\(\Leftrightarrow\left\{{}\begin{matrix}4\left|x+1\right|-10y=6\\5\left|x+1\right|+10y=-3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}9\left|x+1\right|=3\\5\left|x+1\right|+10y=-3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left|x+1\right|=\dfrac{1}{3}\\5.\dfrac{1}{3}+10y=-3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x+1=\dfrac{1}{3}\\x+1=-\dfrac{1}{3}\end{matrix}\right.\\y=-\dfrac{7}{15}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=-\dfrac{2}{3}\\x=-\dfrac{4}{3}\end{matrix}\right.\\y=-\dfrac{7}{15}\end{matrix}\right.\)
Vậy hệ có 2 cặp nghiệm: \(\left(x;y\right)=\left(-\dfrac{2}{3};-\dfrac{7}{15}\right);\left(-\dfrac{4}{3};-\dfrac{7}{15}\right)\)
