{
x
2
+y
2
=5
x(1+y)+4=−4y
\left \{ {{x(1+y)+4+4y=0} \atop {x^2+y^2=5}} \right.{
x
2
+y
2
=5
x(1+y)+4+4y=0
\left \{ {{x(1+y)+4(1+y)=0} \atop {x^2+y^2=5}} \right.{
x
2
+y
2
=5
x(1+y)+4(1+y)=0
\left \{ {{(1+y)(x+4)=0} \atop {x^2+y^2=5}} \right.{
x
2
+y
2
=5
(1+y)(x+4)=0
\left \{ {{1+y=0} \atop {x^2+y^2=5}} \right.{
x
2
+y
2
=5
1+y=0
или \left \{ {{x+4=0} \atop {x^2+y^2=5}} \right.{
x
2
+y
2
=5
x+4=0
\left \{ {{y=-1} \atop {x^2+(-1)^2=5}} \right.{
x
2
+(−1)
2
=5
y=−1
или \left \{ {{x=-4} \atop {(-4)^2+y^2=5}} \right.{
(−4)
2
+y
2
=5
x=−4
\left \{ {{y=-1} \atop {x^2=4}} \right.{
x
2
=4
y=−1
или \left \{ {{x=-4} \atop {(y^2=-11}} \right.{
(y
2
=−11
x=−4
\left \{ {{y=-1} \atop {x=-2}} \right.{
x=−2
y=−1
или \left \{ {{y=-1} \atop {x=2}} \right.{
x=2
y=−1
или нет решения
О т в е т. (-2;-1)(−2;−1) ; (2;-1)(2;−1)
