Question

Решите систему методом замены переменной.

Answer 1

ответ (1;3), (3;1)

Решение

Answer 2

x + y = m      xy = n

Тогда первое уравнение имеет вид :

m² - 5m + 4 = 0

По теореме Виета :

m₁ = 1       m₂ = 4

Из второго уравнения получим :

x + xy + y = 7

(x + y) + xy = 7

m + n = 7

Если m₁ = 1 , то    n₁ = 7 - 1 = 6

Если m₂ = 4 , то   n₂ = 7 - 4 = 3

Следовательно :

$1)\left \{ {{x+y=1} \atop {xy=6}} \right.\\\\\left \{ {{x=1-y} \atop {(1-y)*y=6}} \right.\\\\\left \{ {{x=1-y} \atop {-y^{2}+y-6=0 }} \right. \\\\\left \{ {{x=1-y} \atop {y^{2}-y+6=0 }} \right. \\\\y^{2}-y+6=0\\\\D=(-1)^{2}-4*6=1-24=-23$

Корней нет

$2)\left \{ {{x+y=4} \atop {xy=3}} \right.\\\\\left \{ {{x=4-y} \atop {(4-y)*y=3}} \right.\\\\\left \{ {{x=4-y} \atop {-y^{2}+4y-3=0 }} \right.\\\\\left \{ {{x=4-y} \atop {y^{2}-4y+3=0 }} \right.\\\\\left \{ {{x=4-y} \atop {\left[\begin{array}{ccc}y_{1}=1 \\y_{2}=3 \end{array}\right }} \right.\\\\\\\left[\begin{array}{ccc}\left \{ {{y_{1}=1 } \atop {x_{1}=4-1=3 }} \right. \\\left \{ {{y_{2}=3 } \atop {x_{2}=4-3=1 }} \right. \end{array}\right\\\\Otvet:\boxed{(3;1),(1;3)}$