3x+y=2 x^2-y^2=-12 решите систему, во второй строке должно быть 3x=2-y, ответ должен...

$\left \{ {{3x+y=2} \atop {x^{2}-y^{2}=-12}} \right.\\\\\left \{ {{y=2-3x} \atop {x^{2}-(2-3x)^{2}=-12}} \right.\\\\\left \{ {{y=2-3x} \atop {x^{2}-4+12x-9x^{2}+12=0 }} \right.\\\\\left \{ {{y=2-3x} \atop {-8x^{2} +12x+8=0}} \right.\\\\\left \{ {{y=2-3x} \atop {2x^{2}-3x-2=0 }} \right. \\\\2x^{2} -3x-2=0\\\\D=(-3)^{2}-4*2*(-2)=9+16=25=5^{2} \\\\x_{1}=\frac{3+5}{4} =2\\\\x_{2}=\frac{3-5}{4}=-0,5\\\\y_{1}=2-3x_{1}=2-3*2=2-6=-4\\\\y_{2} =2-3x_{2}=2-3*(-0,5)=2+1,5=3,5$

Ответ : (2 ; - 4) , (- 0,5 , 3,5)

Второй способ:

3x = 2 - y

$x=\frac{2-y}{3} \\\\(\frac{2-y}{3})^{2}-y^{2}=-12\\\\\frac{4-4y+y^{2} }{9}-y^{2}=-12$

4 - 4y + y² - 9y² + 108 = 0

-8y² - 4y + 112 = 0

2y² + y - 28 = 0

D = 1 - 4 * 2 * (- 28) = 1 + 224 = 225 = 15²

$y_{1}=\frac{-1+15}{4}=3,5\\\\y_{2}=\frac{-1-15}{4}=-4\\\\x_{1}=-0,5\\\\x_{2}=2$