1. **State the problem:** We are given a system of three equations: $$z = 2x + 3y - 17$$ $$y = -3z - 7$$ $$x = \frac{z + 2}{2}$$ We want to find the values of $x$, $y$, and $z$ that satisfy all three equations simultaneously. 2. **Substitute $x$ and $y$ in terms of $z$ into the first equation:** From the third equation, $x = \frac{z + 2}{2}$. From the second equation, $y = -3z - 7$. Substitute these into the first equation: $$z = 2\left(\frac{z + 2}{2}\right) + 3(-3z - 7) - 17$$ 3. **Simplify the right side:** $$z = (z + 2) + 3(-3z - 7) - 17$$ $$z = z + 2 - 9z - 21 - 17$$ 4. **Combine like terms:** $$z = z - 9z + 2 - 21 - 17$$ $$z = -8z - 36$$ 5. **Bring all terms to one side:** $$z + 8z = -36$$ $$9z = -36$$ 6. **Solve for $z$:** $$z = \frac{-36}{9}$$ $$z = -4$$ 7. **Find $x$ using $z = -4$:** $$x = \frac{-4 + 2}{2} = \frac{-2}{2} = -1$$ 8. **Find $y$ using $z = -4$:** $$y = -3(-4) - 7 = 12 - 7 = 5$$ **Final answer:** $$x = -1, \quad y = 5, \quad z = -4$$