+160 
my keyboard isn't working at the moment, it's not letting me input operations or equal signs. so i took a screenshot of the question :)
+118690 I am really not sure what the qustion is asking.
There are 3 unknowns, x,y and z so how cha you have sloution pairs?
Here is my take.
\(x+2y+2z=n\\ x+2(y+z)=n\\ \text{Let } \color{red}Q=y+z\\ x+2Q=n\\ \)
Now i will solve for integer values of x+Q
I can see that -1+2=1
so 1(-n)+2(n)=n
If k is an integer then
1(-n+2k)+2(n-k)=n
If x and Q are both positive integers then
\(-n+2k\ge1 \qquad and \qquad n-k\ge1\\ 2k\ge1+n \qquad and \qquad -k\ge1-n\\ k\ge \frac{1+n}{2} \qquad and \qquad k\le n-1\\ \frac{1+n}{2} \le k\qquad and \qquad k\le n-1\\ \frac{1+n}{2} \le k\le n-1\\ \text{Now I want 28 different possible integer values of k so }\\ n-1-\frac{1+n}{2}+1=28\\ n-\frac{1+n}{2}=28\\ \frac{2n-1-n}{2}=28\\ n-1=56\\ n=57 \)
But of course I have made up my own question because yours does not make sense to me.
If n=57 then there are 28 soltions for x an Q [Not for x y and z]
