Question

Ten years ago the age of a father was four times his son. Ten years hence the age of the father will be twice that of his son. The present age of the father and the son are

Answer 1

Answer:

the age of father is 50 years

and the age of son is 20 years

Step-by-step explanation:

let age of father be 'x' and than of his son be 'y'

x-10 = 4(y-10) -----------> eq.1

x+10 = 2(y+10)------------>eq.2

x-10 = 4y-40

x-4y = -30--------------->eq.3

x+10 = 2y+20

x-2y = 10------------------->eq.4

USING ELIMAINATION METHOD

x - 4y = -30

x - 2y =  10            (change of signs)

-2y = -40

  y = -40/-2

  y = 20

  sub. y = 20 in eq.1

x - 10 = 4(y-10)

x - 10 = 80 - 40

x - 10 = 40

       x = 40 + 10

       x = 50

HOPE IT IS HEPLFUL

Answer 2

Answer:

Step-by-step explanation:

let the father's age be 4x and son's age be x

after 10years

4x + 10 = 3 *(x+ 10)

4x+10= 3x +30

4x -3x = 30-10

x= 20

therefore father's present age = 4x

=4 * 20 = 80

son's present age = x=20