资源简介 (共22张PPT)用加减法解二元一次方程组华东师大版七年级数学下册6复习回顾1. 用代入法解二元一次方程组的基本思路是什么?基本思路:消元二元一元转化2. 用代入法解方程组的步骤是什么?变形代入求解回代写解用一个未知数的代数式表示另一个未知数例 4 解方程组:3x + 7y = 9, ①4x-7y = 5. ②解:①+②,得 7x = 14,“二元”变“一元”即 x = 2.将 x = 2 代入①,得 6+7y = 9.37解得 y = .x = 2,y = .所以37新课探究3x + 5y = 5, ①3x - 4y = 23. ②例 3 解方程组:解:①-②得 9y = -18,即 y = -2.把 y = -2代入①,得 3x + 5×(-2) = 5.解得 x = 5.x = 5,y = -2.所以3x + 5y = 5, ①3x - 4y = 23. ②3x + 7y = 9, ①4x -7y = 5. ②方法归纳①-②,得 9y = -18,① + ②,得 7x = 14,“二元”变“一元”通过将两个方程的两边分别相加(或相减)消去一个未知数,将方程组转化为一元一次方程. 这种解法叫做加减消元法,简称加减法.练 习解下列方程组:5x + y = 7,3x - y = 1.(1)①②4x - 3y = 5,4x + 6y = 14.(2)①②直接相加减不能消去一个未知数,怎么办呢?例 5 解方程组:2x- 7y = 30. ②3x + 5y = 76, ①直接相加减不能消去一个未知数,怎么办呢?3 和 2 的最小公倍数是 6例 5 解方程组:2x- 7y = 30. ②3x + 5y = 76, ①①×2,得6x + 10y = 152②×3,得6x- 21y = 90例 5 解方程组:解 由①×2,②×3,得6x- 21y = 90. ④6x + 10y = 152, ③③-④,得解得 y = 2.将 y = 2 代入①,得 3x = 66,解得 x = 22.x = 6,y = 2.所以变形加减2x- 7y = 30. ②3x + 5y = 76, ①6x + 10y-(6x- 21y)=152-9031y= 62.例 5 解方程组:试试用同样的方式把y消掉2x- 7y = 30. ②3x + 5y = 76, ①试一试在解上节课例 2 的方程组时是用代入法解的,现在用加减法试试,看哪种方法比较简便.2x -7y = 8,3x -8y -10 = 0.2x -7y = 8, ①3x -8y -10 = 0. ②解 由①×3,②×2,得6x - 16y -20 = 0. ④6x -21y = 24, ③③ - ④,得 -5y = 4,解得 y = -0.8 .将 y = -0.8 代入①,得 2x-7×(-0.8) = 8,解得 x = 1.2 .x = 1.2,y = -0.8.所以练 习解下列方程组:3x -2y = 6,2x + 3y = 17.(1)①②解 由①×3,②×2,得4x + 6y = 34. ④9x -6y = 18,③③ + ④,得 13x = 52.解得 x = 4.将 x = 4 代入②,得 2×4 + 3y = 17,解得 y = 3.x = 4,y = 3.所以4x-2y = 14,5x + y = 7.(2)①②解 由②×2,得10x + 2y = 14. ③③ + ①,得 14x = 28.解得 x = 2.将 x = 2 代入②,得 5×2 + y = 7,解得 y = -3.x = 2,y = -3.所以x-3y = -20,3x + 7y = 100.(3)①②解 由①×3,得3x -9y = -60, ③② - ③ ,得 16y = 160.解得 y = 10.将 y = 10 代入①,得 x-3×10 = -20,解得 x = 10.x = 10,y = 10.所以2x-3y = 8,5y - 7x = 5.(4)①②解 由①×5,②×3,得15y - 21x = 15. ④10x -15y = 40, ③③ + ④,得 -11x = 55.解得 x = -5.将 x = -5 代入②,得 5y-7×(-5) = 5,解得 y = –6.x = -5,y = -6.所以课堂小结加减消元法条件:步骤:两个二元一次方程中同一未知数的系数相等或互为相反数变形 加减 求解 回代 写解1. 从课后习题中选取;2. 完成练习册本课时的习题.课后作业下 课Thanks!https://www.21cnjy.com/recruitment/home/fine 展开更多...... 收起↑ 资源预览