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 age of father be x

Age of son be y

5years ago,

X-5=7(y-5)

X-5= 7y-35

X-7y=-35+5

X-7y=-30. _______(1)

5 years later,

(X+5)= 3(y+5)

X+5=3y+15

X-3y=15-5

X-3y=10. _______(2)

Solve (1) and (2) by any method and u will get the answer.

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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!