1. **State the problem:** Solve the system of linear equations: $$4x - 3y = 12$$ $$2x + 8y = 24$$ and solve the inequality: $$\frac{|2x + 4|}{2} \leq 6$$ 2. **Solve the system of equations:** From the first equation, the user substituted $x=6$: $$4(6) - 3y = 12$$ $$24 - 3y = 12$$ Subtract 24 from both sides: $$-3y = 12 - 24$$ $$-3y = -12$$ Divide both sides by $-3$: $$y = \frac{-12}{-3} = 4$$ 3. **Check the second equation with $y=4$:** $$2x + 8(4) = 24$$ $$2x + 32 = 24$$ Subtract 32 from both sides: $$2x = 24 - 32$$ $$2x = -8$$ Divide both sides by 2: $$x = -4$$ 4. **Verify the solution:** Plug $x=-4$ and $y=4$ into the first equation: $$4(-4) - 3(4) = -16 - 12 = -28 \neq 12$$ So the initial substitution $x=6$ was incorrect for the system. Let's solve the system properly using substitution or elimination. 5. **Solve by elimination:** Multiply the first equation by 8 and the second by 3 to align $y$ coefficients: $$8(4x - 3y) = 8(12) \Rightarrow 32x - 24y = 96$$ $$3(2x + 8y) = 3(24) \Rightarrow 6x + 24y = 72$$ Add the two equations: $$32x - 24y + 6x + 24y = 96 + 72$$ $$38x = 168$$ Divide both sides by 38: $$x = \frac{168}{38} = \frac{84}{19}$$ 6. **Find $y$ using $x=\frac{84}{19}$ in the first equation:** $$4\left(\frac{84}{19}\right) - 3y = 12$$ $$\frac{336}{19} - 3y = 12$$ Subtract $\frac{336}{19}$ from both sides: $$-3y = 12 - \frac{336}{19} = \frac{228}{19} - \frac{336}{19} = -\frac{108}{19}$$ Divide both sides by $-3$: $$y = \frac{-\frac{108}{19}}{-3} = \frac{108}{57} = \frac{36}{19}$$ 7. **Solution to the system:** $$x = \frac{84}{19}, \quad y = \frac{36}{19}$$ 8. **Solve the inequality:** $$\frac{|2x + 4|}{2} \leq 6$$ Multiply both sides by 2: $$|2x + 4| \leq 12$$ This means: $$-12 \leq 2x + 4 \leq 12$$ Subtract 4 from all parts: $$-16 \leq 2x \leq 8$$ Divide all parts by 2: $$-8 \leq x \leq 4$$ 9. **Final answers:** - System solution: $$\boxed{x = \frac{84}{19}, y = \frac{36}{19}}$$ - Inequality solution: $$\boxed{-8 \leq x \leq 4}$$