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

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

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