## Precalculus (6th Edition) Blitzer

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