1. **Problem 1:** The larger of two numbers is 5 less than twice the smaller, and their sum is 28. 2. Let $x$ be the smaller number and $y$ be the larger number. 3. The relationship is given by $y = 2x - 5$ and the sum is $x + y = 28$. 4. Substitute $y$ in the sum equation: $$x + (2x - 5) = 28$$ 5. Simplify: $$x + 2x - 5 = 28$$ 6. Combine like terms: $$3x - 5 = 28$$ 7. Add 5 to both sides: $$3x - \cancel{5} + 5 = 28 + 5$$ 8. Simplify: $$3x = 33$$ 9. Divide both sides by 3: $$\frac{3x}{\cancel{3}} = \frac{33}{\cancel{3}}$$ 10. Simplify: $$x = 11$$ 11. Find $y$: $$y = 2(11) - 5 = 22 - 5 = 17$$ 12. Check sum: $$11 + 17 = 28$$ --- 13. **Problem 2:** Shane is twice as old as Annie. Ten years ago, her age was thrice Annie's age. 14. Let $A$ be Annie's present age and $S$ be Shane's present age. 15. Given $S = 2A$ and ten years ago: $$S - 10 = 3(A - 10)$$ 16. Substitute $S$: $$2A - 10 = 3A - 30$$ 17. Rearrange: $$2A - 10 - 3A + 30 = 0$$ 18. Simplify: $$-A + 20 = 0$$ 19. Solve for $A$: $$-A = -20 \Rightarrow A = 20$$ 20. Find $S$: $$S = 2(20) = 40$$ --- 21. **Problem 3:** The legs of an isosceles triangle are 4 meters more than its base. The perimeter is 44 meters. 22. Let $b$ be the base length and $l$ be the leg length. 23. Given: $$l = b + 4$$ and perimeter $$P = b + 2l = 44$$ 24. Substitute $l$: $$b + 2(b + 4) = 44$$ 25. Simplify: $$b + 2b + 8 = 44$$ 26. Combine like terms: $$3b + 8 = 44$$ 27. Subtract 8: $$3b + \cancel{8} - 8 = 44 - 8$$ 28. Simplify: $$3b = 36$$ 29. Divide both sides by 3: $$\frac{3b}{\cancel{3}} = \frac{36}{\cancel{3}}$$ 30. Simplify: $$b = 12$$ 31. Find $l$: $$l = 12 + 4 = 16$$ --- **Final answers:** - Problem 1: Smaller number $x = 11$, Larger number $y = 17$ - Problem 2: Annie's age $A = 20$, Shane's age $S = 40$ - Problem 3: Base $b = 12$ meters, Legs $l = 16$ meters