Question

10 years ago fathers age was 7 times his son's age after 2 years, 2 times fathers age is equal to 5 times of sons age. what is their present age?​

Answer 1

Answer:

Step-by-step explanation:

Let S= age of son at present

S-10= age of son 10 years ago

S+2= age of son 2 years from now

F= age of father at present

F-10= age of father 10 years ago

F+2= age of father 2 years from now

F-10= 7(S-10) = 7S-70, F-7S = -60, F= 7S-60 Eq. 1

2(F+2)= 5(S+2)

2F +4= 5S+10 Eq. 2

2F-5S= 6. but F=7S-60

2(7S-60)-5S=6

14S-120–5S=6

9S=126

S= 126/9=14 years old at present

F=7(14)-60= 98–60= 38 years old at present

Checking:

10 years ago, Father was 28 years old and age of son then was 4 years old on which father age of 28 was 7 times as old as his son who was only 4 years old then. Two years hence father will be 40 years old and son will be 16 years old on which twice of 40 is 80 and 5 times of 16 is 80.

Answer 2

Answer:

let father's age be = x

let son's age be= y

ATQ

x-10=7(y-10) 1

2(x+2)=5(y+2)

Step-by-step explanation:

solve them.

from the image now you can take it out.

by putting eq3 in eq2

hope this helps u