Correct. If $\displaystyle{y}-{4}=\frac{{2}}{{3}}{\left({x}+{1}\right)},\usepackage{cmbright}\Huge\$ then $\displaystyle{3}{\left({y}-{4}\right)}={3}{\left(\frac{{2}}{{3}}{\left({x}+{1}\right)}\right)},\usepackage{cmbright}\Huge\$ so $\displaystyle{3}{y}-{12}={2}{\left({x}+{1}\right)}\usepackage{cmbright}\Huge\$ , so $\displaystyle{3}{y}-{12}={2}{x}+{2},\usepackage{cmbright}\Huge\$ and therefore $\displaystyle-{14}={2}{x}-{3}{y}.\usepackage{cmbright}\Huge\$