You may have used the first two equations to eliminate x but forgot to add the y term. This left you with $\displaystyle-{z}={1}\usepackage{cmbright}\Huge\$ . You then used the value $\displaystyle{z}=-{1}\usepackage{cmbright}\Huge\$ to find the values of x and y in the second and third equations. However, the values do not make the first equation true, so they are not a solution.