Question

нужно очень очень срочно, даю 50 баллов

Answer 1

Ответ:

Два решения: ($6$, $2$) и ($-6$, $-2$)

Объяснение:

$\left \{ {{2xy - \frac{3x}{y} = 15} \atop {xy+\frac{x}{y} = 15|*3}} \right. \\\left \{ {{2xy - \frac{3x}{y} = 15} \atop {3xy+\frac{3x}{y} = 45}} \right.\\\+\\\left \{ {{5xy = 60|:5} \atop {3xy+\frac{3x}{y} = 45}} \right.$

$\left \{ {{xy = 12} \atop {3xy+\frac{3x}{y}=45}} \right. \\\left \{ {{xy = 12} \atop {\frac{3x}{y} = 9|*\frac{y}{3} }} \right. \\\left \{ {{xy=12} \atop {x=3y}} \right. \\\\\left \{ {{3y^2=12|:3} \atop {x=3y}} \right. \\\\\left \{ {{y^2 = 4} \atop {x=3y}} \right. \\\\\left \{ {{y = +-2} \atop {x=+-6}} \right.$

Два решения: ($6$, $2$) и ($-6$, $-2$)

Answer 2

$\displaystyle\bf\\\left \{ {{2xy-\dfrac{3x}{y} }=15 \atop {xy+\dfrac{x}{y} =15}} \right. \\\\\\2xy-\frac{3x}{y} =xy+\frac{x}{y} \\\\\\2xy -xy=\frac{3x}{y} +\frac{x}{y} \\\\\\xy=\frac{4x}{y} \\\\\\\frac{4x}{y} +\frac{x}{y} =15\\\\\\\frac{5x}{y} =15\\\\\\\frac{x}{y} =3 \ \ \Rightarrow \ \ x=3y\\\\\\3y\cdot y+3=15\\\\\\3y^{2} =12\\\\\\y^{2} =4 \ \ \ \Rightarrow \ \ y_{1} =-2 \ \ \ ; \ \ \ y_{2} =2\\\\\\x_{1} =3\cdot(-2)=-6 \ \ \ ; \ \ \ x_{2} =3\cdot 2=6$

$\displaystyle\bf\\Otvet \ : \ (-6 \ ; \ -2) \ , \ (6 \ ; \ 2)$

Второй способ :

$\displaystyle\bf\\\left \{ {{2xy-\dfrac{3x}{y} }=15 \atop {xy+\dfrac{x}{y} =15} \ |\cdot(-2)} \right.\\\\\\+\left \{ {{2xy-\dfrac{3x}{y} }=15 \atop {-2xy-\dfrac{2x}{y} =-30}} \right.\\ ------------\\\\-\frac{5x}{y} =-15\\\\\\\frac{x}{y} =3\\\\\\x=3y\\\\\\3y^{2} +3=15\\\\\\3y^{2} =12\\\\\\y^{2} =4\\\\\\y_{1} =-2 \ \ ; \ \ y_{2} =2\\\\x_{1} =-6 \ \ ; \ \ x_{2} =6$