A father's age now is three times the age that his son was 4 years ago. In 12 years, the father will be twice as old as his son. find their ages now.

Let’s say son’s age = S
Dad’s age = D
4 years ago, the son was (S-4) and dad now is three times that so D=3(S-4)
Expand that to get D=3S-12

In 12 years, the son will be (S + 12)
Dad will also be (D + 12) and he will be two times the sons age at that point so:
D + 12 = 2(S + 12)
Expand the right side:
D +12 = 2S + 24
Subtract 12 from both sides:
D = 2S + 12
So now D = 2S + 12 AND
D = 3S - 12
Since both equations equal “D”, they must then equal each other:
2S + 12 = 3S - 12
Do some algebra and you get S = 24
Plug this back into EITHER “D=“ equations to find D
D=2(24) +12
D=60

Best answer that i found on yahoo ok   set :   x= length of 2nd piece in cm  x- 3= length of 1st piece in cm  x-4= length of 3rd piece in cm  then since you know they all came from the 20 cm ribbon you add them together and set them equal to 20x+ x-3 +x-4 = 20  3x - 7= 20  3x= 27  x= 9  the shortest piece is the third piece which is x- 4  x-4  9-4 = 5  the length of the smallest piece is 5 cm.  the other lengths are  9 cm  6 cm