Question 252077
<pre><font size = 4 color = "indigo"><b>
You typed "he" for "she" in the last part.  There is no solution if
it were "he".  But there is if it is "she".
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John is as old as Cindy will be when John is twice as old as Cindy was when John was half as old as the sum of their current ages. Cindy is as old as John was when Cindy was half as old as <b><font size = 6>s</font></b>he will be in ten years. How old are John and Cindy?

<pre><font size = 4 color = "indigo"><b>
Let's go through it carefully, letting the number of years in the future
or in the past be represented by letters:
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John is as old as Cindy will be 
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J = C + x  (that is, x years in the future)
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when John 
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who will be age J + x
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is twice as old as Cindy was 
<pre><font size = 4 color = "indigo"><b>
J + x = 2(C - y)  (this was y years in the past)
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when John 
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at age J - y
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was half as old as the sum of their current ages. 
<pre><font size = 4 color = "indigo"><b>
J - y = 1/2(J + C)
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Cindy is as old as John was 
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C = J - z  (z years in the past)
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when Cindy 
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at age C - z
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was half as old as she will be in ten years.
<pre><font size = 4 color = "indigo"><b>
C - z = 1/2(C + 10) 

So we have this system of equations: 

J = C + x
J + x = 2(C - y)
J - y = 1/2(J + C)
C = J - z 
C - z = 1/2(C + 10)

Simplifying the first:
J - C - x = 0

Simplifying the second:
J + x = 2C - 2y
J - 2C + x + 2y = 0

Simplifying the third:
J - y = 1/2(J + C)
2J - 2y = J + C
J - C - 2y = 0

Simplifying the fourth:
C = J - z 
-J + C + z = 0

Simplifying the fifth:

C - z = 1/2(C + 10)
2C - 2z = C + 10
C - 2z = 10

The simplified system is:

 J -  C - x           =  0
 J - 2C + x + 2y      =  0
 J -  C     - 2y      =  0
-J +  C          +  z =  0
      C          - 2z = 10

Solve that and get J=40, C=30, x=10, y=5, z=10

So John is 40 and Cindy is 30.

Edwin</pre>