\(\left\{{}\begin{matrix}\dfrac{x}{2}+\dfrac{y}{3}=\dfrac{7}{6}\\3x-2y=-1\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}3x+2y=7\\3x-2y=-1\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}6x=6\\3x-2y=-1\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=1\\3\cdot1-2y=-1\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=1\\2y=4\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)
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Ta có: \(\left\{{}\begin{matrix}\dfrac{x}{2}+\dfrac{y}{3}=\dfrac{7}{6}\\3x-2y=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2}x+\dfrac{1}{3}y=\dfrac{7}{6}\\3x-2y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{2}x+y=\dfrac{7}{2}\\\dfrac{3}{2}x-y=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x=3\\3x-2y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\3-2y=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\2y=3+1=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)
Vậy: (x,y)=(1;2)
