to align coefficients of $x$: $$5(3x - 2y) = 5(2) \Rightarrow 15x - 10y = 10$$ $$2(5x - 5y) = 2(10) \Rightarrow 10x - 10y = 20$$ 4. **Subtract the second equation from the first:** $$\begin{aligned} (15x - 10y) - (10x - 10y) &= 10 - 20 \\ 15x - 10y - 10x + 10y &= -10 \\ (15x - 10x) + (-10y + 10y) &= -10 \\ 5x + \cancel{-10y + 10y} &= -10 \\ 5x &= -10 \end{aligned}$$ 5. **Solve for $x$:** $$x = \frac{-10}{5} = -2$$ 6. **Substitute $x = -2$ into the first original equation:** $$3(-2) - 2y = 2$$ $$-6 - 2y = 2$$ 7. **Solve for $y$:** $$-2y = 2 + 6$$ $$-2y = 8$$ $$y = \frac{8}{-2} = -4$$ 8. **Final solution:** $$x = -2, \quad y = -4$$ This means the solution to the system is the point $(-2, -4)$ where both equations intersect.