Correct. After multiplying both sides of the equation by the common denominator $\displaystyle{2}{\left({x}-{4}\right)}{\left({x}+{1}\right)}\usepackage{cmbright}\Huge\$ , you find $\displaystyle{2}{x}{\left({x}-{4}\right)}={x}{\left({x}+{1}\right)}\usepackage{cmbright}\Huge\$ , so $\displaystyle{2}{{x}}^{{2}}-{8}{x}={{x}}^{{2}}+{x}\usepackage{cmbright}\Huge\$ , and $\displaystyle{{x}}^{{2}}-{9}{x}={0}\usepackage{cmbright}\Huge\$ . This can be factored as $\displaystyle{x}{\left({x}-{9}\right)}={0}\usepackage{cmbright}\Huge\$ , so $\displaystyle{x}={0}\usepackage{cmbright}\Huge\$ and 9.