x^2+y^2=16 y=x-1
Решение
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[TeX]
[pretty]
[text]
2 2 x + y = 16
$$x^{2} + y^{2} = 16$$
y = x - 1
$$y = x - 1$$
Быстрый ответ
[TeX]
$$x_{1} = \frac{1}{2} + \frac{\sqrt{31}}{2}$$
=
$$\frac{1}{2} + \frac{\sqrt{31}}{2}$$
=
$$y_{1} = - \frac{1}{2} + \frac{\sqrt{31}}{2}$$
=
$$- \frac{1}{2} + \frac{\sqrt{31}}{2}$$
=
=
$$\frac{1}{2} + \frac{\sqrt{31}}{2}$$
=
3.28388218141501
$$y_{1} = - \frac{1}{2} + \frac{\sqrt{31}}{2}$$
=
$$- \frac{1}{2} + \frac{\sqrt{31}}{2}$$
=
2.28388218141501
$$x_{2} = - \frac{\sqrt{31}}{2} + \frac{1}{2}$$
=
$$- \frac{\sqrt{31}}{2} + \frac{1}{2}$$
=
$$y_{2} = - \frac{\sqrt{31}}{2} - \frac{1}{2}$$
=
$$- \frac{\sqrt{31}}{2} - \frac{1}{2}$$
=
=
$$- \frac{\sqrt{31}}{2} + \frac{1}{2}$$
=
-2.28388218141501
$$y_{2} = - \frac{\sqrt{31}}{2} - \frac{1}{2}$$
=
$$- \frac{\sqrt{31}}{2} - \frac{1}{2}$$
=
-3.28388218141501
Численный ответ
[pretty]
[text]
x1 = -2.283882181415011 y1 = -3.283882181415011
x2 = 3.283882181415011 y2 = 2.283882181415011