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

let the present age of man and his son be x and y respectively...
case1: 5 years ago
x-5=7(y-5)
x-5=7y-35
7y-x=30
case2: five years hence
x+5=3(y+5)
x+5=3y+15
x-3y=10
by elimination method.....
-x+7y=30
(+) x -3y=10
_________
4y=40
y=10
then, x-3y=10
x-30=10
x=40
hence the present age of father is 40 years and son 10 years.......

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hope it helps........

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!