\(\left\{{}\begin{matrix}3x+2y=-2\\3x-2y=-3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}3x+2y=-2\\3x+2y-3x+2y=-2+3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}3x+2y=-2\\4y=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}3x+2.\dfrac{1}{4}=-2\\y=\dfrac{1}{4}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{5}{6}\\x=\dfrac{1}{4}\end{matrix}\right.\)

\(\left\{{}\begin{matrix}-5x+2y=4\\6x-3y=-7\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}15x-6y=-12\\12x-6y=-14\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}15x-6y=-12\\15x-6y-12x+6y=-12-\left(-14\right)\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}15x-6y=-12\\3x=2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}15.\dfrac{2}{3}-6y=-12\\x=\dfrac{2}{3}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=\dfrac{11}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}15x-6y=-12\\12x-6y=-14\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x=2\\-5x+2y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\2y-\dfrac{10}{3}=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=\dfrac{11}{3}\end{matrix}\right.\)

4: \(\Leftrightarrow\left\{{}\begin{matrix}-3y=9\\x+y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-3\\x=-1-y=-1-\left(-3\right)=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-2y=-1\\x-2y=-1\end{matrix}\right.\Leftrightarrow\left(x,y\right)\in R\)

b: \(\left\{{}\begin{matrix}3x-2y=4\\2x+y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x-2y=4\\4x+2y=10\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-x=-6\\2x+y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=6\\y=5-2x=5-12=-7\end{matrix}\right.\)

 (Nhân hai vế pt 2 với 3 để hệ số của y bằng nhau)

 (Trừ từng vế hai phương trình)

Phương trình 0x = 0 nghiệm đúng với mọi x.

Vậy hệ phương trình có vô số nghiệm dạng  (x ∈ R).