Question

giải các hệ phương trình sau {3x+2y=15 {x+4y =5

Answer 1

\(\left\{{}\begin{matrix}3x+2y=15\\x+4y=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}3x+2y=15\\3x+12y=15\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}3x+2y=15\\10y=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}3x+2.0=15\\y=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}3x=15\\y=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=5\\y=0\end{matrix}\right.\)