a, x, y ∈ C [latex]left { {{ax+y=-1} atop {x-y=2}} ight[/latex] ___________________ [latex]ax + x = 1[/latex] [latex]x cdot (a + 1) = 1 / : (a + 1)[/latex] [latex]x = frac{1}{a + 1}[/latex] [latex]x-y=2[/latex] [latex]-y=2-x / cdot (-1)[/latex] [latex]y = x - 2[/latex] [latex]y = frac{1}{a + 1} - 2[/latex] [latex]y = frac{1}{a + 1} - frac{2 cdot (a + 1)}{a + 1}[/latex] [latex]y = frac{1}{a + 1} - frac{2a + 2}{a + 1}[/latex] [latex]y = frac{1 - (2a + 2)}{a + 1}[/latex] [latex]y = frac{1 - 2a - 2}{a + 1}[/latex] [latex]y = frac{- 2a - 1}{a + 1}[/latex] [latex]left { {{x= frac{1}{a + 1}} atop {y = frac{- 2a - 1}{a + 1}}} ight[/latex] [latex]x in C o frac{1}{a + 1} in C[/latex] Stąd: [latex]a + 1 = 1 vee a + 1 = -1[/latex] [latex]a + 1 = 1[/latex] [latex]a = 1 - 1[/latex] [latex]a =0[/latex] [latex]a + 1 = -1[/latex] [latex]a = -1-1[/latex] [latex]a = - 2[/latex] Zatem: [latex]a = 0 vee a = - 2[/latex] Dla a = 0 [latex]x = frac{1}{a + 1} = frac{1}{0 + 1} = frac{1}{1} = 1[/latex] [latex]y = frac{- 2a - 1}{a + 1} = frac{- 2 cdot 0 - 1}{0 + 1} = frac{0 - 1}{1} = frac{- 1}{1} = - 1[/latex] [latex]a = 0 o left { {{x = 1} atop {y = -1}} ight[/latex] Dla a = - 2 [latex]x = frac{1}{a + 1} = frac{1}{-2 + 1} = frac{1}{-1} = -1[/latex] [latex]y = frac{- 2a - 1}{a + 1} = frac{- 2 cdot (-2) - 1}{-2 + 1} = frac{4 - 1}{-1} = frac{3}{-1} = - 3[/latex] [latex]a = -2 o left { {{x = -1} atop {y = -3}} ight[/latex]
Dla jakich całkowitych wartości a rozwiązaniem danego układu równań jest para liczb całkowitych?(Ale najpierw trzeba rozwiązać ukłąd równań) ax+y=-1 x-y=2...
