[tex]\bf \cfrac{3x-2}{x^2-2x-3}-\cfrac{1}{x-3}\qquad \qquad
\begin{array}{lcclll}
x^2&-2x&-3\\
&\uparrow &\uparrow \\
&-3+1&-3\cdot 1
\end{array}\qquad thus
\\\\\\
\cfrac{3x-2}{(x-3)(x+1)}-\cfrac{1}{x-3}\impliedby
\begin{array}{llll}
\textit{LCD will then be}\\
(x-3)(x+1)\\
\textit{since it contains x-3}\\
already
\end{array}
\\\\\\
\cfrac{(3x-2)-(x+1)1}{(x-3)(x+1)}\implies \cfrac{3x-2-x-1}{(x-3)(x+1)}
\\\\\\
\cfrac{2x-3}{(x-3)(x+1)}[/tex]
