Định giá trị của $m$ sao cho tổng  $ (x + y) $  các nghiệm $x$ và y của hệ
$\left\{ \begin{array}{l}
(5m + 1)x + (3m + 2)y = 15\\
(13m - 14)x - (2m - 5)y = 9
\end{array} \right. $  bằng $3$.   
Từ điều kiện x+y=3, ta có y=3-x
Thay  y=3-x vào hai phương trình đã cho, ta có: $ x =  - \frac{{9(m - 1)}}{{2m - 1}};x = \frac{{6(m - 1)}}{{15m - 19}} $
Với  $ m \ne \frac{1}{2},m \ne \frac{{19}}{{15}} $      (*)
Điều kiện đồng nghiệm cho ta: $  - \frac{{9(m - 1)}}{{2m - 1}} = \frac{{6(m - 1)}}{{15m - 9}} $  $  \Leftrightarrow m = 1\ ;\,m = \frac{{59}}{{49}}, $ thỏa (*)
Với   $ m = 1 \Rightarrow x = 0,y = 3. $
Với  $ m = \frac{{59}}{{49}} \Rightarrow x =  - \frac{{30}}{{23}}; y = \frac{{99}}{{23}} $
Vậy m=-1 và  $ m = \frac{{59}}{{49}} $

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