РЕШИТЕ ПОЖАЛУЙСТА!!!

$\left \{ {{\frac{3x+4y^2}{6}-\frac{2x^2-4y}{5}=-\frac{2}{15} \atop {3x-2y=4}} \right. \\\\1)\ \frac{3x+4y^2}{6}-\frac{2x^2-4y}{5}=-\frac{2}{15}\ |\bullet30,\\\\5(3x+4y^2)-6(2x^2-4y)=-4,\\\\15x+20y^2-12x^2+24y=-4,\\\\\\2)\ 3x-2y=4,\ x=\frac{4+2y}{3}\\\\\\3)\ \frac{15(4+2y)}{3}+20y^2-12 \bullet(\frac{4+2y}{3})^2+24y+4=0,$

$20+10y+20y^2-12\bullet\frac{16+16y+4y^2}{9}+24y+4=0,\\\\24+34y+20y^2+\frac{-64-64y-16y^2}{3}=0\ |\bullet3,\\\\72+102y+60y^2-64-64y-16y^2=0,\\\\44y^2+38y+8=0\ |:2,\\\\22y^2+19y+4=0,\\\\D=19^2-4\bullet4\bullet22=9,\\\\y_1=\frac{-19+\sqrt{9}}{44}=-\frac{16}{44}=-\frac{4}{11},\\y_2=\frac{-19-\sqrt{9}}{44}=-\frac{22}{44}=-\frac{1}{2};$

$4)\ x_1= \frac{4+2(-\frac{4}{11})}{3}=\frac{4}{3}-\frac{8}{33} =\frac{44-8}{33}=\frac{36}{33}=1\frac{3}{33}=1\frac{1}{11},\\x_2=\frac{4+2(-\frac{1}{2})}{3}=\frac{4-1}{3}=1.$

Ответ: $(1\frac{1}{11};-\frac{4}{11}),\ (1;-\frac{1}{2}).$