Question

Решите систему уравнений методом замены переменной, пожалуйста
x2−2xy+y2=4
x2+xy=4

Answer 1

x²-2xy+y²=4⇒(x-y)²=4⇒x-y=-2 U x-y=2
x²+xy=4
1)x-y=-2
y=x+2
x²+x²+2x-4=0
2x²+2x-4=0
x²+x-2=0
x1+x2=-1 U x1*x2=-2
x1=-2⇒y1=-2+2=0
x2=1⇒y2=1+2=3
2)x-y=2
y=x-2
x²+x²-2x-4=0
2x²-2x-4=0
x²-x-2=0
x1=x2=1 U x1*x2=-2
x1=-1⇒y1=-1-2=-3
x2=2⇒y2=2-2=0
(-2;0);(1;3);(-1;-3);(2;0)

Answer 2

$\left \{ {{ x^{2} +2xy+y ^{2}=4 } \atop { x^{2} +xy=4}} \right. \\\\ \left \{ {{(x-y) ^{2}=4 } \atop { x^{2} +xy=4}} \right.\\\\ 1)\left \{ {{x-y=2} \atop { x^{2} +xy=4}} \right.\\\\ \left \{ {{y=x-2} \atop { x^{2} +x(x-2)=4}} \right.\\\\ \left \{ {{y=x-2} \atop { x^{2} + x^{2} -2x-4=0}} \right. \\\\ \left \{ {{y=x-2} \atop { x^{2} -x-2=0}} \right.\\\\ \left \{ {{x _{1}=2 } \atop {y _{1} = 0}} \right. \\\\ \left \{ {{x _{2} =-1} \atop {y _{2}=-3 }} \right.$

$2) \left \{ {{x-y=-2} \atop { x^{2} +xy=4}} \right.\\\\ \left \{ {{y=x+2} \atop { x^{2} +x(x+2)=4}} \right.\\\\ \left \{ {{y=x+2} \atop { x^{2} + x^{2} +2x-4=0}} \right.\\\\ \left \{ {{y=x+2} \atop { x^{2} +x-2=0}} \right.\\\\ \left \{ {{x _{3} =-2} \atop {y _{3} =0}} \right.\\\\ \left \{ {{x _{4}=1 } \atop {y _{4}=3 }} \right.$
Ответ: (2 ; 0), (- 1 ; - 3) , ( - 2 , 0) ,(1 , 3)