Question

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Answer 1

здравствуйте.

Находим y:

$\left \{ {{y^2-xy+x=2} \atop {5y+x=12}} \right. \left \{ {y^2-xy+x=2} \atop {x=12-5y}} \right. \\\\y^2-(12-5y)y+(12-5y)=2\\6y^2-17y+10=0\\(6y-5)(y-2)=0\\\\y1=\frac{5}{6} \\\\y2=2$

Находим х:

$x=12-5y\\\\x1=12-5*\frac{5}{6} \\x1=7\frac{5}{6} \\\\x2=12-5*2\\x2=2$

Проверяем:

$Proverka:\\\boxed{1} \\\\\left \{ {{(\frac{5}{6})^2-\frac{5}{6}*7\frac{5}{6}+7\frac{5}{6}=2} \atop {5*\frac{5}{6}+7\frac{5}{6}=12}} \right. \left \{ {{2=2} \atop {12=12}} \right. - Dokazano\\\\\boxed{2}\\\\\left \{ {{2^2-2*2+2=2} \atop {5*2+2=12}} \right. \left \{ {{2=2} \atop {12=12}} \right. - Dokazano$

Ответ:

$x1=7\frac{5}{6} \:u\: y1=\frac{5}{6} \\\\x2=2 \:u\: y2=2$

удачи в учёбе.