Why can't I divide 2 to the other side in y+1=2square root/y+1?
Why can't I divide 2 to the other side in y+1=2square root/y+1? I would like to get rid of the square root in y+1 by squaring it, but my answer sheet say I can't divide the 2 out but I just have to square it with the SR/y+1 to get y.
Original problem: sr/2y+3-sr/y+1=1
I don't know where to find the symbol for square root so I say "sr/"
Re: Why can't I divide 2 to the other side in y+1=2square root/y+1?
To make square roots in LaTeX, click "Go Advanced" then click on the TeX tags (which will look like a sigma 
). The square root of y+1 is denoted by \sqrt{y+1}, which yields 
.
As of right now, your question's too unclear...it seems like you are saying 
but I don't want to guess.
Re: Why can't I divide 2 to the other side in y+1=2square root/y+1?
You guess correctly! Why can't I divide the 2?
Re: Why can't I divide 2 to the other side in y+1=2square root/y+1?
Quote:
Originally Posted by
EJdive43 
You guess correctly! Why can't I divide the 2?
You could divide both sides by 2 to obtain 
, but you don't really have to.
Re: Why can't I divide 2 to the other side in y+1=2square root/y+1?
Ok, how would you solve sr/2y+3-sr/ y+1= 1? Please write it in steps.
Re: Why can't I divide 2 to the other side in y+1=2square root/y+1?
You should try to use the typesetting here. Or at least use sqrt(2y+3) - sqrt(y+1) = 1.
I think you're on the right track, I did it myself by rearranging and squaring both sides twice, obtaining
From here should be easy. Square both sides (again) to get
^2 = 4(y+1))
^2 - 4(y+1) = 0)
. Here, note that the LHS factors to
((y+1) - 4) = 0)
(if you don't see this immediately, you could perhaps let u = y+1 and factor).
Solving, we get 
and 
. Both solutions work.
