1. Problem 1: One number is 12 more than another number. Their sum is 48. Find the two numbers. 2. Problem 2: A father is 26 years older than his son. In 4 years, the father will be three times as old as his son. Find their current ages. --- ### Problem 1 Solution: 1. Let the smaller number be $x$. 2. Then the other number is $x + 12$. 3. Their sum is given as $x + (x + 12) = 48$. 4. Simplify the equation: $2x + 12 = 48$. 5. Subtract 12 from both sides: $2x = 48 - 12 = 36$. 6. Divide both sides by 2: $x = \frac{36}{2} = 18$. 7. The other number is $18 + 12 = 30$. **Answer:** The two numbers are 18 and 30. --- ### Problem 2 Solution: 1. Let the son's current age be $s$. 2. Then the father's current age is $s + 26$. 3. In 4 years, son's age will be $s + 4$ and father's age will be $s + 26 + 4 = s + 30$. 4. According to the problem, in 4 years, father will be three times as old as son: $$s + 30 = 3(s + 4)$$ 5. Expand the right side: $s + 30 = 3s + 12$. 6. Rearrange terms: $30 - 12 = 3s - s$ which gives $18 = 2s$. 7. Divide both sides by 2: $s = \frac{18}{2} = 9$. 8. Father's current age is $9 + 26 = 35$. **Answer:** The son is 9 years old and the father is 35 years old now.