36А. При каких значениях параметра a, для каждого из которых числа x и y, удовлетворяющие системе уравнений \(\left\{ {\begin{array}{*{20}{c}} {3x + y = a,\,\,\,\,\,\,\,\,\,\,} \\ {x + 2y = 2a + 1} \end{array}} \right.\) удовлетворяют также неравенству \(x > 3y\)?
Ответ
ОТВЕТ: \(a \in \left( {-\infty ;-\frac{2}{3}} \right).\)
Решение
\(\left\{ {\begin{array}{*{20}{c}}{3x + y = a,\,\,\,\,\,\,\,\,\,\,}\\{x + 2y = 2a + 1}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{y = a-3x,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{x + 2a-6x = 2a + 1}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{y = a + \frac{3}{5},}\\{x = -\frac{1}{5}.\,\,\,\,\,}\end{array}} \right.\)
По условию \(x > 3y,\) поэтому:
\(-\frac{1}{5} > 3a + \frac{9}{5}\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,-3a > 2\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,a < -\frac{2}{3}\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,a \in \left( {-\infty ;-\frac{2}{3}} \right).\)
Ответ: \(\left( {-\infty ;-\frac{2}{3}} \right)\).