Substitute $\displaystyle{5}-{2}{y}\usepackage{cmbright}\Huge\$ for x in the second equation: $\displaystyle{4}{y}={15}-{2}{\left({5}-{2}{y}\right)}\usepackage{cmbright}\Huge\$

Simplify: $\displaystyle{4}{y}={15}-{10}+{4}{y}\usepackage{cmbright}\Huge\$ or $\displaystyle{4}{y}={5}+{4}{y}\usepackage{cmbright}\Huge\$ , so $\displaystyle{0}={5}\usepackage{cmbright}\Huge\$

Both equations being true would imply that $\displaystyle{0}={5}\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.