Question

Three years hence a father will four times as old as his son will be. Before two years he was seven times as old as his son was. Find their present ages.

Answer 1

Answer: The father is 37 years and the son is 7 years

Step-by-step explanation:

Let x represent the present age of the father.

Let y represent the present age of the son.

Three years hence a father will four times as old as his son will be. This means that

x + 3 = 4(y + 3)

x + 3 = 4y + 12

x - 4y = 12 - 3

x - 4y = 9- - - - - - - - - - - - - - - -1

Before two years he was seven times as old as his son was. This means that

(x - 2) = 7(y - 2)

x - 2 = 7y - 14

x - 7y = - 14 + 2

x - 7y = - 12- - - - - - - - - - - - - - -2

Subtracting equation 2 from equation 1, it becomes

3y = 21

y = 21.3 = 7

Substituting y = 7 into equation 1, it becomes

x - 4 × 7 = 9

x - 28 = 9

x = 9 + 28

x = 37

Answer 2

Answer:

Father 9 years

son 1 year

Step-by-step explanation:

knowing that the father is X and the son is Y we have the equation

X+3=4Y

X-2=7Y