х^2+y^2=40
y=3х
решить систему?
х^2+(3x)^2=40
х^2=4
х=+-2
x=-2, y=-6
x=2, y=6
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х^2+y^2=40
y=3х
х^2+(3x)^2=40
y=3х
х^2=4
y=3х
х=+-2
y=3х
x=-2, y=-6
x=2, y=6