1. **Problem 1:** Determine which equation is NOT equivalent to $9 |x - 1| = 27$. 2. Start by simplifying the given equation: $$9 |x - 1| = 27$$ Divide both sides by 9: $$|x - 1| = \frac{27}{9} = 3$$ 3. The equivalent equations are those that represent the same condition $|x - 1| = 3$. 4. Check each option: - $|x - 1| = 3$ is equivalent. - $x = 4$ and $x = -2$ come from solving $|x - 1| = 3$ because $x - 1 = 3$ or $x - 1 = -3$ gives $x = 4$ or $x = -2$. - $|x + 1| = -3$ is NOT possible because absolute values cannot be negative. - $|x - 1| = \frac{27}{9} = 3$ is equivalent. 5. **Answer for Problem 1:** $|x + 1| = -3$ is NOT equivalent. 6. **Problem 2:** Find the value of $x + y$ given the system: $$3x - 2y = 21$$ $$-4x + 5y = -42$$ 7. Use the elimination method: Multiply the first equation by 5: $$15x - 10y = 105$$ Multiply the second equation by 2: $$-8x + 10y = -84$$ 8. Add the two equations: $$(15x - 10y) + (-8x + 10y) = 105 + (-84)$$ $$7x = 21$$ $$x = 3$$ 9. Substitute $x=3$ into the first equation: $$3(3) - 2y = 21$$ $$9 - 2y = 21$$ $$-2y = 12$$ $$y = -6$$ 10. Calculate $x + y$: $$3 + (-6) = -3$$ 11. **Answer for Problem 2:** $x + y = -3$.