\(\left\{{}\begin{matrix}x-3y=1\\3x+2y=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1+3y\\3x+2y=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1+3y\\3\left(1+3y\right)+2y=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1+3y\\3+9y+2y=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1+3y\\9y+2y=4-3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1+3y\\11y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1+3y\\y=\dfrac{1}{11}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1+3.\dfrac{1}{11}\\\dfrac{1}{11}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{14}{11}\\y=\dfrac{1}{11}\end{matrix}\right.\)
Vậy hệ phương trình có nghiệm `(x;y)` là `((14)/(11); 1/(11))`
\(\left\{{}\begin{matrix}x-3y=1\\3x+2y=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1+3y\\3\left(1+3y\right)+2y=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1+3y\\3+9y+2y=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1+3y\\11y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{1}{11}\\x=1+3.\dfrac{1}{11}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{1}{11}\\x=\dfrac{14}{11}\end{matrix}\right.\)
Vậy hpt có n0 \(\left(x;y\right)=\left(\dfrac{14}{11};\dfrac{1}{11}\right)\)
