Question

Решите систему уравнений

Answer 1

Ответ:

$\bigg ( \dfrac{3}{16} ~~ ; ~~ \dfrac{1}{16 } \bigg )$

Пошаговое объяснение:

$\displaystyle \left \{ \begin{array}{l} \dfrac{2}{x+y} - \dfrac{3x}{y} = -1 \\\\ \dfrac{1}{x+y} + \dfrac{8x}{y} = 28 \end{array}$

ОДЗ :

y ≠ 0  ;  x ≠ - y  

Сделаем замену

$\dfrac{1}{x+y} = a ~~ ; ~~ \dfrac{x}{y} = b$

$\displaystyle \left \{ \begin{array}{l} 2a - 3b = -1 \\\\ a + 8b = 28 ~ ~\big |\cdot 2 \end{array} \Leftrightarrow \ominus \left \{ \begin{array}{l} 2a - 3b = -1 \\\\ 2a + 16b = 56 \end{array} \Leftrightarrow$

$2a - 2a -3b - 16b = -1-56 \\\\ -19b = -57 \\\\ b = 3 \\\\ a=\dfrac{3b -1}{2} = 4$

Тогда

$\displaystyle \left \{ \begin{array}{l} \dfrac{1}{x+y}= 4 \\\\ \dfrac{x}{y} = 3 \end{array} \right. \Leftrightarrow \displaystyle \left \{ \begin{array}{l} 4x +4y = 1 \\\\ x = 3y \end{array} \right. \Leftrightarrow \\\\\\\\ 12 y + 4y = 1 \\\\ y = \frac{1}{16} \\\\\\ x = \frac{1}{16}\cdot 3 = \frac{3}{16}$