2x+y=-4
\(\Leftrightarrow2x=-4-y\)
\(\Leftrightarrow x=\dfrac{-y-4}{2}\)
Ta có: 3x+|y|=-1
\(\Leftrightarrow\dfrac{-3y-12}{2}+\left|y\right|=-1\)
\(\Leftrightarrow\left|y\right|=\dfrac{-2+3y+12}{2}=\dfrac{3y+10}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}y=\dfrac{3}{2}y+5\left(y\ge0\right)\\y=-\dfrac{3}{2}y+5\left(y< 0\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{-1}{2}y=5\\\dfrac{5}{2}y=5\end{matrix}\right.\Leftrightarrow y\in\varnothing\)