Question

Two years ago, a father was five times as old as his son. Two years later, his age will be 8 more than three times the age of the son. Find the present age of father and son.

Answer 1

Answer:

x is present age of father

y is the present age of son

Making eqns acc to question

x-2=(y-2)x5

x+2={(y+2)x3}+8

x-2=5y-10

x+2=(3y+6)+8

x=5y-8

(x=3y+12)x-1

add above two eqns

2y=20

y=10

x=5y-8

so,x=42

So age of son=10years

and age of father=42 years

Answer 2

Answer:

let father's age be x years and son's age be y years

ATQ,

x-2=5(y-2)

=>x-2=5y-10

=>x=5y-8 (equation 1)

Also,

(x+2)=3(y+2)+8

=>x+2=3y+6+8

=>x-3y=12(equation 2)

Now, by substitution method, put the value of x from equation 1 in equation 2

=>5y-8-3y=12

=>2y=20

=> y=10

now, put the value of y in equation 1,

x=5(10)-8

=>x=50-8

=>x=42

hence, the present age of father is 42 years and the present age of son is 10 years