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

The answer is $x=3$ and $x=-3$.
The given system of linear equations is $2x+y=3$ $x-3y=12$ The augmented matrix is $\left[\begin{array}{cc|c} 2& 1 &3 \\ 1&-3 &12 \\ \end{array}\right]$ Perform $R_2\rightarrow R_2-\frac{1}{2}\cdot R_1$. Multiply row one by $-\frac{1}{2}$ and subtract from row two. $\left[\begin{array}{cc|c} 2& 1 &3 \\ 1-\frac{1}{2}\cdot 2&-3-\frac{1}{2}\cdot1 &12-\frac{1}{2}\cdot3 \\ \end{array}\right]$ Simplify. $\left[\begin{array}{cc|c} 2& 1 &3 \\ 1-1&-3-\frac{1}{2} &12-\frac{3}{2} \\ \end{array}\right]$ $\left[\begin{array}{cc|c} 2& 1 &3 \\ 0&\frac{-6-1}{2} &\frac{24-3}{2} \\ \end{array}\right]$ $\left[\begin{array}{cc|c} 2& 1 &3 \\ 0&\frac{-7}{2} &\frac{21}{2} \\ \end{array}\right]$ Perform $R_2\rightarrow \frac{2}{-7}\cdot R_2$. Multiply row two by $\frac{2}{-7}$. $\left[\begin{array}{cc|c} 2& 1 &3 \\ \frac{2}{-7} \cdot 0&\frac{2}{-7}\cdot \frac{-7}{2} &\frac{2}{-7}\cdot \frac{21}{2} \\ \end{array}\right]$ Simplify. $\left[\begin{array}{cc|c} 2& 1 &3 \\ 0&1 &-3 \\ \end{array}\right]$ Perform Perform $R_1 \rightarrow R_1- R_2$. Subtract row two from row one. $\left[\begin{array}{cc|c} 2-0& 1-1 &3+3 \\ 0&1 &-3 \\ \end{array}\right]$ Simplify. $\left[\begin{array}{cc|c} 2& 0 &6 \\ 0&1 &-3 \\ \end{array}\right]$ Use back substitution to solve the linear system. $2x=6$ $x=3$ and $y=-3$.