Question

Решите систему уравнений введением новых переменных

Answer 1

Объяснение:

$\left \{ {{\frac{1}{2x-y^2}+y=1 } \atop {\frac{y}{2x+y^2}=-6 }} \right. \ \ \ \ \left \{ {{\frac{1}{(-\frac{y}{6} )}+y=1 } \atop {2x-y^2=-\frac{y}{6} }} \right. \ \ \ \ \left \{ {{-\frac{6}{y} +y=1} \atop {2x-y^2=-\frac{y}{6} }} \right.\ \ \ \ \left \{ {{-6+y^2=y} \atop {2x-y^2=-\frac{y}{6}}} \right. \ \ \ \ \left \{ {{y^2-y-6=0} \atop {2x-y^2=-\frac{y}{6}}} \right.\\$

$\left \{ {{D=25\ \ \ \ \sqrt{D}=5 } \atop {2x-y^2=-\frac{y}{6}}} \right.\ \ \ \ \ \left \{ {{y_1=-2\ \ \ \ y_2=3} \atop {2x-y^2=-\frac{y}{6} }} \right. .$

1) y₁=-2.

$2x-(-2)^2=-\frac{-2}{6} \\2x-4=\frac{1}{3} \ |*3\\6x-12=1\\6x=13\ |:6\\x_1=\frac{13}{6}=2\frac{1}{6} .$

2) y₂=3.

$2x-3^2=-\frac{3}{6}\\2x-9=-\frac{1}{2}\ |*2\\4x-18=-1\\4x=17\ |:4\\x_2=\frac{17}{4} =4\frac{1}{4}.\\$

ответ: (2¹/₆;-2), (4¹/₄;3).