Answered

Find the best solutions to your questions at Westonci.ca, the premier Q&A platform with a community of knowledgeable experts. Get expert answers to your questions quickly and accurately from our dedicated community of professionals. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform.

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.

Sagot :

goodingenes
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

Answer:

the dad is 60 im not sure about the son

Step-by-step explanation:

Thank you for visiting our platform. We hope you found the answers you were looking for. Come back anytime you need more information. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.