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

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

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

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