Correct. If $\displaystyle{y}-{2}={3}{\left({x}-{4}\right)},\usepackage{cmbright}\Huge\$ then $\displaystyle{y}-{2}={3}{x}-{12},\usepackage{cmbright}\Huge\$ so $\displaystyle-{2}={3}{x}-{y}-{12},\usepackage{cmbright}\Huge\$ and therefore $\displaystyle{10}={3}{x}-{y}.\usepackage{cmbright}\Huge\$