(5y+1)除以6等于(9y+1)除以8-(1-y)除以3 y等于多少

问题描述:

(5y+1)除以6等于(9y+1)除以8-(1-y)除以3 y等于多少

(5y+1)÷6=(9y+1)÷8-(1-y)÷3
[(5y+1)÷6]×24=[(9y+1)÷8-(1-y)÷3]×24
4×(5y+1)=3×(9y+1)-8×(1-y)
20y+4=27y+3-8+8y
20y-27y-8y=3-8-4
-15y=-9
y=3/5