Question

10道关于一元一次方程的练习题，要答案！！~~~ p.s：只是方程！！！！不要填空题什么之类的.......

Answer 1

1）x+7=26
解：两边减7得x+7-7=26-7,x=19
2)-5x=20
解：两边同除以-5得-5x/-5=20/-5,x=-4
3)-1/3x-5=4
解：两边加5得-1/3x-5+5=4+5，-1/3x=9
两边同乘以-3得x=-27
4)x-4=29
解：两边加4得x-4+4=29+4，x=33
5)3x+1=4
解：两边减1得3x+1-1=4-1,3x=3
两边同除以3得x=1
6)4x-2=2
解：两边加2得4x-2+2=2+2,4x=4
两边同除以4得x=1
7)5x-2x=9
解：把含有x的项合并得3x=9，两边同除以3得x=3
8)3x+20=4x-25
解：两边同时减4x得3x-4x+20=4x-4x-25,-x+20=-25
两边同时减20得-x+20-20=-25-20,-x=-45
两边同除以-1得x=45
9)3x+5(138-x)=540
解：去括号得3x+690-5x=540
移项得3x-5x=540-690,-2x=-150
两边同除以-2得x=75
10)2(x+3)=2.5(x-3)
解：去括号得2x+6=2.5x-7.5
移项及合并得0.5x=13.5,x=27

Answer 2

7(2x-1)-3(4x-1)=4(3x+2)-1;

(5y+1)+ (1-y)= (9y+1)+ (1-3y);

2(x-2)+2=x+1

2(x-2)-3(4x-1)=9(1-x)

x/3 -5 = (5-x)/2

2(x+1) /3=5(x+1) /6 -1

(1/5)x +1 =(2x+1)/4

(5-2)/2 - (4+x)/3 =1

x/3 -1 = (1-x)/2

(x-2)/2 - (3x-2)/4 =-1

11x+64-2x=100-9x

15-(8-5x)=7x+(4-3x)

3(x-7)-2[9-4(2-x)]=22

3/2[2/3(1/4x-1)-2]-x=2

2(x-2)-3(4x-1)=9(1-x)

11x+64-2x=100-9x

15-(8-5x)=7x+(4-3x)

3(x-7)-2[9-4(2-x)]=22

3/2[2/3(1/4x-1)-2]-x=2

2(x-2)+2=x+1

1.7(2x-1)-3(4x-1)=4(3x+2)-1

2.(5y+1)+ (1-y)= (9y+1)+ (1-3y)

3.[ (- 2)-4 ]=x+2

5.2(x-2)+2=x+1

6.2(x-2)-3(4x-1)=9(1-x)

7.11x+64-2x=100-9x

8.15-(8-5x)=7x+(4-3x)

9.3(x-7)-2[9-4(2-x)]=22

10.3/2[2/3(1/4x-1)-2]-x=2

Answer 3

回答：1）7（2x-1)-3(4x-1)=4(3x+2）-1
解：去括号：14x-7-12x+3=12x+8-1
移项：14x-12x-12x=8-1+7-3
合并同类项：-10x=11
两边同除-10：x=1.1