1. **State the problem:** Solve the system of linear equations using the elimination method: $$\begin{cases} x + y + z = 3 \\ 13x + 2z = 2 \\ -x - 5z = -5 \end{cases}$$ 2. **Goal:** Eliminate variables step-by-step to find values of $x$, $y$, and $z$. 3. **Step 1: Use the third equation to express $x$ in terms of $z$:** $$-x - 5z = -5 \implies -x = -5 + 5z \implies x = 5 - 5z$$ 4. **Step 2: Substitute $x = 5 - 5z$ into the second equation:** $$13x + 2z = 2 \implies 13(5 - 5z) + 2z = 2$$ Simplify: $$65 - 65z + 2z = 2 \implies 65 - 63z = 2$$ 5. **Step 3: Solve for $z$:** $$65 - 63z = 2 \implies -63z = 2 - 65 = -63 \implies z = \frac{-63}{-63} = 1$$ 6. **Step 4: Substitute $z = 1$ back into $x = 5 - 5z$ to find $x$:** $$x = 5 - 5(1) = 5 - 5 = 0$$ 7. **Step 5: Substitute $x = 0$ and $z = 1$ into the first equation to find $y$:** $$x + y + z = 3 \implies 0 + y + 1 = 3 \implies y = 3 - 1 = 2$$ **Final answer:** $$\boxed{(x, y, z) = (0, 2, 1)}$$ This means the solution to the system is $x=0$, $y=2$, and $z=1$.