Substitute $\displaystyle{7}+{3}{y}\usepackage{cmbright}\Huge\$ for x in the second equation: $\displaystyle{18}{y}={6}{\left({7}+{3}{y}\right)}-{40}\usepackage{cmbright}\Huge\$

Simplify: $\displaystyle{18}{y}={42}+{18}{y}-{40}\usepackage{cmbright}\Huge\$ or $\displaystyle{18}{y}={2}+{18}{y}\usepackage{cmbright}\Huge\$ , so $\displaystyle{0}={2}\usepackage{cmbright}\Huge\$

Both equations being true would imply that $\displaystyle{0}={2}\usepackage{cmbright}\Huge\$ . Since this is false, you cannot find numbers that make both equations true. In other words, the system does not have a solution.