Question

Решите систему уравнений
у +2х=5
2х-ху= -1

Answer 1

Ответ:

(1; 3), (0.5; 4)

Объяснение:

$y + 2x = 5 \ 2x - xy = - 1 \ \ y + xy = 6 \ y(x + 1) = 6 \ \ y = 5 - 2x \ = > (5 - 2x)(x + 1) = 6 \ 3x - 2 {x}^{2} - 1 = 0 \ 2 {x}^{2} - 3x + 1 = 0 \ 2 {x}^{2} - 2x - x + 1 = 0 \ 2x(x - 1) - (x - 1) = 0 \ (x - 1)(2x - 1) = 0 \ x = frac{1}{2} \ x = 1 \ = > y = 3 \ y = 4$

Answer 2

$left { {{y+2x=5} atop {2x-xy=-1}} right. \\y+2x=5\y = 5-2x\\2x-xy=-1\2x-x*(5-2x)=-1\2x-5x+2x^{2} =-1\2x^{2} -3+1=0\\x1=frac{3+sqrt{9-4*2*1} }{2*2} =frac{3+1}{4}=1\\x2=frac{3-sqrt{9-4*2*1} }{2*2} =frac{3-1}{4}=0.5$

$y1+2x1=5\y1+2=5\y1=3\\y2+2x2=5\y2+1=5\y2=4$