Add the first two equations to eliminate z: $\displaystyle{2}{x}+{y}={10}\usepackage{cmbright}\Huge\$

Add the third equation, $\displaystyle{2}{x}-{y}={2}\usepackage{cmbright}\Huge\$ , to this: $\displaystyle{4}{x}={12}\usepackage{cmbright}\Huge\$

Solve for x: $\displaystyle{x}={3}\usepackage{cmbright}\Huge\$

Substitute into the third equation: $\displaystyle{2}{\left({3}\right)}-{y}={2}\usepackage{cmbright}\Huge\$ , or $\displaystyle{6}-{y}={2}\usepackage{cmbright}\Huge\$ , so $\displaystyle{y}={4}\usepackage{cmbright}\Huge\$

Substitute into the second equation: $\displaystyle{3}+{z}={4}\usepackage{cmbright}\Huge\$ , so $\displaystyle{z}={1}\usepackage{cmbright}\Huge\$