Question

пожалуйста помогите!! мне уже надо сдавать
*на фото
Який розв'язок системи нерівностей?

​

Answer 1

$\begin{cases} {x}^{2} + {y}^{2} = 16 \\ y = {x}^{2} - 4 \end{cases}$

$\begin{cases} {x}^{2} + {y}^{2} = 16 \\ y - {x}^{2} = - 4 \end{cases}$

${x}^{2} + {y}^{2} + y - {x}^{2} = 16 - 4$

${y}^{2} + y = 12$

${y}^{2} + y - 12 = 0$

${y}^{2} + 4y - 3y - 12 = 0$

$y(y + 4) - 3(y + 4) = 0$

$(y + 4)(y - 3) = 0$

$y + 4 = 0 \\ y - 3 = 0$

$y = - 4 \\ y = 3$

$- 4 - {x}^{2} = - 4 \\ 3 - {x}^{2} = - 4$

$- 4 - {x}^{2} = - 4 \\ - {x}^{2} = 0 \\ {x}^{2} = 0 \\ x = 0$

$3 - {x}^{2} = - 4 \\ - {x}^{2} = - 4 \\ x = - \sqrt{7} - 3 \\ - {x}^{2} = - 7 \\ {x}^{2} = 7 \\ x = ± \sqrt{7} \\ x = - \sqrt{7} \\ x = \sqrt{7}$

$x = 0 \\ x = - \sqrt{7} \\ x = \sqrt{7}$

$({x}_{1}, {y}_{1})=(0, - 4) \\ ({x}_{2}, {y}_{2})=(-\sqrt{7}, 3) \\ ({x}_{3}, {y}_{3})=(\sqrt{7}, 3)$