Chapter 1.1, Problem 7E

The two systems

$\begin{array}{l}2x_1+x_2=3\\ 4x_1+3x_2=5\end{array}$ and $\begin{array}{l}2x_1+x_2=-1\\ 4x_1+3x_2=1\end{array}$

have the same coefficient matrix but different right-hand sides. Solve both systems simultaneously by eliminating the first entry in the second row of the augmented matrix

$\left(\begin{array}{cc}2& 1\\ 4& 3\end{array}\begin{array}{cc}3& -1\\ 5& 1\end{array}\right)$

and then performing back substitutions for each of the columns corresponding to the right-hand sides.