Question

Помогите с алгеброй с решениями (пожалуйста не предлагайте делать в photomath)

Answer 1

Ответ:

1) (-5;12) и (4;3)

2) (-6;-22) и (3;5)

5) (2;-1) и (5;2)

$5)x - {y}^{2} = 1 \\ x - y = 3 > > x = 3 + y \\ \\ 3 + y - {y}^{2} = 1 \\ {y}^{2} - y - 2= 0 \\ d = {1}^{2} - 4 \times ( - 2) = 9 \\ y1 = \frac{1 - \sqrt{9} }{2} = \frac{ - 2}{2} = - 1 \\ y2 = \frac{1 + \sqrt{9} }{2} = \frac{4}{2} = 2 \\ x1 = 3 + ( -1) = 2 \\ x2 = 3 + 2 = 5$

Объяснение:

$1)x + y = 7 \\ {x}^{2} - y = 13 > > y = {x}^{2} - 13 \\ \\ x + {x}^{2} - 13 = 7 \\ x + {x}^{2} - 20 \\ d = 1 {}^{2} - 4 \times ( - 20) = 81 \\ x1 = \frac{ - 1 - \sqrt{81} }{2} = \frac{ - 10}{2} = - 5 \\ x2 = \frac{ - 1 + \sqrt{81} }{2} = \frac{8}{2} = 4 \\ \\ y1 > ( - 5) {}^{2} - 13 = 25 - 13 = 12 \\ y2 = {4}^{2} - 13 = 16 - 13 = 3$

$2)3x - y = 4 > > y = 3x - 4 \\ {x}^{2} + y = 14 \\ \\ {x }^{2} + 3x - 4 = 14 \\ {x}^{2} + 3x - 18 = 0 \\ d = {3}^{2} - 4 \times ( - 18) = 9 + 72 = 81 \\ x1 = \frac{ - 3 - \sqrt{81} }{2} = \frac{ - 12}{2} = - 6 \\ x2 = \frac{ - 3 + \sqrt[]{81} }{2} = \frac{6}{2} = 3 \\ \\ y 1 = 3 \times ( -6) - 4 = - 18 - 4 = - 22 \\ y2 = 3 \times 3 - 4 = 9 - 4 = 5$