Решите систему уравнений:
[tex] \left \{ {{x-y=1} \atop {x^2-y=3}} \right. [/tex]
[tex] \left \{ {{x-y=1} \atop {x^2-y=3}} \right. [/tex]
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x²-y=3 (1+y)²-y=3
Находим y
(1+y)²-y=3
1+2y+y²-y=3
y²+2y-y+1-3=0
y²+y-2=0
D=1+8=9
y₁=1
y₂=-2
Находим x
x₁=1+1=2
x₂=1-2=-1
Ответ: x₁=2 y₁=1
x₂=-1 y₂=-2