#### Answer

The system of equations can be solved to obtain $A=5,B=-2,C=3$.

#### Work Step by Step

We have the system of equations
$\begin{align}
& A+B=3\text{ }.......\text{ }\left( 1 \right) \\
& 2A-2B+C=17\text{ }.......\text{ }\left( 2 \right) \\
& 4A-2C=14\text{ }.......\text{ }\left( 3 \right)
\end{align}$
So, by multiplying equation (1) by 2 and adding to equation (2), we get
$4A+C=23$
And subtracting equation (3) from the above obtained equation, we get
$\begin{align}
& 3C=9 \\
& C=3
\end{align}$
Putting the value in equation (3), we get
$\begin{align}
& 4A-6=14 \\
& 4A=20 \\
& A=\frac{20}{4} \\
& A=5
\end{align}$
Then put this in equation (1) to get
$\begin{align}
& 5+B=3 \\
& B=3-5 \\
& B=-2
\end{align}$
Hence, $A=5,B=-2,C=3$.