Question

(x + y = 6,
y = x² ;

Answer 1

$\displaystyle\bf\\\left \{ {{x+y=6} \atop {y=x^{2} }} \right. \\\\\\\left \{ {{x^{2} +x-6=0} \atop {y=x^{2} }} \right. \\\\\\x^{2} +x-6=0\\\\D=1^{2} -4\cdot(-6)=1+24=25=5^{2} \\\\\\x_{1} =\frac{-1-5}{2} =\frac{-6}{2} =-3\\\\\\x_{2} =\frac{-1+5}{2} =\frac{4}{2} =2\\\\\\1)\\\\x=-3 \ \ \ \Rightarrow \ \ \ y=(-3)^{2} =9\\\\2)\\\\x=2 \ \ \ \Rightarrow \ \ \ y=2^{2} =4\\\\\\Otvet \ . \ (-3 \ ; \ 9) \ , \ (2 \ ; \ 4)$