## Intermediate Algebra for College Students (7th Edition)

The solution is $x=3$ and $y=4$.
The given system of linear equation is $x+2y=11$ $x-y=-1$ The augmented matrix is $\left[\begin{array}{cc|c} 1& 2 &11 \\ 1&-1 &-1 \\ \end{array}\right]$ Perform $R_2\rightarrow R_2- R_1$. Subtract row one from row two as shown below. $\left[\begin{array}{cc|c} 1& 2 &11 \\ 1-1&-1-2 &-1-11 \\ \end{array}\right]$ Simplify. $\left[\begin{array}{cc|c} 1& 2 &11 \\ 0&-3 &-12 \\ \end{array}\right]$ Perform $R_2\rightarrow \frac{R_2}{-3}$. Divide row two by $-3$. $\left[\begin{array}{cc|c} 1& 2 &11 \\ \frac{0}{-3}&\frac{-3}{-3} &\frac{-12}{-3} \\ \end{array}\right]$ Simplify. $\left[\begin{array}{cc|c} 1& 2 &11 \\ 0&1 &4 \\ \end{array}\right]$ Use back substitution to solve the linear system. $x+2y=11$ $y=4$ Substitute into above equation. $x+2(4)=11$ $x+8=11$ $x=11-8$ $x=3$.
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