Question

Срочно дам 100 балов за решение(алгебра)​

Answer 1

Ответ: x=2; y=1

Объяснение:

$\left \{ {{x^{3}y^{3}+x^{2}y^{4}=12} \atop {x^{4}y^{2}+x^{3}y^{3}=24}} \right.$

$\left \{ {{x^{2}y^{3}(x+y)=12} \atop {x^{3}y^{2}(x+y)=24}} \right.$

$\left \{ {{(x+y)=\frac{12}{x^{2}y^{3}}} \atop {(x+y)=\frac{24}{x^{3}y^{2}}} \right.$

$\left \{ {\frac{24}{x^{3}y^{2}}=\frac{12}{x^{2}y^{3}}} \atop {(x+y)=\frac{12}{x^{2}y^{3}}} \right.$

$\left \{ {{\frac{x^{3}}{x^{2}}=\frac{24y^{3}}{12y^{2}}} \atop {(x+y)=\frac{12}{x^{2}y^{3}}}} \right.$

$\left \{ {x=2y} \atop {(2y+y)=\frac{12}{4y^{2}y^{3}}}} \right.$

$\left \{ {x=2y} \atop {3y=\frac{3}{y^{5}}} \right.$

$\left \{ {x=2y} \atop {y^{6}=1} \right.$

$\left \{ {{x=2y} \atop {y=1}} \right.$

$\left \{ {{x=2} \atop {y=1}} \right.$