My Daughter's 4th grade math question got me thinking Given a number of 3in squares and 2in squares, how many of each are needed to get a total area of 35 in^2?
Through quick trial and error (the method they wanted I believe) you find that you need 3 3in squares and 2 2in squares, but I got to thinking on how to solve this exactly.
You have 2 unknowns and the following info:
4x + 9y = 35
x >= 0, y >= 0, x and y are both integers.
It also follows then that x <= 8 and y <= 3
I'm not sure how to use the inequalities or the integer only info to form a direct 2nd equation in order to solve the system of equations.  How would you do this without trial and error?

A quick way to see the answer is to convert both sides of the equation mod 4. So the left hand side is y  (mod 4) (because 4=0, 9=1 mod 4), and the right hand side is 3 (mod 4). So y=3 mod 4. Since $y\le 3$ as you observed, the only solution (if there is any ) is $y=3$. Then you check that $4x+27=35$ and hence $x=2$. 
