\(\left\{ {\begin{array}{*{20}{c}}{x + y = a,\,}\\{2\,x-y = 3}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{x = a-y,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{2\,\left( {a-y} \right)-y = 3}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{x = a-y,\,}\\{y = \frac{{2a-3}}{3}}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{x = a-\frac{{2a-3}}{3},\,}\\{y = \frac{{2a-3}}{3}\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{x = \frac{{a + 3}}{3},\,\,}\\{y = \frac{{2a-3}}{3}.}\end{array}} \right.\)
По условию \(x > y,\) поэтому:
\(\frac{{a + 3}}{3} > \frac{{2a-3}}{3}\,\left| { \cdot 3} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,a + 3 > 2a-3\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,a < 6.\)
Ответ: \(\left( {-\infty ;6} \right)\).