# Devise System of Equations to Solve Age Problem

Hi this may be simple silly problem but it is bugging me as I am not able to devise a system of equations to solve it.

My husband's age," remarked a lady the other day, "is represented by the figures of my own age reversed. He is my senior, and the difference between our ages is one-eleventh of their sum.

The answer is 54 and 45, but not able to find way to get it.

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Let the lady's age be 10a+b, and the husband's age be 10b+a... – J. M. May 13 '11 at 6:40
Also, (594,495), (5454,4545), (5994,4995), etc... – user4143 May 13 '11 at 8:51

Not sure if this totally solves what you want, but here's a way to get a stage where the numbers should be easier to guess and check. I assume the ages of the two people are two digit numbers. So we can represent the husband's age as $10x+y$, where $x$ and $y$ are the decimal digits, and the wife's age is then $10y+x$, with $10x+y\gt 10y+x$.
Since the difference of their ages is one eleventh of the sum, this translates to $$(10x+y)-(10y+x)=\frac{1}{11}(10x+y+10y+x)$$ but this implies $$9x-9y=\frac{1}{11}(11x+11y)=x+y.$$
So you have $9(x-y)=x+y$. Since $0\leq x,y\leq 9$, it's not hard to experiment with these numbers to find that $x=5$ and $y=4$.
The next step is $8x =10y$, i.e. $4x=5y$, so $x$ is a multiple of $5$ which means it is $0$ or $5$ ($10$ would be too big for a decimal digit) so $y=0$ or $4$, but $0$ would make them the same age and unable to talk. – Henry May 13 '11 at 6:52
Actually, you don't need to do much trial and error, since you know that $x+y$ must be divisible by $9$, so either $x+y=9$ or $x+y=18.$ But $x+y=18$ only happens when $x=9, y=9$ which is obviously not a solution. If $x+y=9$, then $x-y=1$, and you get $45/54$ – Thomas Andrews May 13 '11 at 6:52