\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2x}{3}-\dfrac{y}{2}=4\\\dfrac{2x}{3}+\dfrac{4y}{15}=\dfrac{2}{3}\end{matrix}\right.\)
Trừ phương trình dưới cho pt trên\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{23y}{30}=-\dfrac{10}{3}\\\dfrac{2x}{3}+\dfrac{4y}{15}=\dfrac{2}{3}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}y=-\dfrac{100}{23}\\x=\dfrac{63}{23}\end{matrix}\right.\)