Chapter 5, Systems of Equations and Inequalities - Section 5.2 - Systems of Linear Equations in Several Variables - 5.2 Exercises - Page 455: 13

$\begin{cases} 3x+y+z=4\\ 0x-y+z=0\\ x-2y-z=-0 \end{cases}$

$\begin{cases} 3x+y+z=4\\ -x+y+2z=0\\ x-2y-z=-1 \end{cases}$ Adding Equation 2 and Equation 3. results in a new Equation 2. $\begin{cases} -x+y+2z=0\\ x-2y-z=-1\\ -- -- -- -\\ -y+z=-1 \end{cases}$. Therefore, The new equivalent system is... $\begin{cases} 3x+y+z=4\\ 0x-y+z=0\\ x-2y-z=-0 \end{cases}$