Question

Решить систему уравнений

(x-5)(y+2)=0

x(в кв)-2xy=5

Answer 1

$begin{cases} (x-5)(y+2)=0\ x^2-2xy=5 end{cases}\ x-5=0Rightarrow x=5:\ 25-10y=5\ 10y=20\ y=2\ y+2=0Rightarrow y=-2\ x^2+4x-5=0\ D=16+4cdot5=36\ x_1=-5,;x_2=1\ begin{cases} x=5\y=2 end{cases}quad begin{cases} x=-5\y=-2 end{cases}quad begin{cases} x=1\y=-2 end{cases}$

Answer 2

1)

(x-5)(y+2)=0

xy + 2x - 5y - 10 = 0

x = 5

y = -2