Since the denominators are equal, the numerators must be equal: $\displaystyle{{x}}^{{2}}-{1}={24}\usepackage{cmbright}\Huge\$

Subtract 24 and factor: $\displaystyle{{x}}^{{2}}-{25}={0}\usepackage{cmbright}\Huge\$ , so $\displaystyle{\left({x}+{5}\right)}{\left({x}-{5}\right)}={0}\usepackage{cmbright}\Huge\$

Solve: $\displaystyle{x}=-{5},{5}\usepackage{cmbright}\Huge\$

In the original equation, the denominator $\displaystyle{x}+{5}\usepackage{cmbright}\Huge\$ makes $\displaystyle-{5}\usepackage{cmbright}\Huge\$ an extraneous solution.