(5y-1/0.3)-(0.3y-0.12/0.02)=1 求y

问题描述:

(5y-1/0.3)-(0.3y-0.12/0.02)=1 求y

(5y-1/0.3)-(0.3y-0.12/0.02)=1
(5y-10/3)-(0.3y-6)=1
5y-10/3-0.3y+6=1
5y-0.3y=1+10/3-6
4.7y= -5/3
y= -5/3÷4.7
y= - 50/141
y= 负141分之50