[tex]2x+2y+z=20\\
z=\dfrac{20}{2x+2y}\\
z=\dfrac{10}{x+y}[/tex]
Now, for z to be an integer, the sum x+y must be a divisor of 10.
It has to be a positive divisor since z≥0. Also x≠y.
[tex]x+y=1 \\
x=1-y\\
[/tex]
x≥0 and y≥0 so y can be equal to either 0 or 1. There are 2 solution in this case.
[tex]x+y=2\\
x=2-y\\[/tex]
In this case, y can be equal to 0,1, or 2, but for y=1 ⇒ x=1, so there are two solutions.
[tex]x+y=5\\
x=5-y[/tex]
y can be 0,1,2,3,4 or 5 - 6 solutions
[tex]x+y=10\\
x=10-y[/tex]
y can be 0,1,2,3,4,5,6,7,8,9,10, but for y=5 ⇒ x=5, so 10 solutions.
2+2+6+10=20 solutions in total.
