How to find values from 1 equation with 3 variables Is there any way to get something out of: $50x + 5y - 18z = 0$ ?
I found this trying to solve this:
\begin{equation}
xyz+xyz+xyz=zzz
\end{equation}
in which xyz are not multiplying themselves. They're just variable of a number; for example, let's say that x=2;y=1;z=8, then xyz=218. As you can see in the example before, x is in the hundreds place, y in the tens and z in the ones.
Then:
\begin{equation}
xyz + xyz + xyz = zzz
\end{equation}
\begin{equation}
3 (xyz) = zzz
\end{equation}
\begin{equation}
3 [(x*100)+(y*10)+(z*1)] = (z*111)
\end{equation}
\begin{equation}
50x + 5y - 18z = 0
\end{equation}
From here I couldn't get nothing more than separating x,y or z; nothing more than variables.
Is there any way to extract values from it? I was thinking about solving it by linear algebra, but it's only 1 equation (and 3 variables).

The answer is:

 x=1; y=8; z=5


Thanks advance, 
Luiz.
 A: Well, we need that $0 <x \leq 9, 0\leq y,z \leq 9$ as they are all the digits in a three-digit number. Thus, we can rearrange our equation as: $$z = \frac {50x+5y}{18} = \frac {5 (10x+y)}{18}$$
It is obvious that the numerator's last digit is always $5$ or $0$, the former of which will not make the numerator to be divisible by $18$. So, the last digit of the numerator is $0$, and the first number satisfying this condition divisible by $18$ is $90$. Thus, $z=5$. Now, it will bring easy to see what values can be taken by $x $ and $y $.
Hope it helps. 
A: This question is a logic problem, rather than a straight algebra problem.
You have the equation $xyz+xyz+xyz=zzz$, where $0<x\leq9$ and $0\leq y,z\leq9$.
Combining the left side gives $3(xyz)=zzz$. Expanding as you did gives $300x+30y+3z=100z+10z+z$.
Notice the last "digit" of each side. We need $3z=z$ thinking in terms of last digit. The only digits for which this is true are $z=0,5$.
If $z=0$, then you will find all digits must be 0. No good!
If $z=5$, then we have $3(xy5)=555$. Dividing by 3 gives $xy5=185$, which gives the solution.
