a: Thay a=-2 vào hệ phương trình, ta được:
\(\left\{{}\begin{matrix}2x-y=1\\-2x+2y=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2x-y-2x+2y=1+2\\2x-y=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=3\\2x=y+1=3+1=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\)
b: Để hệ có nghiệm duy nhất thì \(\dfrac{2}{a}\ne\dfrac{-1}{2}\)
=>\(a\ne-4\)
\(\left\{{}\begin{matrix}2x-y=1\\ax+2y=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}4x-2y=2\\ax+2y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\left(a+4\right)=4\\2x-y=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{4}{a+4}\\y=2x-1=\dfrac{8}{a+4}-1=\dfrac{8-a-4}{a+4}=\dfrac{4-a}{a+4}\end{matrix}\right.\)
2x-3y=1
=>\(\dfrac{8}{a+4}-\dfrac{3\left(4-a\right)}{a+4}=1\)
=>\(\dfrac{8-12+3a}{a+4}=1\)
=>3a-4=a+4
=>2a=8
=>a=4(nhận)
