Question

{х^2-у^2=-3,х-у=3;
{х^2-4ху+у^2=6,у-2х=3;

Помогите пожалуйста

Answer 1

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Answer 2

№1

$left { {{ x^{2}- y^{2}=-3 } atop {x-y=3}} right. left { {{ x^{2} - y^{2}=-3 } atop {x=y+3}} right.$

$(y+3)^{2} - y^{2} =-3$
$y^{2} +6y+9- y^{2}=-3$
$6y=-12$
$y = -12:6$
$y = -2$

$x = -2 +3=1$

№2

$left { {{ x^{2} -4xy+ y^{2}=6 } atop {y-2x=3}} right. left { {{ x^{2} -4xy+ y^{2}=6 } atop {y=2x+3}} right.$

$x^{2} -4x*(2x+3)+ (2x+3)^{2}=6$
$x^{2} -8x^{2}-12x+4 x^{2} +12x+9=6$
$-3 x^{2} =-3$
$x^{2} =-3:(-3)$
$x^{2} =1$
$x=1$

$y =2*1+3=5$