Question

Giải hệ phương trình

\(\left\{{}\begin{matrix}7x-3y=5\\\frac{x}{2}+\frac{y}{3}=2\end{matrix}\right.\)

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Answer 1

Ta có :

\(\left\{{}\begin{matrix}7x-3y=5\\\frac{x}{2}+\frac{y}{3}=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}7x-3y=5\\\frac{3x+2y}{6}=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}7x-3y=5\\3x+2y=12\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}14x-6y=10\\9x+6y=36\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}23x=46\\7x-3y=5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\)

Vậy...

Answer 2

\(\left\{{}\begin{matrix}7x-3y=5\\\frac{x}{2}+\frac{y}{3}=2\end{matrix}\right.\) \(\left\{{}\begin{matrix}7x-3y=5\\3x+2y=12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}14x-6y=10\\9x+6y=36\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}23x=46\\7x-3y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\)

Vậy hpt có 1 nghiệm duy nhất (x;y) =(2;3)