\(\left(d\right)//\left(d'\right)\Rightarrow\left\{{}\begin{matrix}a=3\\b\ne-8\end{matrix}\right.\\ \Rightarrow\left(d'\right)y=3x+b\)
\(\left(d'\right)\) đi qua \(A\left(2;5\right)\)
\(\Rightarrow5=3.2+b\\ \Rightarrow b=5-6=-1\left(t/m\right)\)
Vậy \(a=3\\ b=-1\)