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1、第第7 7章章 二元一次方程組二元一次方程組七年級數(shù)學七年級數(shù)學(下下)加減法加減法(2)解方程組解方程組:7x-2y=3,9x+2y= -19. 解解y= -5.即即即即9(-1) +2y = -19,x= -1.所以所以x=-1,y= + ,得得將將x= -1代入代入,得得(1)7x+9x=3+(-19),16x= -16,2y = -19+9-5.2y = -10,消去消去y解方程組解方程組:5x+2y =1,3x+2y = 3.解解y= 3.即即即即3(-1) +2y = 3,x = -1.所以所以x = -1,y = - ,得得將將x= -1代入代入,得得(2)5x-3x = 1-3
2、,2x = -2,2y = 3+33.2y = 6,消去消去y解方程組解方程組:3x-5y = 6,x+4y = -15. - ,得得解解3,得得17y = -51,y = -3.即即即即x+4 ( )= -15,-3x = -3所以所以x = -3,y = -3.3x+12y = -453x - 5y = 612y-(-5y) = -45-6,將將y=-3代入代入,得得x-12= -15,(3)消去消去x解方程組解方程組:x-3y = -20,3x+7y =100 - ,得得解解3,得得16y = 160,y =10.即即即即x-3 = -2010 x =10.所以所以x = 10,y =
3、10.3x-9y = -60,3x +7y = 100.7y-(-9y) = 100-(-60),將將y=10代入代入,得得x-30 = -20,(4)消去消去x解方程組解方程組:4x-2y =14,5x+y = 7.+,得得解解2,得得14x = 28,x = 2.即即即即52+y = 7,y = -3.所以所以x = 2,y = -3.10 x+2y = 14,4x - 2y = 14.10 x+4x = 14+14,將將x=2代入代入,得得10+y = 7,(5)消去消去y做一做做一做:用加減法解方程組用加減法解方程組18223) 1 (yxyx. 634, 02)2(yxyx. 2,
4、8yx. 6, 3yx解方程組解方程組:3x - 4y = 10,5x+6y = 42. 解解x = 6.即即即即所以所以x = 6,y = 2.19x = 114,把把x=6代入代入,得得y = 2. 3,得得 2,得得(6)9x - 12y = 30,10 x+12y = 84.+ ,得得56+6y = 42,30+6y = 42,6y = 42-30,6y = 12,消去消去y解方程組解方程組:3x - 2y = 6,2x+3y = 17. 解解x = 4.即即即即所以所以x = 4,y = 3.13x = 52,把把x=3代入代入,得得y = 3. 3,得得 2,得得(7)9x - 6
5、y = 18,4x+6y = 34.+ ,得得24+3y = 17,8+3y = 17,3y = 17-8,3y = 9,消去消去y解方程組解方程組:2x - 3y =8,5y-7x = 5. 解解x = -5.即即即即所以所以x = -5,y = -6.-11x = 55,把把x=-5代入代入,得得y = -6. 5,得得 3,得得(8)10 x - 15y =40,15y-21x = 15.+,得得5y-7(-5 ) = 5,5y+35 = 5,5y = 5-35,5y = -30,消去消去y解方程組解方程組:2x-7y =10,3x-8y- 10 = 0. 解解2x-7( ) = 10,5y = -10,即即所以所以x = -2,y = -2.2x+14 = 10,把把y=-2代入代入,得得y = -2.(9)3x-8y = 10.由由,得得2,得得3,得得6x-16y = 20,6x-21y = 30.- ,得得 -16y-(-21y) = 20-30,-2 2x = 10-14,2x = -4,x = -2.消去消去x做一做做一做:用加減法解方程組用加減法解方程組.1123, 642) 1 (yxyx. 153, 732)2(yxyx. 1, 5yx. 1, 2yx. 010073, 0203)3(yxyx.10,10yx(附加題附加題)