已知1-2y-(y-11)/3=0
3-6y-y+11=0
14=7y
y=2
所以(3y-1)/4+(2y-1)/3
=(3*2-1)/4+(2*2-1)/3
=5/4+1=9/43-6y-y+11=0是如何算出14=7y，求解。3-6y-y+11=014-7y=07y=14y=2