Question

х+у-ху=1 ху+(х+у)=9
Способом замены прошу помогите

Answer 1

Ответ: (x, y) = (1, 4)

Обьяснение: {х+у-ху=1, ху+(х+у)=9

{х = 1, ху+(х+у)=9

1y+(1+y) = 9

y = 4

(x, y) = (1, 4)

{1+4-1*4 = 1, 1*4+(1+4) = 9

{1 = 1, 9 = 9

(x, y) = (1, 4)

Answer 2

Объяснение:

$\displaystyle\\\left \{ {{x+y-xy=1} \atop {xy+(x+y)=9}} \right.$

Пусть х+у=t, а ху=v.         ⇒

$\displaystyle\\\left \{ {{t-v=1} \atop {v+t=9}} \right.$

Суммируем эти уравнения:

$\displaystyle\\2t=10\ |:2\\\\t=5\ \ \ \ \Rightarrow\\\\v+5=9\\\\v=4\ \ \ \ \Rightarrow\\\\\left \{ {{x+y=5} \atop {xy=4}} \right. \ \ \ \ \left \{ {{y=5-x} \atop {x*(5-x)=4}} \right.\ \ \ \ \left \{ {{y=5-x} \atop {5x-x^2=4}} \right.\ \ \ \ \left \{ {{y=5-x} \atop {x^2-5x+4=0} \right. \\\\\\\left \{ {{y=5-x} \atop {x^2-4x-x+4=0}} \right.\ \ \ \ \left \{ {{y=5-x} \atop {x*(x-4)-(x-4)=0}} \right. \ \ \ \ \left \{ {{y=5-x} \atop {(x-4)(x-1)=0}} \right.$

$\displaystyle\\\left \{ {{y_1=1\ \ \ \ y_2=4} \atop {x_1=4\ \ \ \ x_2=1}} \right. .$

Ответ: (4;1),  (1;4).