Question

\(\left\{{}\begin{matrix}\left|x\right|+4\left|y\right|=18\\3\left|x\right|+\left|y\right|=10\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\left|x\right|+4\left|y\right|=18\\3\left|x\right|+\left|y\right|=10\end{matrix}\right.\)

Answer 1

a) \(\left\{{}\begin{matrix}\left|x\right|+4\left|y\right|=18\\3\left|x\right|+\left|y\right|=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3\left|x\right|+12\left|y\right|=54\\3\left|x\right|+\left|y\right|=10\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}11\left|y\right|=44\\3\left|x\right|+\left|y\right|=10\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}y=\pm4\\x=\pm2\end{matrix}\right.\)

Vậy ...

Answer 2

\(\left\{{}\begin{matrix}\left|x\right|+4\left|y\right|=18\\3\left|x\right|+\left|y\right|=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-3\left|x\right|-12\left|y\right|=-54\left(1\right)\\3\left|x\right|+\left|y\right|=10\left(2\right)\end{matrix}\right.\)

Cộng (1) và (2), ta được phương trình: \(-11\left|y\right|=-44\Leftrightarrow\left[{}\begin{matrix}y=4\\y=-4\end{matrix}\right.\)

Thay $y=4$ vào $(2)$, ta được:

\(3\left|x\right|+\left|4\right|=10\\ \Leftrightarrow3\left|x\right|=6\\ \Leftrightarrow\left|x\right|=2\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

Thay $y=-4$ vào $(2)$, ta được:

\(3\left|x\right|+\left|-4\right|=10\\ \Leftrightarrow3\left|x\right|+4=10\\ \Leftrightarrow3\left|x\right|=6\\ \Leftrightarrow\left|x\right|=2\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

Vậy...