Question

Помогите! Реши систему способом подстановки

Answer 1

$\left \{ {{(x + 1)^2 - y = x (x + 3)} \atop {x + (y - 1)^2 = y^2 - 10 }} \right.$
$\left \{ {x^2 + 2x + 1 - y = x^2 + 3x} \atop {x + y^2 - 2y = 1 = y^2 - 10 }} \right.$
$\left \{ {x^2 + 2x + 1 - x^2 - 3x =y} \atop {x + y^2 - 2y = 1 = y^2 - 10 }} \right.$
$\left \{ {1 - x=y} \atop {x + y^2 - 2y = 1 = y^2 - 10 }} \right.$
x + (1 - x)² - 2(1 - x) = (1 - x)² - 10
x + 1 - 2x + x² - 2 + 2x = 1 - 2x + x² - 10
x + 1 - 2x + x² - 2 + 2x - 1 + 2x - x² + 10 = 0
3x + 8 = 0
3x = - 8
x = - 8/3
x = - 2 2/3
$\left \{ {{x = - 2 \frac{2}{3} } \atop {y=1 - (- 2 \frac{2}{3}) }} \right.$
$\left \{ {{x = - 2 \frac{2}{3} } \atop {y= 3 \frac{2}{3} }} \right.$