\(\Leftrightarrow\left\{{}\begin{matrix}3\left|x\right|+12\left|y\right|=54\\3\left|x\right|+\left|y\right|=10\end{matrix}\right.\)
Trừ 2 vế của hpt, ta được: \(11\left|y\right|=44\Leftrightarrow y=\pm4\)
*Với y=4, thay vào pt dưới, ta được:
\(3\left|x\right|=6\Leftrightarrow x=\pm2\)
*Với y=-4, thay vào pt dưới, ta được:
\(3\left|x\right|=14\Leftrightarrow\left|x\right|=\frac{14}{3}\)\(\Leftrightarrow x=\pm\frac{14}{3}\)
Vậy (x;y)=\(\left(-2;4\right);\left(2;4\right);\left(\frac{14}{3};-4\right);\left(\frac{-14}{3};-4\right)\)
\(\left\{{}\begin{matrix}\left|x\right|+4\left|y\right|=18\\3\left|x\right|+\left|y\right|=10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3\left|x\right|+12\left|y\right|=54\\3\left|x\right|+\left|y\right|=10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}11\left|y\right|=44\\3\left|x\right|+\left|y\right|=10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=4\\3\left|x\right|+4=10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=4\\3\left|x\right|=10-4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=4\\3\left|x\right|=6\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=4\\x=2\end{matrix}\right.\)
