1. **State the problem:** We have two equations: $$x = 4y + 8$$ and $$y = 2x + 12$$ We want to find the values of $x$ and $y$ that make both equations true at the same time. 2. **Rewrite the first equation to express $y$ in terms of $x$:** $$x = 4y + 8$$ Subtract 8 from both sides: $$x - 8 = 4y$$ Divide both sides by 4: $$y = \frac{\cancel{x - 8}}{\cancel{4}} = \frac{x - 8}{4}$$ 3. **Substitute this expression for $y$ into the second equation:** $$y = 2x + 12$$ Replace $y$ with $\frac{x - 8}{4}$: $$\frac{x - 8}{4} = 2x + 12$$ 4. **Solve for $x$:** Multiply both sides by 4 to get rid of the denominator: $$x - 8 = 4(2x + 12)$$ Simplify the right side: $$x - 8 = 8x + 48$$ 5. **Get all $x$ terms on one side and constants on the other:** Subtract $x$ from both sides: $$-8 = 7x + 48$$ Subtract 48 from both sides: $$-8 - 48 = 7x$$ $$-56 = 7x$$ 6. **Divide both sides by 7 to solve for $x$:** $$\frac{\cancel{-56}}{\cancel{7}} = x$$ $$x = -8$$ 7. **Find $y$ by substituting $x = -8$ back into $y = \frac{x - 8}{4}$:** $$y = \frac{-8 - 8}{4} = \frac{-16}{4} = -4$$ **Final answer:** $$x = -8, \quad y = -4$$ These values satisfy both equations.