h/số y=(a-1)x+2⇒ a=a-1,b=2
h/số y=(3-a)x+1⇒a'=3-a,b'=1
a, d//d' ⇒\(\left\{{}\begin{matrix}a=a'\\b\ne b'\end{matrix}\right.=\left\{{}\begin{matrix}a-1=3-a\\2\ne1\end{matrix}\right.=\left\{{}\begin{matrix}a=4\\2\ne1\end{matrix}\right.\)
b, d cắt d' ⇒ a≠a' =a-1≠3-a=a≠2