消元法解二元一次方程组-课件教学文案.ppt

单击此处编辑母版标题样式,单击此处编辑母版文本样式,第二级,第三级,第四级,第五级,*,Evaluation only.,Created with Aspose.Slides for.NET 3.5 Client Profile 5.2.0.0.,Copyright 2004-2011 Aspose Pty Ltd.,y=2 x 1,3 x 4 y=2,(1),(2),x y=5,2 x+4 y=,1,(3),3 x=5 y,2 x 7 y=3,(4),3x+5 y=21,2x -5 y=-11,Evaluation only.,Created with Aspose.Slides for.NET 3.5 Client Profile 5.2.0.0.,Copyright 2004-2011 Aspose Pty Ltd.,和,互为相反数,按照这种思路，你能消去,一个未知数吗？,？,（,3x,5y,）,+,（,2x,5y,）,21 +(,11),分析：,3X+5y+2,x,5y,10,左边,+,左边,=,右边,+,右边,5x+0y,10,5,x=10,Evaluation only.,Created with Aspose.Slides for.NET 3.5 Client Profile 5.2.0.0.,Copyright 2004-2011 Aspose Pty Ltd.,解方程组,3x +5y =5,3x -4y =23,解：,把,-,得,9y =-18,y =-2,把,y =-2,代入,，得,3x +5 (-2)=5,解得,x =5,所以原方程组的解是,x =5,y =-2,Evaluation only.,Created with Aspose.Slides for.NET 3.5 Client Profile 5.2.0.0.,Copyright 2004-2011 Aspose Pty Ltd.,当二元一次方程组中两个方程的某个未知数的系数,互为相反数或相等,时，将两个方程的两边分别,相加或相减,，就能消去这个,未知数,，得到一个,一元一次方程,，进而求出二元一次方程组的解。,加减消元法,像这样,通过对方程组中的两个方程进行加或减的运算就可以消去一个未知数,得到一个一元一次方程,这种方法叫做,加减消元法,简称,加减法,.,Evaluation only.,Created with Aspose.Slides for.NET 3.5 Client Profile 5.2.0.0.,Copyright 2004-2011 Aspose Pty Ltd.,试一试,7x-2y=3,9x+2y=-19,6x-5y=3,6x+y=-15,用加减消元法解下列方程 组,.,(,你可以,选择你喜欢的一题解答,),相信自己,Evaluation only.,Created with Aspose.Slides for.NET 3.5 Client Profile 5.2.0.0.,Copyright 2004-2011 Aspose Pty Ltd.,1,、选择题,1.,用加减法解方程组,6x+7y=-19,6x-5y=17,应用（）,A.-,消去,y,B.-,消去,x,C.-,消去常数项,D.,以上都不对,B,2.,方程组,3x+2y=13,3x-2y=5,消去,y,后所得的方程是（）,B,A.6x=8,B.6x=18,C.6x=5,D.x,=18,Evaluation only.,Created with Aspose.Slides for.NET 3.5 Client Profile 5.2.0.0.,Copyright 2004-2011 Aspose Pty Ltd.,二、指出下列方程组求解过程中有错误步骤，并给予订正：,7x,4y,4,5x,4y,4,解,:,，得,2x,4,4,，,x,0,3x,4y,14,5x,4y,2,解：,，得,2x,12,x,6,解,:,，得,2x,4,4,，,x,4,解,:,，得,8x,16,x,2,练一练,Evaluation only.,Created with Aspose.Slides for.NET 3.5 Client Profile 5.2.0.0.,Copyright 2004-2011 Aspose Pty Ltd.,分别相加,y,1,、已知方程组,x+3y=17,2x-3y=6,两个方程只要两边,就可以消去未知数,分别相减,2,、已知方程组,25x-7y=16,25x+6y=10,两个方程只要两边,就可以消去未知数,x,已知,a,、,b,满足方程组,a+2b=8,2a+b=7,则,a+b,=,5,训练场,3,、,Evaluation only.,Created with Aspose.Slides