25А. Решите систему уравнений при всех значениях параметра a: \(\left\{ {\begin{array}{*{20}{c}} {5x + a\,y = 2a + 1,} \\ {\left( {a + 1} \right)\,\,x + 4y = 9.}\end{array}} \right.\)
ОТВЕТ: нет решений при \(a = -5\); бесконечное множество решений при \(a = 4\); \(x = \frac{1}{{a + 5}},\,\,\,y = \frac{{2a + 11}}{{a + 5}}\) при \(a \ne -5,\,\,\,a \ne 4.\)
Определим при каких значениях параметра a прямые соответствующие уравнениям системы будут параллельны или совпадать. Для этого коэффициенты перед x и y должны быть пропорциональны. \(\frac{5}{{a + 1}} = \frac{a}{4}\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,{a^2} + a-20 = 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{a = 4,\,\,\,}\\{a = -5.}\end{array}} \right.\) Если \(a = 4\), то система примет вид: \(\left\{ {\begin{array}{*{20}{c}}{5x + 4y = 9,}\\{5x + 4y = 9.}\end{array}} \right.\) Уравнения совпадают, то есть система имеет бесконечное множество решений. Если \(a = -5\), то система примет вид: \(\left\{ {\begin{array}{*{20}{c}}{5x-5y = -9,}\\{-4x + 4y = 9.}\end{array}} \right.\) Так как \(\frac{5}{{-4}} = \frac{{-5}}{4} \ne \frac{{-9}}{9},\) то прямые параллельны, поэтому система не имеет решений. Если \(a \ne 4\) и \(a \ne -5\), то: \(\left\{ {\begin{array}{*{20}{c}}{x = \frac{{2a + 1-ay}}{5},\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{\frac{{\left( {a + 1} \right)\left( {2a + 1-ay} \right)}}{5} + 4y = 9}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{x = \frac{{2a + 1-ay}}{5},\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{\left( {-{a^2}-a + 20} \right)y = -2{a^2}-3a + 44}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \) \( \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{x = \frac{{2a + 1-ay}}{5},\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{y = \frac{{2{a^2} + 3a-44}}{{{a^2} + a-20}} = \frac{{\left( {a-4} \right)\left( {2a + 11} \right)}}{{\left( {a + 5} \right)\left( {a-4} \right)}} = \frac{{2a + 11}}{{a + 5}}}\end{array}} \right.\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \) \( \Leftrightarrow \,\,\,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{x = \frac{1}{5}\left( {2a + 1-\frac{{2{a^2} + 11a}}{{a + 5}}} \right) = \frac{{2{a^2} + a + 10a + 5-2{a^2}-11a}}{{5\left( {a + 5} \right)}} = \frac{1}{{a + 5}},}\\{y = \frac{{2a + 11}}{{a + 5}}.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\) ОТВЕТ: нет решений при \(a = -5\); бесконечное множество решений при \(a = 4\); \(x = \frac{1}{{a + 5}},\,\,\,y = \frac{{2a + 11}}{{a + 5}}\) при \(a \ne -5,\,\,\,a \ne 4.\)