If the solution of the equations 4x + y = K + 1x + 4Y = 3 satisfies the condition 0 < x + y < 1, then the value range of K is () A. -4＜k＜1B. -4＜k＜0C. 0＜k＜9D. k＞-4

Method 1: (1) - (2) × 4 is obtained, - 15y = k-11, y = 11 − K15; (1) × 4 - (2) is obtained, 15x = 4K + 1, x = 4K + 115, ∵ 0 ＜ x + y ＜ 1, ∵ 11 − K15 + 4K + 115 ＞ 011 − K15 + 4K + 115 ＜ 1, the solution is, - 4 ＜ K ＜ 1

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