a) \(\left\{{}\begin{matrix}3x+2y=15\\x+4y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x+2y=15\\3x+12y=15\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x=15-2y\\10y=0\Leftrightarrow y=0\end{matrix}\right.\)
Vì \(y=0\Leftrightarrow2y=0\) nên ta có: \(3x=15-2y=15\\\Leftrightarrow x=\dfrac{15}{3}=5 \)
➤\(\left\{{}\begin{matrix}x=5\\y=0\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}\dfrac{2}{x-2}+\dfrac{1}{y+1}=3\\\dfrac{4}{x-2}-\dfrac{3}{y+1}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{4}{x-2}+\dfrac{2}{y+1}=6\\\dfrac{4}{x-2}-\dfrac{3}{y+1}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{y+1}=5\Leftrightarrow y+1=\dfrac{5}{5}=1\Leftrightarrow y=0\\\dfrac{4}{x-2}=1+\dfrac{3}{y+1}\end{matrix}\right.\)
+)Vì \(y+1=1\Leftrightarrow\dfrac{3}{y+1}=3\) nên ta có: \(\dfrac{4}{x-2}=1+\dfrac{3}{y+1}=1+3=4\\\Leftrightarrow x-2=\dfrac{4}{4}=1\\\Leftrightarrow x=1+2=3\)
➤\(\left\{{}\begin{matrix}y=0\\x=3\end{matrix}\right.\)
