Question

Решить систему уравнений
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Answer 1

$\left \{ {{x+2y=4} \atop {x^2+xy=y-5}} \right.$

1) $x+2y=4$

$x = 4-2y$

2) $(4-2y)^2+(4-2y)*y=y-5$

$16-16y+4y^2+4y-2y^2=y-5$

$16-16y+4y^2+4y-2y^2-y+5=0$

$2y^2-13y+21=0$

$D=(-13)^2-4*2*21=169-168=1$

$y_{1} = \frac{13+ \sqrt[]{1} }{4} =\frac{13+1}{4} =\frac{14}{4} = \frac{7}{2}\\y_{2} = \frac{13- \sqrt[]{1} }{4} =\frac{13-1}{4} =\frac{12}{4} = 3$

3) $x_{1} = 4-2y_{1}= 4-2*\frac{7}{2} =4-7=-3\\x_{2} = 4-2y_{2}= 4-2*3 =4-8=-2$

Ответ: $(-3;\frac{7}{2})$    $(-2;3)$