Answer
$A=5,B=-2,C=3$
Work Step by Step
We are given the system:
$\begin{cases}
A+B=3\\
2A-2B+C=17\\
4A-2C=14
\end{cases}$
We will use the addition method. Multiply Equation 2 by 2 and add it to Equation 3 to eliminate $C$:
$\begin{cases}
4A-4B+2C=34\\
4A -2C=14
\end{cases}$
$-- -- -- -- $
$\begin{cases}
8A-4B=48
\end{cases}$
Multiply Equation 1 by 4 and add it to new found Equation to eliminate $B$.
$\begin{cases}
8A-4B=68\\
4A+4B=12
\end{cases}$
$-- -- -- --$
$12A=60$,
$A=5$.
Substitute $A$ into the original Equation 1 to get $B$.
$5+B=3$,
$B=-2$
Substitute $A$ and $B$ into the original Equation 3 to get $C$.
$4(5)-2C=14$,
$20-14=2C$,
$C=3$.
Therefore, the answer we get is:
$A=5,B=-2,C=3$
