Question

19. Five years ago, a man was five times as old as his son. Eight years hense , the
man's age will be two years less than three times his son age. Find their present ages.
please solve it is urgent please ​

Answer 1

Step-by-step explanation:

Let Five years ago the age of son be x years and age of father be 7x years

Present age of son =x+5

Present age of father =7x+5

5 years later their age will (x+10) and (7x+10)

∴7x+10=3(x+10)

7x−3x=20

4x=20

x=5

So, the present age of son=x+5=5+5=10years

and the present age of father=7x+5=35+5=40years

Answer 2

Step-by-step explanation:

Son age be x

father age 7x

present son age x+5

present father age 7x+5

age after 5years (x+10) (7x+10)

7x+10=3(x+10)

7x-3x=20

4x = 20

x=5

present age of son=5+5=10 years

present father age 7×5+5=40years