Question

решить системы квадратных уравнений

x^2+4xy=-1
x+2y=0

и

4x+y=4
x-y=-4

Answer 1

$\displaystyle\bf\\1) \ \left \{ {{x^{2} +4xy=-1} \atop {x+2y=0}} \right. \\\\\\\left \{ {{x^{2} +4xy=-1} \atop {x^{2} +4xy+4y^{2} =0}} \right. \\\\\\\left \{ {{x^{2} +4xy=-1} \atop {-1+4y^{2}=0 }} \right.\\\\\\\left \{ {{x^{2} +4xy=-1} \atop {4y^{2} =1}} \right.\\\\\\\left \{ {{x^{2} +4xy=-1} \atop {y^{2}=\frac{1}{4} }} \right. \\\\\\1) \ \left \{ {{x+2y=0} \atop {y=-\frac{1}{2} }} \right. \\\\\\\left \{ {{x=-2\cdot(-\frac{1}{2} )} \atop {y=-\frac{1}{2} }} \right.$

$\displaystyle\bf\\\left \{ {{x_{1} =1} \atop {y_{1} =-\frac{1}{2} }} \right. \\\\\\2) \ \ \left \{ {{x+2y=0} \atop {y=\frac{1}{2} }} \right. \\\\\\\left \{ {{x=-2\cdot\frac{1}{2} } \atop {y=\frac{1}{2} }} \right.\\\\\\\left \{ {{x_{2} =-1} \atop {y_{2} =\frac{1}{2} }} \right. \\\\\\Otvet:(1 \ ; \ -\frac{1}{2} ) \ , \ (-1 \ ; \ \frac{1}{2} )\\\\\\\\2) \ \\+\left \{ {{4x+y=4} \atop {x-y=-4}} \right. \\--------\\5x=0\\\\x=0\\\\y=x+4=0+4=4\\\\Otvet:(0 \ ; \ 4)$