1. **State the problem:** We are given the system of equations: $$\begin{cases} x + 3y = 10 \\ -2x - 2y = 4 \end{cases}$$ We need to find which of the given systems has the same solution as this system. 2. **Solve the original system:** From the first equation: $$x = 10 - 3y$$ Substitute into the second equation: $$-2(10 - 3y) - 2y = 4$$ Simplify: $$-20 + 6y - 2y = 4$$ $$4y - 20 = 4$$ Add 20 to both sides: $$4y = 24$$ Divide both sides by 4: $$\cancel{4}y = \cancel{4}6$$ $$y = 6$$ Substitute back to find $x$: $$x = 10 - 3(6) = 10 - 18 = -8$$ So the solution is: $$(x, y) = (-8, 6)$$ 3. **Check each system to see if it has the same solution:** **Option 1:** $$\begin{cases} -x + y = 6 \\ 2x + 6y = 20 \end{cases}$$ Substitute $x = -8$, $y = 6$: $$-(-8) + 6 = 8 + 6 = 14 \neq 6$$ So option 1 does not have the same solution. **Option 2:** $$\begin{cases} -x + y = 14 \\ 2x + 6y = 20 \end{cases}$$ Substitute $x = -8$, $y = 6$: $$-(-8) + 6 = 8 + 6 = 14$$ $$2(-8) + 6(6) = -16 + 36 = 20$$ Both equations are satisfied, so option 2 has the same solution. **Option 3:** $$\begin{cases} x + y = 6 \\ 2x + 6y = 20 \end{cases}$$ Substitute $x = -8$, $y = 6$: $$-8 + 6 = -2 \neq 6$$ So option 3 does not have the same solution. **Option 4:** $$x + y = 14$$ Substitute $x = -8$, $y = 6$: $$-8 + 6 = -2 \neq 14$$ So option 4 does not have the same solution. **Final answer:** Option 2 has the same solution as the original system.