1. **State the problem:** Kyle is adding the two equations: $$4x - 2y = 18$$ $$-3x + 2y = -11$$ 2. **Add the equations:** When adding, combine like terms: $$ (4x - 3x) + (-2y + 2y) = 18 + (-11) $$ Simplify: $$ x + 0 = 7 $$ So, $$ x = 7 $$ 3. **Answer for question 1:** The sum of the two equations is $x = 7$, which corresponds to option b. --- 4. **State the problem:** Vivi wants to solve the system: $$4x + 6y = -14$$ $$-x + 3y = -10$$ 5. **Goal:** Use elimination to eliminate one variable by making coefficients opposites. 6. **Check options:** - Multiply the 2nd equation by 2: $$2(-x + 3y) = 2(-10) \\ -2x + 6y = -20$$ Now the system is: $$4x + 6y = -14$$ $$-2x + 6y = -20$$ - Multiply the 2nd equation by -4: $$-4(-x + 3y) = -4(-10) \\ 4x - 12y = 40$$ - Multiply the 2nd equation by 4: $$4(-x + 3y) = 4(-10) \\ -4x + 12y = -40$$ 7. **Conclusion:** Both multiplying by 2 and by 4 are valid first steps to create coefficients that can be eliminated. So option d (both a and c) is correct. --- 8. **Solve system 3:** $$-x + 5y = -29$$ $$2x - 5y = 38$$ Add the two equations: $$(-x + 2x) + (5y - 5y) = -29 + 38$$ $$x + 0 = 9$$ $$x = 9$$ Substitute $x=9$ into first equation: $$-9 + 5y = -29 \\ 5y = -20 \\ y = -4$$ Answer: $(9, -4)$ --- 9. **Solve system 4:** $$3x + 3y = 21$$ $$-3x + 2y = 19$$ Add the two equations: $$(3x - 3x) + (3y + 2y) = 21 + 19$$ $$0 + 5y = 40$$ $$5y = 40 \\ y = 8$$ Substitute $y=8$ into first equation: $$3x + 3(8) = 21 \\ 3x + 24 = 21 \\ 3x = -3 \\ x = -1$$ Answer: $(-1, 8)$ --- 10. **Solve system 5:** $$8x - 2y = -10$$ $$4x - y = -5$$ Multiply second equation by -2: $$-2(4x - y) = -2(-5) \\ -8x + 2y = 10$$ Add to first equation: $$(8x - 8x) + (-2y + 2y) = -10 + 10$$ $$0 = 0$$ This means the system has infinitely many solutions. --- 11. **Solve system 6:** $$3x - 5y = -9$$ $$2x + 2y = 10$$ Multiply second equation by 5: $$5(2x + 2y) = 5(10) \\ 10x + 10y = 50$$ Multiply first equation by 2: $$2(3x - 5y) = 2(-9) \\ 6x - 10y = -18$$ Add the two equations: $$(6x + 10x) + (-10y + 10y) = -18 + 50$$ $$16x + 0 = 32 \\ 16x = 32 \\ x = 2$$ Substitute $x=2$ into second original equation: $$2(2) + 2y = 10 \\ 4 + 2y = 10 \\ 2y = 6 \\ y = 3$$ Answer: $(2, 3)$ --- 12. **Solve system 7:** $$3(4x - 6y) = 3(-2) \\ 12x - 18y = -6$$ $$-4(3x + 5y) = -4(8) \\ -12x - 20y = -32$$ Add the two equations: $$(12x - 12x) + (-18y - 20y) = -6 + (-32)$$ $$0 - 38y = -38 \\ -38y = -38 \\ y = 1$$ Substitute $y=1$ into first equation: $$12x - 18(1) = -6 \\ 12x - 18 = -6 \\ 12x = 12 \\ x = 1$$ Answer: $(1, 1)$ --- **Summary of answers:** 1. b. $x=7$ 2. d. Both a and c could work 3. $(9, -4)$ 4. $(-1, 8)$ 5. Infinitely many solutions 6. $(2, 3)$ 7. $(1, 1)$