Question

Розв'яжіть систему рівнянь методом заміни.
x+y+xy=9
x^2+y^2=17

Answer 1

$\displaystyle\bf\\\left \{ {{x+y+xy=9} \atop {x^{2} + y^{2} }=17} \right. \\\\\\\left \{ {{x+y+xy=9} \atop {(x^{2} + 2xy +y^{2} )-2xy}=17} \right.\\\\\\\left \{ {{(x+y)+xy=9} \atop {(x + y)^{2} -2xy}=17} \right. \\\\\\x+y=m \ \ ; \ \ xy=n\\\\\\\left \{ {{m+n=9} \ |\cdot 2 \atop {m^{2} -2n=17}} \right.\\\\\\+\left \{ {{2m+2n=18} \atop {m^{2}-2n=17 }} \right. \\----------\\m^{2} +2m=35\\\\m^{2} +2m-35=0\\\\D=2^{2} -4\cdot(-35)=4+140=144=12^{2}$

$\displaystyle\bf\\m_{1} =\frac{-2-12}{2} =-7\\\\\\m_{2} =\frac{-2+12}{2} =5\\\\\\n_{1} =9-(-7)=9+7=16\\\\n_{2} =9-5=4\\\\\\1)\\\\\left \{ {{x+y=-7} \atop {xy=16}} \right. \\\\\\\left \{ {{x=-y-7} \atop {(-y-7)\cdot y-16=0}} \right. \\\\\\\left \{ {{x=-y-7} \atop {-y^{2} -7y-16=0}} \right.\\\\\\y^{2} +7y+16=0\\\\D=7^{2} -4\cdot 16=49-64=-15 < 0\\\\kornei \ \ net\\\\\\2)\\\\\left \{ {{x+y=5} \atop {xy=4}} \right.\\\\Teorema \ Vieta:$

$\displaystyle\bf\\x_{1} =1 \ \ \ ; \ \ \ y_{1} =4\\\\x_{2} =4 \ \ \ ; \ \ \ y_{2} =1\\\\Otvet:(1 \ ; \ 4) \ \ , \ \ (4 \ ; \ 1)$