1/3(3y-10--7y/2)-1/6(2y-2y+2/3)=y/2-1三分之一(3y减二分之十减七y)-六分之一(二y减三分之2y+2)=二分之y减一

问题描述:

1/3(3y-10--7y/2)-1/6(2y-2y+2/3)=y/2-1
三分之一(3y减二分之十减七y)-六分之一(二y减三分之2y+2)=二分之y减一

1/3*[3y-(10--7y)/2]-1/6*[2y-(2y+2)/3]=y/2-1
1/3*(-y-10)/2-1/6*(4y-2)/3=y/2-1
1/6(-y-10)-1/18*(4y-2)=y/2-1
3(-y-10)-(4y-2)=9y-18
-3y-30-4y+2=9y-18
16y=-10
y=-5/8

1/3[3y-(10-7y)/2]-1/6[2y-(2y+2)/3]=y/2-1
[y-(10-8y)/6]-[1/3y-(2y+2)/18]=y/2-1
y-(10-8y)/6-1/3y+(2y+2)/18]=y/2-1
18y-3(10-8y)-6y+(2y+2)=9y-18
18y-30-24y-6y+2y+2=9y-18
-10y-28=9y-18
9y+10y=-28+18
19y=-10
y=-10/19

两边乘6
6y-10+7y-2y+(2y+2)/3=3y-6
8y+(2y+2)/3=4
两边乘3
24y+2y+2=12
26y=10
y=5/13