1. **Problem:** Find the number of solutions for the equation $x + 5.5 + 8 = 5x - 13.5 - 4x$. 2. **Step 1:** Simplify both sides. $$x + 13.5 = 5x - 13.5 - 4x$$ $$x + 13.5 = (5x - 4x) - 13.5$$ $$x + 13.5 = x - 13.5$$ 3. **Step 2:** Subtract $x$ from both sides. $$\cancel{x} + 13.5 = \cancel{x} - 13.5$$ $$13.5 = -13.5$$ 4. **Step 3:** Since $13.5 \neq -13.5$, this is a contradiction. 5. **Conclusion:** No solution exists for this equation. 2. **Problem:** Find the number of solutions for the equation $4 \left(\frac{1}{2} x + 3\right) = 3x + 12 - x$. 6. **Step 1:** Distribute 4 on the left side. $$4 \times \frac{1}{2} x + 4 \times 3 = 3x + 12 - x$$ $$2x + 12 = 3x + 12 - x$$ 7. **Step 2:** Simplify the right side. $$2x + 12 = (3x - x) + 12$$ $$2x + 12 = 2x + 12$$ 8. **Step 3:** Subtract $2x$ from both sides. $$\cancel{2x} + 12 = \cancel{2x} + 12$$ $$12 = 12$$ 9. **Step 4:** This is always true, so there are infinitely many solutions. 3. **Problem:** Find the number of solutions for the equation $2 (6x + 9 - 3x) = 5x + 21$. 10. **Step 1:** Simplify inside the parentheses. $$2 (3x + 9) = 5x + 21$$ 11. **Step 2:** Distribute 2. $$6x + 18 = 5x + 21$$ 12. **Step 3:** Subtract $5x$ from both sides. $$6x - 5x + 18 = \cancel{5x} + 21 - \cancel{5x}$$ $$x + 18 = 21$$ 13. **Step 4:** Subtract 18 from both sides. $$x + \cancel{18} = 21 - \cancel{18}$$ $$x = 3$$ 14. **Conclusion:** There is exactly one solution, $x=3$. 4. **Problem:** Will Abe's dog and Karen's dog ever be the same weight? Abe's dog weight: $2(x + 3)$ Karen's dog weight: $3(x + 1)$ 15. **Step 1:** Set the weights equal. $$2(x + 3) = 3(x + 1)$$ 16. **Step 2:** Distribute. $$2x + 6 = 3x + 3$$ 17. **Step 3:** Subtract $2x$ from both sides. $$\cancel{2x} + 6 = 3x - \cancel{2x} + 3$$ $$6 = x + 3$$ 18. **Step 4:** Subtract 3 from both sides. $$6 - 3 = x + \cancel{3}$$ $$3 = x$$ 19. **Conclusion:** Yes, the dogs will be the same weight after 3 weeks. **Final answers:** 1. No solution. 2. Infinitely many solutions. 3. One solution: $x=3$. 4. Yes, same weight at $x=3$ weeks.