1. **State the problem:** Solve the system of linear equations: $$x = 16 - 4y$$ $$3x + 4y = 8$$ 2. **Use substitution method:** Since $x$ is already expressed in terms of $y$ from the first equation, substitute $x = 16 - 4y$ into the second equation. 3. **Substitute and simplify:** $$3(16 - 4y) + 4y = 8$$ $$48 - 12y + 4y = 8$$ $$48 - 8y = 8$$ 4. **Isolate $y$:** $$48 - 8y = 8$$ Subtract 48 from both sides: $$\cancel{48} - 8y = 8 - \cancel{48}$$ $$-8y = -40$$ Divide both sides by $-8$: $$\frac{-8y}{\cancel{-8}} = \frac{-40}{\cancel{-8}}$$ $$y = 5$$ 5. **Find $x$ using $y=5$:** $$x = 16 - 4(5) = 16 - 20 = -4$$ 6. **Final answer:** $$x = -4, \quad y = 5$$ This means the solution to the system is the point $(-4, 5)$ where both lines intersect.