\(\left\{ {\begin{array}{*{20}{c}}{3x-6y = 1,}\\{5x-a\,y = 2}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{x = \dfrac{{6y + 1}}{3},\,\,\,\,\,\,\,\,\,\,\,\,}\\{\dfrac{{30y + 5}}{3}-ay = 2}\end{array}} \right.\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{x = \dfrac{{6y + 1}}{3},\,\,}\\{y = \dfrac{1}{{30-3a}}}\end{array}} \right.\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{x = \dfrac{1}{3}\left( {\dfrac{6}{{30-3a}} + 1} \right),}\\{y = \dfrac{1}{{30-3a}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{x = \dfrac{{12-a}}{{30-3a}},}\\{y = \dfrac{1}{{30-3a}}.\,}\end{array}} \right.\)
По условию \(x < 0,\,\,\,\,y < 0,\) поэтому:
\(\left\{ {\begin{array}{*{20}{c}}{\dfrac{{12-a}}{{30-3a}} < 0,}\\{\dfrac{1}{{30-3a}} < 0}\end{array}\,} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{a \in \left( {10;12} \right),}\\{a \in \left( {10;\infty } \right)\,\,\,}\end{array}} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,a \in \left( {10;12} \right).\)
Ответ: \(\left( {10;12} \right)\).