Question

Решите систему уравнений
x+xy+3y=3
2x-xy-y=2

Answer 1

$\left \{ {{x+xy+3y=3} \atop {2x-xy-y=2}} \right. \; \left \{ {{3x+2y=5} \atop {x+xy+3y=3}} \right. \; \left \{ {{x=\frac{5-2y}{3}=\frac{5}{3}-\frac{2}{3}y} \atop {(\frac{5}{3}-\frac{2}{3}y)+(\frac{5}{3}y-\frac{2}{3}y^2)+3y=3}} \right. \\\\-\frac{2}{3}y^2+4y-\frac{4}{3}=0\, |\cdot (-3)\\\\2y^2-12y+4=0\, |:2\\\\y^2-6y+2=0\\\\(y-3)^2-7=0\\\\y-3=\pm \sqrt7,\\\\y_1=3-\sqrt7,\; y_2=3+\sqrt7\\\\x_1=\frac{5}{3}-\frac{2}{3}(3-\sqrt7)=\frac{2\sqrt7-1}{3}\\\\x_2=\frac{5}{3}-\frac{2}{3}(3+\sqrt7)=$

$x_2=\frac{-2\sqrt7-1}{3}\\\\Otvet;\; \; (\frac{2\sqrt7-1}{3};\, 3-\sqrt7),\; (-\frac{2\sqrt7+1}{3};\, 3+\sqrt7)$

Answer 2

Складываем уравнения:
x+xy+3y+2x-xy-y=5
3x+2y=5
y=2,5-1,5x
x+x*(2,5-1,5x)+(2,5-1,5x)*3=3
x+2,5x-1,5x²+7,5-4,5x-3=0
-1,5x²-x+4,5=0   I*(-2)
3x²+2x-9=0  D=112
x₁=(-1+2√7)/3  x₂=(-1-2√7)/3
y₁=3-√7     y₂=3+√7.