Since the solution of the system must be a solution to both of the equations in the system, check the point in each equation. Substitute $\displaystyle-{2}\usepackage{cmbright}\Huge\$ for x and 3 for y.

$\displaystyle{3}={5}{\left(-{2}\right)}+{13}\usepackage{cmbright}\Huge\$ $\displaystyle{3}{\left(-{2}\right)}-{2}{\left({3}\right)}=-{12}\usepackage{cmbright}\Huge\$

$\displaystyle{3}=-{10}+{13}\usepackage{cmbright}\Huge\$ $\displaystyle-{6}-{6}=-{12}\usepackage{cmbright}\Huge\$

$\displaystyle{3}={3}\usepackage{cmbright}\Huge\$ $\displaystyle-{12}=-{12}\usepackage{cmbright}\Huge\$

Since both equations are true, $\displaystyle{\left(-{2},{3}\right)}\usepackage{cmbright}\Huge\$ is a solution to the system.

$\displaystyle{3}={5}{\left(-{2}\right)}+{13}\usepackage{cmbright}\Huge\$	$\displaystyle{3}{\left(-{2}\right)}-{2}{\left({3}\right)}=-{12}\usepackage{cmbright}\Huge\$
$\displaystyle{3}=-{10}+{13}\usepackage{cmbright}\Huge\$	$\displaystyle-{6}-{6}=-{12}\usepackage{cmbright}\Huge\$
$\displaystyle{3}={3}\usepackage{cmbright}\Huge\$	$\displaystyle-{12}=-{12}\usepackage{cmbright}\Huge\$