代入消元法&加減消元法－yffuhxy的部落格

標題:

代入消元法&加減消元法

發問:

如題......幫幫忙！我要詳細步驟！！！！ 1)4x-2y=2(x+y)+y=5(代入消元法) 2)2x-3=6(y+3)﹜加減消元法 9x+10y=2 3)x-y/2=x/2+y/4=2

最佳解答:

您好，我是 lop，高興能解答您的問題。 (1) 4x - 2y= 2(x+y) + y --- (1) 2(x+y) + y = 5 --- (2) 從 (1) : 4x - 2y = 2x + 2y + y 4x - 2y = 2x + 3y 4x - 2x = 2y + 3y 2x = 5y x = 5y/2 --- (3) 代 (3) 入 (2) , 2(x+y) + y = 5 2(5y/2+y) + y = 5 5y + 2y + y = 5 8y = 5 y = 5/8 --- (4) 代 (4) 入 (3) x = 5(5/8)/2 x = (25/8)/2 x = 25/16 答：x = 25/16，y = 5/8。 (2) 2x - 3 = 6(y+3) --- (1) 9x + 10y = 2 --- (2) 從 (1) : 2x - 3 = 6y + 18 2x - 6y = 18 + 3 2x - 6y = 21 --- (3) (3) × 5 : 10x - 30y = 105 --- (4) (2) × 3 : 27x + 30y = 6 --- (5) (4) + (5) : 37x = 111 x = 3 --- (6) (6) × 2 - (3) : 6y = -21 y = -2.5 答：x = 3，y = -2.5。 (3) x - y/2 = x/2 + y/4 --- (1) x/2 + y/4 = 2 --- (2) (1) × 4 : 4x - 2y = 2x + y 4x - 2x = y + 2y 2x = 3y x = 3y/2 --- (3) 代 (3) 入 (2) , (3y/2)/2 + y/4 = 2 3y/4 + y/4 = 2 4y/4 = 2 y = 2 --- (4) 代 (4) 入 (3) , x = 3(2)/2 x = 3 答：x = 3，y = 2。

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