$$3x^{2}+2xy+y^{2}=80; 2x-y=4$$
$x=\frac{12-4\sqrt{53}}{11}\approx -1.556403596\text{, }y=\frac{-8\sqrt{53}-20}{11}\approx -7.112807192$
$x=\frac{4\sqrt{53}+12}{11}\approx 3.738221778\text{, }y=\frac{8\sqrt{53}-20}{11}\approx 3.476443556$
