Question

3x+y=2 x^2-y^2=-12 решите систему, во второй строке должно быть 3x=2-y, ответ должен получиться (2;-4) и (-0,5;3,5)


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Answer 1

{3x+y=2

{y^2-x^2=12

{y=2-3x

{(2-3x)^2-x^2=12

4-12x+8x^2=12

8x^2-12x-8=0

D=144+4*8*8=20^2

x1=(12+20)/16=2

x2=(12-20)/16=-1/2

y1=2-6=-4

y2=2+3/2=7/2

Ответ (2,-4) U (-1/2,7/2)

Answer 2

$\left \{ {{3x+y=2} \atop {x^{2}- y^{2}=-12}} \right. \\\\\left \{ {{y=2-3x} \atop {x^{2}-(2-3x)^{2}=-12}} \right.\\\\\left \{ {{y=2-3x} \atop {x^{2}-4+12x-9x^{2}+12=0}} \right.\\\\\left \{ {{y=2-3x} \atop {-8x^{2}+12x+8=0 }} \right.\\\\\left \{ {{y=2-3x} \atop {2x^{2}-3x-2=0 }} \right.$

$1)\left \{ {{x_{1}=2 } \atop {y_{1}=2-3*2}} \right.\\\\\left \{ {{x_{1}=2 } \atop {y_{1} =-4}} \right. \\\\2)\left \{ {{x_{2}=-0,5 } \atop {y_{2}=2-3*(-0,5) }} \right.\\\\\left \{ {{x_{2}=-0,5 } \atop {y_{2}=3,5 }} \right.\\\\Otvet:(2;-4),(-0,5;3,5)$