1. **Problem 6:** Solve the system using substitution or elimination method. Given: $$6x + 4y = 62$$ $$3x + 4y = 47$$ 2. We will use the elimination method here because both equations have the same coefficient for $y$. 3. Subtract the second equation from the first to eliminate $y$: $$\cancel{6x} + 4y - \cancel{3x} - 4y = 62 - 47$$ $$3x = 15$$ 4. Solve for $x$: $$x = \frac{15}{3} = 5$$ 5. Substitute $x=5$ into the second equation to find $y$: $$3(5) + 4y = 47$$ $$15 + 4y = 47$$ 6. Solve for $y$: $$4y = 47 - 15$$ $$4y = 32$$ $$y = \frac{32}{4} = 8$$ 7. **Answer for problem 6:** $x=5$, $y=8$ --- 8. **Problem 4:** Solve the system using substitution or elimination method. Given: $$2x + y = -3$$ $$2x + 3y = 3$$ 9. We will use the substitution method here. 10. From the first equation, express $y$ in terms of $x$: $$y = -3 - 2x$$ 11. Substitute this expression for $y$ into the second equation: $$2x + 3(-3 - 2x) = 3$$ 12. Simplify and solve for $x$: $$2x - 9 - 6x = 3$$ $$\cancel{2x} - 9 - \cancel{6x} = 3$$ $$-4x - 9 = 3$$ 13. Add 9 to both sides: $$-4x = 3 + 9$$ $$-4x = 12$$ 14. Divide both sides by $-4$: $$x = \frac{12}{-4} = -3$$ 15. Substitute $x = -3$ back into the expression for $y$: $$y = -3 - 2(-3) = -3 + 6 = 3$$ 16. **Answer for problem 4:** $x = -3$, $y = 3$