Multiply both sides by the common denominator 15:

$\displaystyle{15}\cdot\frac{{{x}+{4}}}{{5}}=\frac{{x}}{{15}}\cdot{15}\usepackage{cmbright}\Huge\$

$\displaystyle\frac{{15}}{{5}}\cdot{\left({x}+{4}\right)}={x}\cdot\frac{{15}}{{15}}\usepackage{cmbright}\Huge\$

$\displaystyle{3}{\left({x}+{4}\right)}={x}\usepackage{cmbright}\Huge\$ or $\displaystyle{3}{x}+{12}={x}\usepackage{cmbright}\Huge\$

$\displaystyle{2}{x}+{12}={0}\usepackage{cmbright}\Huge\$

$\displaystyle{2}{x}=-{12}\usepackage{cmbright}\Huge\$ , so $\displaystyle{x}=-{6}\usepackage{cmbright}\Huge\$