While it is true that $\displaystyle{\left({4},-{3}\right)}\usepackage{cmbright}\Huge\$ is not a solution to the system, it is actually a solution to the second equation. You may have incorrectly substituted. Make sure that you replace x by 4 and y by $\displaystyle-{3}\usepackage{cmbright}\Huge\$ . The correct answer is: No, because $\displaystyle{\left({4},-{3}\right)}\usepackage{cmbright}\Huge\$ is a solution to only one equation.