Question

five years ago a man was seven times as old as his son. five years hence, the Father will be three times as old as his son. find their present ages.

Answer 1

five years ago
let son's age=x
man's age=7x
present
son's age =x+5
Man's age=7x+5

Answer 2

$\huge\sf{Answer:-}$

Let y be the age of son.

Let 7y be the age of father.

y + 5 , 7y + 5 are the present ages

= y + 10 future age of son.

7y + 10 is father.

= (7y + 10 = 3) (y + 10)

= (y = 5)

= (7y = 35)

(y + 5 = 5 + 5 = 10)

.•. 10 years is age of son and

(7y + 5 = 7) ×( 5 + 5 = 40) .•. 40 is age of father.

Hope it helps!