3 equations with 4 unknowns and information about the unknowns' types My dad just sent me a math problem on email and asked if I could solve it, but frankly, I have no idea where to start. I've got 3 equations with 4 unknowns and some information about the unknowns.
First, the equations:
$$2(z-1)-x=55$$
$$4xy-8z=12$$
$$a(y+z)=11$$
I am supposed to determine the 2 greatest real number values of $a$, where $x,y,z$ are positive natural numbers.
The problem is driving me nuts, so any help will be greatly appreciated (even just a hint). Thanks in advance.

The thing that jumps out at me is
$4xy - 8z = 12$ means
$xy -2z = 3$  .... oh, for a second there, I thought that bounded things.
So well...
it gives a relation between $z$ and $xy$ as $z = \frac{xy -3}2$ which means $xy-3$ is even which means $xy$ is odd so $x$ and $y$ are odd.
We also have $2(z-1)-x=55$ so $2(\frac{xy-3}2 - 1) - x = 55$ so $xy-3 -2 - x = 55$
so $xy - x = 60$ so $x(y-1) = 60 = 4*3*5$ but both $x,y$ are odd $y-1$ is even. 
So we have:
$(x,y-1) = \{(15,4)(5,12)(3,20)(1,60)\}$
And $(x,y,z=\frac{xy-3}2) = \{(15,5,36),(5,13,31),(3,21,30),(1,61,29)\}$
So $y + z = \{41, 44, 51, 90\}$
So $a(y+z) = 11 \implies a = \frac{11}{y+z} = \{\frac{11}{41},\frac{11}{44},\frac{11}{51},\frac{11}{90}\}$
so the to greatest are $a = 11/41$ and $x,y,z =15,5,36$ and $a= 1/4$ and $x,y,z = 5,13,31$.
Check: 
$2(z−1)−x=55; 2(36 - 1) - 15 = 70-15=55;2(31-1)-5=60-5=55$
$4xy−8z=12;4*15*5 -8*36 = 300 - 288 =12; 4*5*13-8*31= 260 - 248 = 12$
$a(y+z)=11;\frac{11}{41}*(5+36) = 11; \frac 14*(13+31) = \frac{44}4 = 11$. 
