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

The answer is $x=4$ and $2$.
The given equations are $x+y=6$ $x-y=2$ The augmented matrix is $\left[\begin{array}{cc|c} 1& 1 &6 \\ 1&-1 &2 \\ \end{array}\right]$ Use matrix row operations to simplify. Perform $R_2\rightarrow R_2-1\cdot R_1$ Multiply row one by $-1$ and subtract from row 2 as shown below. $\left[\begin{array}{cc|c} 1& 1 &6 \\ 1-1 \cdot 1&-1-1 \cdot 1 &2-1 \cdot 6 \\ \end{array}\right]$ Simplify. $\left[\begin{array}{cc|c} 1& 1 &6 \\ 1-1 &-1-1 &2- 6 \\ \end{array}\right]$ $\left[\begin{array}{cc|c} 1& 1 &6 \\ 0 &-2 &-4 \\ \end{array}\right]$ $R_2\rightarrow \frac{R_2}{-2}$ Multiply row two by $\frac{1}{-2}$ $\left[\begin{array}{cc|c} 1& 1 &6 \\ 0 \cdot \frac{1}{-2}&-2\cdot \frac{1}{-2} &-4\cdot \frac{1}{-2} \\ \end{array}\right]$ Simplify. $\left[\begin{array}{cc|c} 1& 1 &6 \\ 0 &1 &2 \\ \end{array}\right]$ Use back substitution to find the system's solution. $x+y=6$ $y=2$ Substitute into above equation. $x+2=6$ $x=4$.