Решить систему уравнений

$\left \{ {{x^2-xy-20y^2=0} \atop {x^2+2xy-3y^2=32}} \right.; \left \{ {{x^2-5xy+4xy-20y^2=0} \atop {x^2+2xy+y^2-4y^2=32}} \right. ; \left \{ {{x(x-5y)+4y(x-5y)=0} \atop {(x+y)^2-(2y)^2=32}} \right.; \\ \left \{ {{(x+4y)(x-5y)=0} \atop {(x+y-2y)(x+y+2y)=32}} \right.; \left \{ {{x=-4y\ or\ x=5y} \atop {(x-y)(x+3y)=32}} \right. ;\\ \left \{ {{x=-4y} \atop {(-4y-y)(-4y+3y)=32}} \right. \ or\ \left \{ {{x=5y} \atop {(5y-y)(5y+3y)=32}} \right.;\\$

$\left \{ {{x=-4y} \atop {5y^2=32}} \right. \ or\ \left \{ {{x=5y} \atop {32y^2=32}} \right.;\\ \left \{ {{x=-4y} \atop {y=\pm 4\sqrt{\frac{2}{5}} }} \right. \ or\ \left \{ {{x=5y} \atop {y=\pm1}} \right.;\\ \left \{ {{x=\mp16 \sqrt{ \frac{2}{5} } } \atop {y=\pm 4\sqrt{\frac{2}{5}} }} \right. \ or\ \left \{ {{x=\pm5} \atop {y=\pm1}} \right..\\$

Answer: $(-16 \sqrt{ \frac{2}{5}};\ 4 \sqrt{ \frac{2}{5}}),\ (+16 \sqrt{ \frac{2}{5}};\ -4 \sqrt{ \frac{2}{5}}),\ (5;\ 1),\ (-5;\ -1)$