Question

Решите систему уравнении x^2+y^2=26,x-y=4

Answer 1

$\displaystyle \tt \left \{ {{x^2+y^2=26} \atop {x-y=4}} \right. \to \left \{ {{x^2+y^2=26} \atop {x=4+y}} \right. \to \left \{ {{(4+y)^2+y^2=26} \atop {x=4+y}} \right. \\4^2+2\cdot4\cdot y+y^2+y^2=26\\16+8y+y^2+y^2=26\\2y^2+8y+16-26=0\\2y^2+8y-10=0 \: \: \: \: | \div 2\\y^2+4y-5=0\\D=4^2-4\cdot1\cdot(-5)=16+20=36\\\sqrt{D}=\sqrt{36}=6\\x_1=\frac{-4+6}{2}=\frac{2}{2}=1\\x_2=\frac{-4-6}{2}=\frac{-10}{2}=-5$

$\displaystyle \tt \left \{ {{y_1=1} \atop {x_1=4+1}} \right. \to \left \{ {{y_1=1} \atop {x_1=5}} \right. \\\left \{ {{y_2=-5} \atop {x_2=4-5}} \right. \to\left \{ {{y_2=-5} \atop {x_2=-1}} \right.$

Ответ: $(5; \: 1);(-1; \: -5)$