You correctly found the value for y to be 2. However, you may have made a mistake when substituting this value into $\displaystyle{2}{y}+{2}{z}={2}\usepackage{cmbright}\Huge\$ , and found $\displaystyle{z}={3}\usepackage{cmbright}\Huge\$ instead of $\displaystyle{z}=-{1}\usepackage{cmbright}\Huge\$ . You then used the values $\displaystyle{z}={3}\usepackage{cmbright}\Huge\$ and $\displaystyle{y}={2}\usepackage{cmbright}\Huge\$ to solve for x in the third equation. To be a solution, the values must make each of the equations in the system true. However, these values do not make the first or second equation true.