a/ Bạn tự giải
b/ \(a=0\Rightarrow\left\{{}\begin{matrix}x=\frac{7}{2}\\y=-5\end{matrix}\right.\) (t/m)
\(a\ne0\Rightarrow\left\{{}\begin{matrix}3a^2x-3ay=15a\\2x+3ay=7\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\frac{15a+7}{3a^2+2}\\y=\frac{7a-10}{3a^2+2}\end{matrix}\right.\)
Để \(\left\{{}\begin{matrix}x>0\\y< 0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\frac{15a+7}{3a^2+2}>0\\\frac{7a-10}{3a^2+2}< 0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}15a+7>0\\7a-10< 0\end{matrix}\right.\)
\(\Rightarrow\frac{-7}{15}< a< \frac{10}{7}\)