1. **State the problem:** Solve the simultaneous equations: $$\begin{cases} x + y = 2 \\ y + 2z = 12 \\ 5x + 7y + z = 13 \end{cases}$$ 2. **Use substitution or elimination method:** We will use substitution here. 3. From the first equation, express $x$ in terms of $y$: $$x = 2 - y$$ 4. From the second equation, express $z$ in terms of $y$: $$y + 2z = 12 \implies 2z = 12 - y \implies z = \frac{12 - y}{2}$$ 5. Substitute $x$ and $z$ into the third equation: $$5x + 7y + z = 13$$ Substitute $x = 2 - y$ and $z = \frac{12 - y}{2}$: $$5(2 - y) + 7y + \frac{12 - y}{2} = 13$$ 6. Simplify the equation: $$10 - 5y + 7y + \frac{12 - y}{2} = 13$$ Combine like terms: $$10 + 2y + \frac{12 - y}{2} = 13$$ 7. Multiply both sides by 2 to clear the denominator: $$2 \times \left(10 + 2y + \frac{12 - y}{2}\right) = 2 \times 13$$ $$2 \times 10 + 2 \times 2y + \cancel{2} \times \frac{12 - y}{\cancel{2}} = 26$$ $$20 + 4y + 12 - y = 26$$ 8. Combine like terms: $$20 + 12 + 4y - y = 26$$ $$32 + 3y = 26$$ 9. Solve for $y$: $$3y = 26 - 32$$ $$3y = -6$$ $$y = \frac{-6}{3} = -2$$ 10. Substitute $y = -2$ back into expressions for $x$ and $z$: $$x = 2 - (-2) = 2 + 2 = 4$$ $$z = \frac{12 - (-2)}{2} = \frac{12 + 2}{2} = \frac{14}{2} = 7$$ **Final answer:** $$x = 4, \quad y = -2, \quad z = 7$$