for.NET 3.5 Client Profile 5.2.0.0.,Copyright 2004-2011 Aspose Pty Ltd.,(2),已知,3a,3x,b,2x-y,和,-7a,8-y,b,7,是同类项,求,xy,解：根据题意：得,3x=8-y,2x-y=7,转化为,3x+y=8,2x-y=7,x=3,y=-1,即,xy,=-3,Evaluation only.,Created with Aspose.Slides for.NET 3.5 Client Profile 5.2.0.0.,Copyright 2004-2011 Aspose Pty Ltd.,总结,总结评价,1,、,易错点：在用加减法消元时，,符号,易,出现错误。,2,、,用加减法解二元一次方程组的条件：某一未知数系数的,绝对值相等,Evaluation only.,Created with Aspose.Slides for.NET 3.5 Client Profile 5.2.0.0.,Copyright 2004-2011 Aspose Pty Ltd.,基本思路,:,主要步骤,：,加减消元,:,二元,一元,加减,消去一个元,求解,分别求出两个未知数的值,3,、加减消元法解方程组基本思路是什么？,主要步骤有哪些？,变形,同一个未知数的系,数相同或互为相反数,4,、,.,二元一次方程组解法有,.,代入法、加减法,写解,写出方程组的解,Evaluation only.,Created with Aspose.Slides for.NET 3.5 Client Profile 5.2.0.0.,Copyright 2004-2011 Aspose Pty Ltd.,当方程组中的一个未知数系数 的绝对值是,1,或一个方程的常数项为,0,时用代入消元法较方便。,当两个方程中同一个未知数的系数绝对值相等时，用加减消元法较方便。,Evaluation only.,Created with Aspose.Slides for.NET 3.5 Client Profile 5.2.0.0.,Copyright 2004-2011 Aspose Pty Ltd.,知识拓展：,1,、,3x,2a+b+2,+5y,3a-b+1,=8,是关于,x,、,y,的二元一次方程,求,a,、,b,解：根据题意：得,2a+b+2=1,3a-b+1=1,得：,a=,b=,1,5,-,3,5,-,Evaluation only.,Created with Aspose.Slides for.NET 3.5 Client Profile 5.2.0.0.,Copyright 2004-2011 Aspose Pty Ltd.,已知（,3m+2n-16),2,与,|3m-n-1|,互为相反数,求：,m+n,的值,解：根据题意：得,3m+2n-16=0,3m-n-1=0,解得：,m=2,n=5,即：,m+n,=7,Evaluation only.,Created with Aspose.Slides for.NET 3.5 Client Profile 5.2.0.0.,Copyright 2004-2011 Aspose Pty Ltd.,问题,3,：解方程组,2,（,2x+1,）,=6-5y,3,（,y+1,）,=3-4x,解原方程组变形为,4x+5y=4,4x+3y=0,得：,2y=4,把,y=2,代入,得,x=-1.5,X=-1.5,Y=2,y=2,Evaluation only.,Created with Aspose.Slides for.NET 3.5 Client Profile 5.2.0.0.,Copyright 2004-2011 Aspose Pty Ltd.,问题,4,：已知方程组,ax-by=4,ax+by=2,与方程组,4x+3y=4,4x-5y=6,的解相同，求,a,，,b,解：解方程组,4x+3y=4,4x-5y=6,得,X=,？,Y=,？,将,X=,？,Y=,？代入,ax-by=4,ax+by=2,由此可求出,a=,？,b=,？,Evaluation only.,Created with Aspose.Slides for.NET 3.5 Client Profile 5.2.0.0.,Copyright 2004-2011 Aspose Pty Ltd.,知 识 的 升 华,独立,作业,1.,当,m,为何值时,关于,x,、,y,的,方程组,得的解的和为,12?,2x+3y=m,3x+5y=m+2,Evaluation only.,Created with Aspose.Slides for.NET 3.5 Client Profile 5.2.0.0.,Copyright 2004-2011 Aspose Pty Ltd.,