Question

已知代数式-3y-(y-11)\2+1的值为0,求代数式（3y+1）\2+(2y+1）\4的值

Answer 1

-3y-(y-11)/2+1=0

-6y-y+11+2=0

-7y=-13

y=13/7

(3y+1)/2+(2y+1)/4

=(6y+2+2y+1)/4

=(8y+3)/4

=(8*13/7+3)/4

=125/28