# Chapter 3 - Section 3.4 - Matrix Solutions to Linear Systems - Exercise Set - Page 229: 18

The answer is $x=-1$ and $y=-2$

#### Work Step by Step

The given system of linear equations is $3x-5y=7$ $x-y=1$ The augmented matrix is $\left[\begin{array}{cc|c} 3 & -5 & 7\\ 1 & -1 & 1 \end{array}\right]$ Perform $R_2\rightarrow R_2-\frac{1}{3}\cdot R_1$. $\left[\begin{array}{cc|c} 3 & -5 & 7\\ 1-\frac{1}{3}\cdot 3 & -1-\frac{1}{3}\cdot (-5) & 1-\frac{1}{3}\cdot 7 \end{array}\right]$ Simplify. $\left[\begin{array}{cc|c} 3 & -5 & 7\\ 1-1 & -1+\frac{5}{3} & 1-\frac{7}{3} \end{array}\right]$ $\left[\begin{array}{cc|c} 3 & -5 & 7\\ 0 & \frac{-3+5}{3} & \frac{3-7}{3} \end{array}\right]$ $\left[\begin{array}{cc|c} 3 & -5 & 7\\ 0 & \frac{2}{3} & \frac{-4}{3} \end{array}\right]$ Perform $R_1\rightarrow R_1+\frac{15}{2}\cdot R_2$. $\left[\begin{array}{cc|c} 3 & -5+5 & 7-10\\ 0 & \frac{2}{3} & \frac{-4}{3} \end{array}\right]$ Simplify. $\left[\begin{array}{cc|c} 3 & 0 & -3\\ 0 & \frac{2}{3} & \frac{-4}{3} \end{array}\right]$ Use back substitution to solve the linear system. $3x=-3$ $x=-1$ and $\frac{2}{3}\cdot y=\frac{-4}{3}$ $y=-2$

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