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中师数学课本《代数与初等函数》第二册83页例2需要简化。例题是这样的: 求方程407x-2816y=33的一个整数解。解:将方程化简为 37x-256y=3即 37x+256(-y)=3∵ 256=6×37+34 37=1×34+3(1) 34=11×3+1∴ 1=34-11×3 =34-11×(37-34) =(256-6×37) =11×[37-(256-6×37)] =37(-6-11-66)+256(1+11) =37×(-83)+256×12上式各项乘以3得 37×(-249)+256×36=3 原方程的一个整数解是 x_0=-249 y_0=-36 这道题应该怎样简化呢?我认为(1)式以前不变,根据(1)式的特点,我们完全有理由把(1)式直接变成
MATHMASTER TEXTBOOK Algebraic and Elementary Functions Volume 2 Page 83 Example 2 needs to be simplified. The example is this: Find an integer solution to equation 407x-2816y = 33. Solution: The equation is reduced to 37x-256y = 3 ie 37x + 256 (-y) = 3∵ 256 = 6 × 37 + 34 37 = 1 × 34 + 3 34 = 11 × 3 + 1∴ 1 = 34-11 × 3 = 34-11 × 37-34 = 256-6 × 37 = 11 × 37- 256-6 × 37 = 37 -6-11-66 + 256 1 + 11) = 37 × (-83) + 256 × 12 The above equation is multiplied by 3 37 × (-249) + 256 × 36 = 3 The original equation of an integer solution is x_0 = -249 y_0 = -36 How can I simplify this question? I think that (1) has not changed before. Based on the characteristics of (1), we have every reason to turn (1) into