# Why can't I divide 2 to the other side in y+1=2square root/y+1?

• Aug 13th 2012, 06:59 PM
EJdive43
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/"
• Aug 13th 2012, 07:18 PM
richard1234
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 $\Sigma$). The square root of y+1 is denoted by \sqrt{y+1}, which yields $\sqrt{y+1}$.

As of right now, your question's too unclear...it seems like you are saying $y+1 = 2\sqrt{y+1}$ but I don't want to guess.
• Aug 13th 2012, 07:24 PM
EJdive43
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?
• Aug 13th 2012, 07:25 PM
richard1234
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 $\frac{y+1}{2} = \sqrt{y+1}$, but you don't really have to.
• Aug 13th 2012, 07:43 PM
EJdive43
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.
• Aug 13th 2012, 10:00 PM
richard1234
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

$y+1 = 2 \sqrt{y+1}$

From here should be easy. Square both sides (again) to get

$(y+1)^2 = 4(y+1)$

$(y+1)^2 - 4(y+1) = 0$. Here, note that the LHS factors to

$(y+1)((y+1) - 4) = 0$ (if you don't see this immediately, you could perhaps let u = y+1 and factor).

Solving, we get $y = -1$ and $y = 3$. Both solutions work.