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复习提问：下列等式（ 2 ）是根据等式的哪一条性质从等式（ 1 ）变形得到的？

复习提问：下列等式（ 2 ）是根据等式的哪一条性质从等式（ 1 ）变形得到的？. (1) 4x = 3x +2 (2) x = 2. (1) 5x = -15 (2) x = -3. 解方程： x - 7 = 5. 算术解法 x = 5 + 7 x = 12. 代数解法 x - 7 + 7 = 5 + 7 x = 12. 一元一次方程和它的解法 第一课时. x - 7 = 5. x - 7 + 7 = 5 + 7. x = 5 + 7. x - 7 = 5 x = 5 + 7. 7x = 6x + 4.

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复习提问：下列等式（ 2 ）是根据等式的哪一条性质从等式（ 1 ）变形得到的？

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  1. 复习提问：下列等式（2）是根据等式的哪一条性质从等式（1）变形得到的？复习提问：下列等式（2）是根据等式的哪一条性质从等式（1）变形得到的？ (1) 4x = 3x +2 (2) x = 2 (1) 5x = -15 (2) x = -3

  2. 解方程： x - 7 = 5 算术解法 x = 5 + 7 x = 12 代数解法 x - 7 + 7 = 5 + 7 x = 12

  3. 一元一次方程和它的解法 第一课时

  4. x - 7 = 5 x - 7+ 7= 5+ 7 x = 5 + 7

  5. x - 7 = 5 x = 5 + 7

  6. 7x = 6x + 4 7x - 6x = 6x + 4 - 6x 7x - 6x = 4

  7. 7x = 6x - 4 7x - 6x = -4

  8. 下面的变形对不对？如果不对，错在哪里？应当怎样改正？下面的变形对不对？如果不对，错在哪里？应当怎样改正？ （1）由7+x=13，得x=13+7 x=13-7 （2）由5x=4x+8，得5x+4x=8 5x-4x=8

  9. （3）由-6x+7x=-8，得-7x+6x=-8 7x-6x=-8 （4）由3x=2x-1，得3x-2x=1 3x-2x=-1

  10. 填空：解方程 5x - 3 = 4x 解：移项，得5x____=__ __=3 检验：把____代入______，左边=____，右边=____，左边=右边, 所以x=___是原方程的解。 -4x 3 x x=3 原方程 12 12 3

  11. 通过移项解下列方程并检验： (1) x-15=74 (2) 3x=2x-5 (3) 7x-3=6x (4) 2x+3=x-2 (5) 3x-4+2x=4x-3 (6) 10.5x+7=12.5x-5-3x

  12. (2) 3x=2x-5 (1) x-15=74 解：3x-2x=-5 解：x=74+15 (3) 7x-3=6x (4) 2x+3=x-2 解：7x-6x=3 解：2x-x=-2-3

  13. (5) 3x-4+2x=4x-3 解：3x+2x-4x=-3+4 另解：5x-4=4x-3 5x-4x=-3+4 (6) 10.5x+7=12.5x-5-3x 解：10.5x+7=9.5x-5 10.5x-9.5x=-5-7

  14. 本节课我们主要学习了 ____法则，这个法则的理 论依据是_________，运用 这个法则时要注意______ _____。 移项 等式性质1 移项要 变号

  15. 作业： 1. P205 1 2.试解方程： 5x - 3 = 2x - 9

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