Question

(2y-1)/3-(10y+1)/6=(2y+1),5x/12=(2x-1)/4-1 1/3[3y-(10-7y)/2]

Answer 1

(2y-1)/3-(10y+1)/6=(2y+1)

2(2y-1)-10-1=12y+6

-8y=19

y=-19/8

5x/12=(2x-1)/4-1

5x=4(2x-1)

-3x=-4

x=4/3

1/3[3y-(10-7y)/2]-1/2[2y-(2y+2)/3]=y/2-1

y-(10-7y)/6-y+(2y+2)/6=y

-(10-7y)+2y+2=6y

3y=8

y=8/3