Ta có: \(10x=23y\Rightarrow\dfrac{x}{23}=\dfrac{y}{10}=\dfrac{2y}{20}\)
AD...., có: \(\dfrac{x}{23}=\dfrac{2y}{20}=\dfrac{x-2y}{23-20}=\dfrac{6}{3}=2\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{23}=2\\\dfrac{y}{10}=2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=46\\y=20\end{matrix}\right.\)
\(\left\{{}\begin{matrix}10x=23y\\x-2y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}10\left(2y+6\right)=23y\\x=2y+6\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}20y+60=23y\\x=2y+6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3y=60\\x=2y+6\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=20\\x=2\cdot20+6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=20\\x=46\end{matrix}\right.\)
Vậy \(x=46\) và \(y=20\).
