Question

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

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

Answer 1

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

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

<=> \(\left\{{}\begin{matrix}14x-6y=10\\9x+6y=18\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}23x=28\\y=\dfrac{6-3x}{2}\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}x=\dfrac{28}{23}\\y=\dfrac{27}{23}\end{matrix}\right.\)

Vậy S = {28/23; 27/23}

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

<=> \(\left\{{}\begin{matrix}12x\sqrt{2}-4y=12\\\sqrt{2}x-4y=-10\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}11\sqrt{2}x=22\\y=3x\sqrt{2}-3\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}x=\sqrt{2}\\y=3\end{matrix}\right.\)

Vậy ...

Answer 2

1) Ta có: \(\left\{{}\begin{matrix}7x-3y=5\\\dfrac{x}{2}+\dfrac{y}{3}=2\end{matrix}\right.\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}y=3\\7x-9=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=3\\7x=14\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\)

Vậy: (x,y)=(2;3)