Question

2
(x² - y = 3
у
l x + y = 3
=
(4; -1)
(2; -1)
36:32
Які з пар чисел є розв'язком системи рiвнянь?
(5; -2)
(-3; 6)
Во

Answer 1

$\left \{ {{x^{2} -y=3} \atop {x+y=3}} \right. \: \: \: \: \left \{ {{x^{2} -(3-x)=3} \atop {y=3-x}} \right. \\\\x^{2} +x-3-3=0\\x^{2} +x-6=0\\D = 1^{2} -4*1*(-6)=1+24=25\\x_{1} =\frac{-1-5}{2} =-\frac{6}{2} =-3\\x_{2} =\frac{-1+5}{2} =\frac{4}{2} =2\\\\y_{1} =3-(-3)=3+3=6\\x_{2} =3-2=1$

Ответ: ( -3 ; 6 ) и ( 2 ; 1 )