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
问题描述:
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
答
-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