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?

There is an algorithmic way to solve this which works when you have two types of squares.
if $\displaystyle \text{gcd}(a,b) = 1$, then for any integer $c$ the linear diophantine equation $\displaystyle ax + by = c$ has an infinite number of solution, with integer $\displaystyle x,y$.
In fact if $\displaystyle x_0, y_0$ are such that $\displaystyle  a x_0 - b y_0 = 1$, then all the solutions of $\displaystyle ax + by = c$ are given by
$\displaystyle x = -tb + cx_0$, $\displaystyle y = ta - cy_0$, where $\displaystyle t$ is an arbitrary integer.
$\displaystyle x_0 , y_0$ can be found using the Extended Euclidean Algorithm.
Since you also need $\displaystyle x \ge 0$ and $\displaystyle y \ge 0$ you must pick a $\displaystyle t$ such that
$\displaystyle c x_0 \ge tb$ and $ta \ge cy_0$.
If there is no such $\displaystyle t$, then you do not have a solution.
In your case, $\displaystyle a= 9, b= 4$, we need a solution of $\displaystyle ax + by = 35$.
We can easily see that $\displaystyle x_0 = 1, y_0 = 2$ gives us $\displaystyle a x_0 - by_0 = 1$.
Thus we need to find a $\displaystyle t$ such that $ 35 \ge t\times 4$ and $ t\times 9 \ge 35\times 2$.
i.e.
$\displaystyle 35/4 \ge t \ge 35\times 2/9$
i.e.
$\displaystyle  8.75 \ge t \ge 7.77\dots$ 
Thus $t = 8$.
This gives us $\displaystyle x = cx_0 - tb  = 3$, $\displaystyle y = ta- cy_0 = 2$.
(Note: I have swapped your x and y).
