Решите систему неравенств. 1)4х-2y=-6 3x-3y=9 2)x+y=5 x(в кв.)+y( в кв.)=6

$\left \{{{4x-2y=-6|:2}\atop {3x-3y=9|:3}}\right.\\\\\left\{{{2x-y=-3}\atop {x-y=3}}\right.\\\\\left \{{{y=2x+3} \atop {x-2x-3=3}}\right.\\\\\left\{{{y=2x+3}\atop{-x-3=3}}\right.\\\\\left\{{{y=2x+3}\atop{x=-6}} \right.\\\\\left\{{{y=2(-6)+3} \atop {x=-6}} \right.\\\\\left \{{{y=-12+3}\atop{x=-6}} \right.\\\\\left\{ {{y=9} \atop {x=-6}} \right.\\\\(-6;9)$

$\left \{{{x+y=5}\atop {x^2+y^2=6}} \right.\\\\\left\{{{x=5-y}\atop {(5-y)^2+y^2=6}}\right.\\\\(5-y)^2+y^2=6\\25-10y+y^2+y^2-6=0\\2y^2-10y+19=0\\D=(-10)^2-4*2*19=100-152=-52<0$
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