решите систему уравнений(2x+y)*(x+3y)=48 (2x+y) / (x+3y)=3/4

$\left \{ {{(2x+y)(x+3y)=48} \atop { \frac{2x+y}{x+3y}= \frac{3}{4} }} \right. \\ \\ 2x+y= \frac{48}{x+3y} \\ \\ \frac{48}{ (x+3y)^{2}}= \frac{3}{4} \\ \\ x+3y= \sqrt{48* \frac{4}{3} } = 8 \\ 2x+y=48/8=6 \\ \left \{ {{2x+y=6} \atop {x+3y=8}} \right. \\ \\ \left \{ {{2x+y=6} \atop {-2x-6y=-16}} \right. \\ \\ 5y=10 \\ y=2 \\ x=2$
Ответ: (2; 2)