When adding the first two equations to eliminate z, you forgot to add the term y. This gave you $\displaystyle{2}{x}={10}\usepackage{cmbright}\Huge\$ . You then used the value $\displaystyle{x}={5}\usepackage{cmbright}\Huge\$ to find the values of y and z in the second and third equations. However, the values of x, y, and z must make each of the equations in the system true to be a solution. These values do not make the first equation true.