You may have thought you needed to solve $\displaystyle-{6}{c}-{3}={15}\usepackage{cmbright}\Huge\$ , but you put the negative sign in the wrong place. If $\displaystyle{\left|{X}\right|}={15}\usepackage{cmbright}\Huge\$ , then $\displaystyle{X}={15}\usepackage{cmbright}\Huge\$ or $\displaystyle{X}=-{15}\usepackage{cmbright}\Huge\$ . In this problem, X is replaced by $\displaystyle{6}{c}-{3}\usepackage{cmbright}\Huge\$ . You should be solving two equations, $\displaystyle{6}{c}-{3}={15}\usepackage{cmbright}\Huge\$ and $\displaystyle{6}{c}-{3}=-{15}\usepackage{cmbright}\Huge\$ .