How to solve $Ax+By+z=C$ for $x$, $y$ and $z$? This is my first post in this forum! So let me know if anything is not clear.
Given a series of 10 different Values for $A$, $B$ & $C$.  How to Solve for minimum value of $x$, $y$, $z$ respectively.
$$Ax + B  y + z = C$$
where $A$, $B$, $C$ are constants in the range $[0,255]$. Please provide me with some ideas to solve this with some samples.
Edit:
Note: Find the minimum values of $x$, $y$, $z$. You will be given $10$ sets of values for $A$, $B$, $C$ like these 
$[20,39,39]$,
$[138,149,215]$,
$[185,139,95]$,
$[41,130,87]$,
$[242,171,158]$,
$[29,119,13]$,
$[201,84,59]$,
$[99,78,246]$,
$[141,40,99]$,
$[111,253,175]$.
 A: Your question, I guess, would be that if we are given 10 equations,
$$A_i x + B_i y + z = C_i \;\;(i=1,2,\cdot\cdot\cdot,10)$$ 
How to find the best solution for $(x,y,z)$?
Since we have more equations than unknowns $(10>3)$, we need to use the least square method to get the best solution for $(x,y,z)$. 
Let,$$A_ix + B_i y + z - C_i = \delta_i \;\;(i=1,2,\cdot\cdot\cdot,10)$$
And,we need to find $(x,y,z)$ so that the following function $f(x,y,z)$ be minimized.
$$f(x,y,z)=\sum_{i=1}^{10} \delta_i^2$$ 
$$\left\{\begin{matrix}
 \frac{\partial f}{\partial x}=0 \\ 
 \frac{\partial f}{\partial y}=0 \\ 
 \frac{\partial f}{\partial z}=0 \\ 
\end{matrix}\right.$$
It gives the following system of 3 equations with 3 unknowns.
$$\left\{\begin{matrix}
\sum_{i=1}^{10} (A_ix + B_i y + z - C_i)A_i =0\\ 
\sum_{i=1}^{10} (A_ix + B_i y + z - C_i)B_i =0\\ 
\sum_{i=1}^{10} (A_ix + B_i y + z - C_i)\cdot 1=0 \\ 
\end{matrix}\right.$$
This will lead to the solution of $(x,y,z)$
A: You have a matrix equation $$\left[\begin{array}{ccc}A1&B1&1\\A2&B2&1\\...\\A10&B10&1\end{array}\right]\left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{c}C1\\C2\\...\\C10\end{array}\right]$$
or $MX=C$.  This can't usually be solved exactly.
A minimum-error solution is 
$$X=(M^TM)^{-1}M^TC$$
Here, $(M^TM)$ is a $3\times3$ matrix, and $M^TC$ is a $3\times1$ vector.
NOTE: This is the same as @PdotWang's solution
