Đáp án:
Đặt $\dfrac{x-1}{2005}=$ $\dfrac{3-y}{2006}=k$
$\rightarrow \left\{\begin{matrix}
x-1=2005k & \\
3-y=2006k &
\end{matrix}\right.$
$\rightarrow \left\{\begin{matrix}
x=2005k +1& \\
3-2006k =y&
\end{matrix}\right.$
Mà $2x+3y-2=-14$
$→2x+3y=-12$
$→2(2005k+1)+3(3-2006k)=-12$
$→-2008k+11=-12$
$→k=$ $\dfrac{23}{2008}$
$→x=23,96563745;y=-19,97709163$