数学：7.2《二元一次方程组的解法》(加减法2)课件（华师大版七年级下）

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七年级数学(下)
加减法(2)
解方程组:
7x-2y=3,
9x+2y= -19.
②
①
解
y= -5.
即
即
9×(-1) +2y = -19,
x= -1.
所以
x=-1,
y=
①+ ②,得
将x= -1代入②,得
(1)
7x+9x=3+(-19),
16x= -16,
2y = -19+9
-5.
2y = -10,
消去y
解方程组:
5x+2y =1,
3x+2y = 3.
②
①
解
y= 3.
即
即
3×(-1) +2y = 3,
x = -1.
所以
x = -1,
y =
①- ②,得
将x= -1代入②,得
(2)
5x-3x = 1-3,
2x = -2,
2y = 3+3
3.
2y = 6,
消去y
解方程组:
3x-5y = 6,
x+4y = -15.
②
①
③ - ①,得
解
②×3,得
③
17y = -51,
y = -3.
即
即
x+4× ( )= -15,
-3
x = -3
所以
x = -3,
y = -3.
3x+12y = -45
3x - 5y = 6
①
12y-(-5y) = -45-6,
将y=-3代入②,得
x-12= -15,
(3)
消去x
解方程组:
x-3y = -20,
3x+7y = 100.
②
①
② - ③,得
解
①×3,得
③
16y = 160,
y =10.
即
即
x-3× = -20
10
x =10.
所以
x = 10,
y = 10.
3x-9y = -60,
3x +7y = 100.
②
7y-(-9y) = 100-(-60),
将y=10代入①,得
x-30 = -20,
(4)
消去x
解方程组:
4x-2y =14,
5x+y = 7.
②
①
③+①,得
解
②×2,得
③
14x = 28,
x = 2.
即
即
5×2+y = 7,
y = -3.
所以
x = 2,
y = -3.
10x+2y = 14,
4x - 2y = 14.
①
10x+4x = 14+14,
将x=2代入②,得
10+y = 7,
(5)
消去y
做一做:用加减法解方程组
解方程组:
3x - 4y = 10,
5x+6y = 42.
②
①
解
x = 6.
即
即
所以
x = 6,
y = 2.
19x = 114,
把x=6代入②,得
y = 2.
① ×3,得
② ×2,得
(6)
9x - 12y = 30,
10x+12y = 84.
③
④
③+ ④,得
5×6+6y = 42,
30+6y = 42,
6y = 42-30,
6y = 12,
消去y
解方程组:
3x - 2y = 6,
2x+3y = 17.
②
①
解
x = 4.
即
即
所以
x = 4,
y = 3.
13x = 52,
把x=3代入②,得
y = 3.
① ×3,得
② ×2,得
(7)
9x - 6y = 18,
4x+6y = 34.
③
④
③+ ④,得
2×4+3y = 17,
8+3y = 17,
3y = 17-8,
3y = 9,
消去y
解方程组:
2x - 3y =8,
5y-7x = 5.
②
①
解
x = -5.
即
即
所以
x = -5,
y = -6.
-11x = 55,
把x=-5代入②,得
y = -6.
① ×5,得
② ×3,得
(8)
10x - 15y =40,
15y-21x = 15.
③
④
③+④,得
5y-7×(-5 ) = 5,
5y+35 = 5,
5y = 5-35,
5y = -30,
消去y
解方程组:
2x-7y =10,
3x-8y- 10 = 0.
②
①
解
2x-7( ) = 10,
5y = -10,
即
所以
x = -2,
y = -2.
2x+14 = 10,
把y=-2代入①,得
y = -2.
(9)
3x-8y = 10.
由②,得
③
③×2,得
①×3,得
6x-16y = 20,
6x-21y = 30.
④
⑤
④- ⑤,得
-16y-(-21y) = 20-30,
-2
2x = 10-14,
2x = -4,
x = -2.
消去x
做一做:用加减法解方程组
(附加题)