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1、二元一次方程組的解法,七年級(jí)數(shù)學(xué)(下),加減法(2),解方程組:,7x-2y=3,9x+2y= -19.,解,y= -5.,即,即,9(-1) +2y = -19,x= -1.,所以,x=-1,y=,+ ,得,將x= -1代入,得,(1),7x+9x=3+(-19),16x= -16,2y = -19+9,-5.,2y = -10,消去y,解方程組:,5x+2y =1,3x+2y = 3.,解,y= 3.,即,即,3(-1) +2y = 3,x = -1.,所以,x = -1,y =,- ,得,將x= -1代入,得,(2),5x-3x = 1-3,2x = -2,2y = 3+3,3.,2y
2、= 6,消去y,解方程組:,3x-5y = 6,x+4y = -15., - ,得,解,3,得,17y = -51,y = -3.,即,即,x+4 ( )= -15,-3,x = -3,所以,x = -3,y = -3.,3x+12y = -45,3x - 5y = 6,12y-(-5y) = -45-6,將y=-3代入,得,x-12= -15,(3),消去x,解方程組:,x-3y = -20,3x+7y = 100., - ,得,解,3,得,16y = 160,y =10.,即,即,x-3 = -20,10,x =10.,所以,x = 10,y = 10.,3x-9y = -60,3x +7
3、y = 100.,7y-(-9y) = 100-(-60),將y=10代入,得,x-30 = -20,(4),消去x,解方程組:,4x-2y =14,5x+y = 7.,+,得,解,2,得,14x = 28,x = 2.,即,即,52+y = 7,y = -3.,所以,x = 2,y = -3.,10 x+2y = 14,4x - 2y = 14.,10 x+4x = 14+14,將x=2代入,得,10+y = 7,(5),消去y,做一做:用加減法解方程組,解方程組:,3x - 4y = 10,5x+6y = 42.,解,x = 6.,即,即,所以,x = 6,y = 2.,19x = 114
4、,把x=6代入,得,y = 2., 3,得, 2,得,(6),9x - 12y = 30,10 x+12y = 84.,+ ,得,56+6y = 42,30+6y = 42,6y = 42-30,6y = 12,消去y,解方程組:,3x - 2y = 6,2x+3y = 17.,解,x = 4.,即,即,所以,x = 4,y = 3.,13x = 52,把x=3代入,得,y = 3., 3,得, 2,得,(7),9x - 6y = 18,4x+6y = 34.,+ ,得,24+3y = 17,8+3y = 17,3y = 17-8,3y = 9,消去y,解方程組:,2x - 3y =8,5y-7x = 5.,解,x = -5.,即,即,所以,x = -5,y = -6.,-11x = 55,把x=-5代入,得,y = -6., 5,得, 3,得,(8),10 x - 15y =40,15y-21x = 15.,+,得,5y-7(-5 ) = 5,5y+35 = 5,5y = 5-35,5y = -30,消去y,
華師大版7.2《二元一次方程組的解法》(加減法2)課件.ppt
