Question

решите систему уравнений методом подстановки а)y=2-x x2+4y=8
Б)x-y=4
X2+xy=6

Answer 1

а)y=2-x
x2+4y=8
x²+4(2-x)-8=0
x²-4x=0
x(x-4)=0
x=0⇒y=2
x=4⇒y=-2
(0;2);(4;-2)

Б)x-y=4⇒y=x-4
X2+xy=6
x²+x(x-4)-6=0
x²+x²-4x-6=0
2x²-4x-6=0
x²-2x-3=0
x1+x2=2 U x1*x2=-3
x1=-1⇒y=-5
x2=3⇒y=-1
(-1;-5);(3;-1)

Answer 2

$left { {{y=2-x} atop {x^2+4y=8}} right. \ \ left { {{y=2-x} atop {x^2+4(2-x)=8}} right. \ \ left { {{y=2-x} atop {x^2+8-4x-8=0}} right. \ \ left { {{y=2-x} atop {x^2-4x=0}} right. \ \ left { {{y=2-x} atop {x(x-4)}} right. \ \ left { {{x=0} atop {y=2}} right. i li left { {{x=4} atop {y=-2}} right.$

$left { {{x-y=4} atop {x^2+xy=6}} right. \ \ left { {{x=y+4} atop {(y+4)^2+(y+4)y=6}} right. \ \ left { {{x=y+4} atop {y^2+8y+16+y^2+4y-6=0}} right. \ \ left { {{x=y+4} atop {2y^2+12y+10=0}} right. \ \ left { {{x=y+4} atop {2(y^2+6y+5)=0}} right. \ \ left { {{x=y+4} atop {y^2+6y+5=0}} right. \ \ left { {{x=y+4} atop {y^2+5y+y+5=0}} right. \ \ left { {{x=y+4} atop {(y+5)(y+1))=0}} right.$

$left { {{x=-1} atop {y=-5}} right. ili left { {{x=3} atop {y=-1}} right.$