SOLUTION: Hello. I have been struggling with a question for a while and I am getting frustrated. The question is: "The sum of the ages of John and Mary is 32. Four years ago, John was twice

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Question 1052923: Hello.
I have been struggling with a question for a while and I am getting frustrated. The question is: "The sum of the ages of John and Mary is 32. Four years ago, John was twice as old as Mary. Find the present age of each."
I have tried sum of 32, then divide by 2 = 16. Then 16 - 4 = 12 and then I get stuck. Please can you help? Thank you :)
Found 3 solutions by ewatrrr, MathTherapy, addingup:
Answer by ewatrrr(24785)   (Show Source): You can put this solution on YOUR website!
x John's age..the (32-x) is Mary's age NOW
4 yrs ago...
(x-4) = 2[(32-x) - 4}
x - 4 = 64 - 2x - 8
3x = 60
x = 20, John's age. Mary's age is 12
And...4 yrs ago...
16 = 2*8 CHECKS!
Answer by MathTherapy(10552)   (Show Source): You can put this solution on YOUR website!

Hello.
I have been struggling with a question for a while and I am getting frustrated. The question is: "The sum of the ages of John and Mary is 32. Four years ago, John was twice as old as Mary. Find the present age of each."
I have tried sum of 32, then divide by 2 = 16. Then 16 - 4 = 12 and then I get stuck. Please can you help? Thank you :)
Let John’s age be J

Then Mary’s is: 32 – J

We then get: J – 4 = 2(32 – J – 4)

J – 4 = 2(28 – J)

J – 4 = 56 – 2J

J + 2J = 56 + 4

3J = 60

J, or John is: 

Mary is:  

Answer by addingup(3677)   (Show Source): You can put this solution on YOUR website!
 The sum of the ages:

J+M = 32 Subtract M on both sides

J = 32-M (1)

Four years ago John was 2*Mary:

J-4 = 2(M-4) (2)

In equation (2) substitute the value of J per (1):

32-M-4 = 2M-8

28-M = 2M-8

36-M = 2M

36 = 3M flip the equation so we have the unknown of the left

3M = 36 divide both sides by 3

M = 12

and J = 32-M = 32-12 = 20

----------------------------------

Check

20-4 = 2(12-4)

16 = 16 correct 

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