Set up the system of equations described in the question, and add the equations:
[tex] x-y=10[/tex]
[tex]+(x+y=18)[/tex]
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[tex]\hspace*{1em} 2x \enspace \enspace \enspace =28 \newline \hspace*{1em} \enspace x \enspace \enspace \enspace =14[/tex]
Now that we have x=14, the other side length can be solved by substituting 14 into either equation to get y=4.
The side length of one square is 14. The side length of the other is 4.
