Question

Twenty years ago the age of a father was four times the age of his son. After four years from now the age of the father will be double that of the son. Find the present ages of the father and son.

Answer 1

4(y-20) =x

4y-80=x

x-4y=-80 ......... (1)

2(y+4)=x

2y+8=x

x-2y=8..............(2)

for I and 2 adding by elimination.

-2y=-88

y=44

put in eq 1

x-4y=-80

x-176=-80

x=96

so the age of father 96 and son 44

Answer 2

Answer:

let's father age is x and the son age is y

ten years ago,the father was 4 time his son

x - 10= 4(y-10)

x- 10= 4y-40

x- 4y = -30. (eq...1)

x. = 4y- 30(eq...2)

10 years from now,the father whould be twice be son age

x+10=2(y+10)

x+10=2y+20

x=2y +10(eq.....3)

eq 2 and eq 3

4y-30=2y+10

collect like terms

4y-2y=10+30

2y=40

y= 20 years

put y as 20 in equestion

x=2(20)+10

x=40+10=50

the son is 20 years and father is 50 years