(không cần "z" nha bn)
ta có : \(\dfrac{x-3}{8}=\dfrac{y-1}{3}\Leftrightarrow3\left(x-3\right)=8\left(y-1\right)\Leftrightarrow3x-9=8y-8\)
\(\Leftrightarrow3x-8y=-8+9\Leftrightarrow3x-8y=1\)
\(\Rightarrow\) ta có hệ phương trình \(\left\{{}\begin{matrix}4x-5y=10\\3x-8y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}12x-15y=30\\12x-32y=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}17y=26\\3x-8y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{26}{17}\\3x-8.\dfrac{26}{17}=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{26}{17}\\3x=1+8.\dfrac{26}{17}=\dfrac{225}{17}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{26}{17}\\x=\dfrac{225}{17.3}=\dfrac{75}{17}\end{matrix}\right.\) vậy \(y=\dfrac{26}{17};x=\dfrac{75}{17}\)
