#### Answer

$-\dfrac{5}{t-4}$

#### Work Step by Step

Adding the numerators and copying the denominator, the given expression, $
\dfrac{1-2t}{t^2-5t+4}+\dfrac{4-3t}{t^2-5t+4}
,$ simplifies to
\begin{array}{l}\require{cancel}
\dfrac{1-2t+4-3t}{t^2-5t+4}
\\\\=
\dfrac{5-5t}{t^2-5t+4}
\\\\=
\dfrac{5(1-t)}{(t-4)(t-1)}
\\\\=
\dfrac{-5(t-1)}{(t-4)(t-1)}
\\\\=
\dfrac{-5(\cancel{t-1})}{(t-4)(\cancel{t-1})}
\\\\=
\dfrac{-5}{t-4}
\\\\=
-\dfrac{5}{t-4}
.\end{array}