Question 199567
A man is twice the age his wife was when he was the age she is now. When she is the age that he is now, the sum of their ages will be ninety-nine. What age are the man and woman? 
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Let {{{M}}} = the man's present age.

Let {{{W}}} = the wife's present age

Let {{{d}}} = the difference in their ages.

So one equation is

{{{M - W = d}}}

So notice these facts:

d years ago, the man was as old as the wife is now,

Also notice that: 

d years from now, the wife will be as old as the 
man is now.
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>>...A man is twice the age his wife was when he was the age she is now...<<
<pre><font size = 4 color = "indigo"><b>
He was her present age {{{d}}} years ago, and she 
was {{{W-d}}} then.

He is now twice that so he is now {{{2(W-d)}}}.  
So the second equation is

{{{M = 2(W-d)}}}
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>>...When she is the age that he is now, 
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She will be his present age {{{d}}} years from now, 
when she will be {{{W+d}}}, and he will be {{{M+d}}}.
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>>...the sum of their ages will be ninety-nine...<<
<pre><font size = 4 color = "indigo"><b>

So the third equation is

{{{(M+d)+(W+d) = 99}}}

Now we have three equations in three unknowns:

{{{system(M - W = d,
M = 2(W-d),
(M+d)+(W+d) = 99)}}}

Simplifying them and setting them up so that like
variables line up vertically:

{{{system(M-W-d=0,M-2W+2d = 0, M+W+2d = 99)}}}

Can you solve that system?  If not post again asking how.

The solution is {{{M=44}}}, {{{W=33}}}, {{{d=11}}}

He is 44, she is 33 and the difference in their ages is
11 years.

So ll years ago he was her present age of 33, she was 22.  
And since he is 44 now he is twice as old as she was then.

And 11 years from now, she will be his present age of 44,
and he will be 55, and the sum of 44 and 55 then will 
indeed be 99.  So it checks.

Edwin</pre>