Question

3(2х+3у)=2 х/5+у/3=1 -7(х-2у)=5
-6х+у=-2 х/4-у\2=1 3х-у=0
Система линейных уравнений с двумя пермеными

Answer 1

$\begin{cases}3(2x+3y)=2\\-6x+y=-2\end{cases}=\ \textgreater \ \begin{cases}6x+9y=2\\+\\-6x+y=-2\end{cases}\\10y=0\\y=0\\-6x+0=-2\\x=\frac{1}{3}\\OTBET:(\frac{1}{3};0)$

$\begin{cases}\frac{x}{5}+\frac{y}{3}=1|*15\\\frac{x}{4}-\frac{y}{2}=1|*12\end{cases}=\ \textgreater \ \begin{cases}3x+5y=15\\-\\3x-6y=12\end{cases}\\11y=3\\y=\frac{3}{11}\\3x-\frac{18}{11}=12\\3x=\frac{150}{11}\\x=\frac{50}{11}\\OTBET:(\frac{50}{11};\frac{3}{11})$

$\begin{cases}-7(x-2y)=5\\3x-y=0=\ \textgreater \ y=3x\end{cases}\\-7x+14*3x=5\\35x=5\\x=\frac{1}{7}\\y=\frac{3}{7}\\OTBET:(\frac{1}{7};\frac{3}{7})$