Why does finding x = (1+y)/(3y) show that R−{0} is in the image of F?

[annoymage]

Homework Statement

F : R-{1/3} ---> R-{0}
F(x) = 1/(3x-1)

to show R-{0} $$\subseteq$$ image of F

for any y in R-{0}, if we try to solve y=1/(3x-1), we find x=(1+y)/3y,

F((1+y)/3y)=y

Homework Equations

R is real number

The Attempt at a Solution

so, if (1+y)/3y are the values for x in R-{0} such that F((1+y)/3y)=y

but i don't understand, how is that showing R-{0} $$\subseteq$$ image of F

explain to me please

 

[radou]

You have shown that if y is in R\{0}, then it is in the image of F, too, since for such a number y there exists a number x = (1 + y) / 3y whose image under F is y. (btw the image of F is defined as {F(x) : x is in R\{1/3}})

 

[annoymage]

since for such a number y there exists a number x = (1 + y) / 3y whose image under F is y.

thanks, i get it now ^^