10道关于一元一次方程的练习题,p.s：只是方程!不要填空题什么之类的.

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

1）x+7=26
两边减7得x+7-7=26-7,x=19
2)-5x=20
两边同除以-5得-5x/-5=20/-5,x=-4
3)-1/3x-5=4
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
两边加4得x-4+4=29+4，x=33
5)3x+1=4
两边减...

回答：1）7（2x-1)-3(4x-1)=4(3x+2）-1
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