The sum of the ages of a man and his daughter is 44. Four years ago, the man was five times as old as his daughter. Find their present ages.

Here's the solution to the age problem. Question: The sum of the ages of a man and his daughter is 44. Four years ago, the man was five times as old as his daughter. Find their present ages. Step 1: Define variables and set up the first equation. Let M be the man's present age and D be the daughter's present age. According to the first statement, the sum of their ages is 44: M + D = 44 (1) Step 2: Set up the second equation based on their ages four years ago. Four years ago, the man's age was M - 4. Four years ago, the daughter's age was D - 4. According to the second statement, the man was five times as old as his daughter four years ago: M - 4 = 5(D - 4) (2) Step 3: Simplify the second equation. Expand the right side of equation (2): M - 4 = 5D - 20 Add 4 to both sides: M = 5D - 16 (3) Step 4: Substitute equation (3) into equation (1) and solve for D. Substitute M = 5D - 16 into M + D = 44: (5D - 16) + D = 44 6D - 16 = 44 Add 16 to both sides: 6D = 44 + 16 6D = 60 Divide by 6: D = (60)/(6) D = 10 The daughter's present age is 10 years. Step 5: Substitute the value of D back into equation (1) to find M. Using M + D = 44: M + 10 = 44 Subtract 10 from both sides: M = 44 - 10 M = 34 The man's present age is 34 years. The man's present age is 34 years and the daughter's present age is 10 years. What's next? 📸