Question

$\frac{50}{x} - \frac{50}{y} = \frac{5}{12} \\ \frac{1}{2} x + \frac{1}{2} y = 50$
Помогите прошу ​

Answer 1

Ответ:

$(40; 60) \quad ; \quad (300; -200) \quad ;$

Пошаговое объяснение:

ОДЗ:

$x \neq 0, \quad y \neq 0;$

Решение:

$\displaystyle \left \{ {{\dfrac{50}{x}-\dfrac{50}{y}=\dfrac{5}{12} \quad \bigg | \quad \cdot 12xy} \atop {\dfrac{1}{2}x+\dfrac{1}{2}y=50} \quad \bigg | \quad \cdot 2} \right. \Leftrightarrow \left \{ {{600y-600x=5xy} \atop {x+y=100}} \right. \Leftrightarrow \left \{ {{120y-120x=xy} \atop {y=100-x}} \right. \Leftrightarrow$

$\displaystyle \Leftrightarrow \left \{ {{120 \cdot (100-x)-120x=x \cdot (100-x)} \atop {y=100-x}} \right. \Leftrightarrow \left \{ {{12000-120x-120x=100x-x^{2}} \atop {y=100-x}} \right. \Leftrightarrow$$\displaystyle \Leftrightarrow \left \{ {{x^{2}-340x+12000=0} \atop {y=100-x}} \right. \Leftrightarrow \left \{ {{x^{2}-300x-40x+12000=0} \atop {y=100-x}} \right. \Leftrightarrow$

$\displaystyle \Leftrightarrow \left \{ {{x \cdot (x-300)-40 \cdot (x-300)=0} \atop {y=100-x}} \right. \Leftrightarrow \left \{ {{(x-40) \cdot (x-300)=0} \atop {y=100-x}} \right. \Leftrightarrow$

$\displaystyle \Leftrightarrow \left \{ {{x-40=0} \atop {y=100-x}} \right. \vee \left \{ {{x-300=0} \atop {y=100-x}} \right. \Leftrightarrow \left \{ {{x=40} \atop {y=60}} \right. \vee \left \{ {{x=300} \atop {y=-200}} \right. ;$

$(40; 60) \quad ; \quad (300; -200) \quad ;$