Question

алгебра пж помогите оч срочно ​

Answer 1

Ответ:

$1)\ \left\{\begin{array}{l}\dfrac{2x}{3}=2+\dfrac{y}{2}\ |\cdot 6\\\ \dfrac{2x}{3}+y=8\ |\cdot 3\end{array}\right\ \ \left\{\begin{array}{l}4x=12+3y\\2x+3y=24\end{array}\right\ \ \left\{\begin{array}{l}4x-3y=12\\2x+3y=24\ |\cdot (-2)\end{array}\right\ \oplus \\\\\\\left\{\begin{array}{l}4x=3y+12\\-9y=-36\end{array}\right\ \ \left\{\begin{array}{l}4x=12+12\\y=4\end{array}\right\ \ \left\{\begin{array}{l}4x=24\\y=4\end{array}\right\ \ \left\{\begin{array}{l}x=6\\y=4\end{array}\right\ \ \ (6;4)$

$2)\ \ \left\{\begin{array}{l}\dfrac{2}{x}-\dfrac{1}{y}=5\\\ \dfrac{2}{x}+\dfrac{1}{y}=7\end{array}\right\ \oplus \ominus \ \left\{\begin{array}{l}2\cdot \dfrac{2}{x}=12\\-2\cdot \dfrac{1}{y}=-2\end{array}\right\ \ \left\{\begin{array}{l}\dfrac{4}{x}=12\\-\dfrac{2}{y}=-2\end{array}\right\\\\\\\left\{\begin{array}{l}x=\dfrac{4}{12}\\y=1\end{array}\right\ \ \left\{\begin{array}{l}x=\dfrac{1}{3}\\y=1\end{array}\right\ \ \ \ \ \ \ \ \ \Big(\ \dfrac{1}{3}\ ;\ 1\ \Big)$