Question

four years ago a father was three times as old as his son. Six years from now he will be twice as old as his son. Find their present ages​

Answer 1

Answer:

Then now they are

(1) 3S = F

and six years ago they were

(2) 5*(S-6) = F-6

calculate 5*(S-6)

5S - 30 = F-6

Subtract left and right side by -F

Add 30 to left and right side

5S - F = 24

Put (2) 3S = F into the equation above

5S - 3S = 24

2S = 24

S = 12

insert result in (2) 3S = F

3*12 = F

36 = F

So the son is 12 years, and the father is 36 years.

Answer 2

Answer:

Let the father's age be x and son's age be y

So, I) x-4=3(y-4) = x-4=3y-12

=x-3y=-8

ii) x+6=2(y+6) = x+6=2y+12

=x-2y=6

Now let's subtract the equations

x-3y=-8

(-) x-2y=6

-y=-14

So, y=14

Now for x

x-2×14=6

x=6+28=34

Hence, father's age is 34 and son's age is 14