You correctly found the value for x to be $\displaystyle-{1}\usepackage{cmbright}\Huge\$ . However, you may have made a mistake when substituting this value into $\displaystyle-{x}+{z}=-{1}\usepackage{cmbright}\Huge\$ , and found $\displaystyle{z}={0}\usepackage{cmbright}\Huge\$ instead of $\displaystyle{z}=-{2}\usepackage{cmbright}\Huge\$ . You then used the values $\displaystyle{z}={0}\usepackage{cmbright}\Huge\$ and $\displaystyle{x}=-{1}\usepackage{cmbright}\Huge\$ to solve for y in the first equation. To be a solution, the values must make each of the equations in the system true. However, these values do not make the second or third equation true.