Question

Решительно систему уравнений (x-2)*(y+3)=160, y-x=1

Answer 1

$left { {{(x-2)(y+3)=160} atop {y-x=1}} right.$
$left { {{(x-2)(y+3)=160} atop {y=1+x}} right.$
$left { {{(x-2)(1+x+3)=160} atop {y=1+x}} right.$
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(x-2)(1+x+3)=160
(x-2)(x+4)=160
$x^{2}$+4x-2x-8=160
$x^{2}$+2x-168=0
D=4+4*168=676
x1=12
x2= -14
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$left { {{x1=12} atop {y=1+12}} right.$
$left { {{x2=-14} atop {y=1-14}} right.$
$left { {{x=12} atop {y=-13}} right.$