Homework Help: Rearrangement - Quickie easy

1. Jan 21, 2010

Toby_Obie

1. The problem statement, all variables and given/known data

Hello, I just need some quick help with the following, rearranging the below to make y the subject

2. Relevant equations

Rearrange x = (y^3+1)/(y^2+1) to find y

3. The attempt at a solution

Not sure , not been doing algebra for long

2. Jan 21, 2010

tiny-tim

Welcome to PF!

Hi Toby_Obie! Welcome to PF!

(try using the X2 tag just above the Reply box )
Are you sure that's the question?

I don't think there's any way of making y the subject, without knowing the solution to a general cubic equation.

3. Jan 21, 2010

Toby_Obie

Question is correct - is it possible then ? There must be some way

4. Jan 21, 2010

story645

Re: Welcome to PF!

You're over complicating a bit. It's probably just solving the equation in terms of y, which itself just means moving somethings to the other side of the equals side.
Remember, what you do to one side of the equals sign, you do to the other side.

hints:
factor
get rid of fractions,
you'll have to move things around in both directions

Last edited: Jan 21, 2010
5. Jan 21, 2010

Toby_Obie

I'm searching for an answer in the following form :

y = f(x)

Can it be done

It may be over complicating but I have to know

Thanks for you help

6. Jan 21, 2010

story645

yes
factor y^3+1, simplify, and you should be able to work it out from there. It's just a lot of shuffling from one side of the equation to the other. Keep stuff in factored form and factor some more.

Last edited: Jan 21, 2010
7. Jan 21, 2010

Toby_Obie

Sorry I don't get it

Could someone help me out ?

8. Jan 21, 2010

tiny-tim

sorry, not me … i don't get it either

9. Jan 21, 2010

Toby_Obie

I'm unsure how to release the x and y

x = (y^3+1)/(y^2+1)

x(y^2+1) = (y^3+1)

(xy^2)+x = y^3+1

x-1 = (y^3) - (xy^2) ?

Factoring (y^3+1) only creates (y+1)(y^2-y+1)

10. Jan 21, 2010

story645

I've spent way too many pages on this, but basically I think you just have to keep rearranging the equations (adding, subtracting, multiplying, dividing, multiplying) until everything simplifies.

Have you learned complex numbers yet? If so, you can use those to simplify things further.

11. Jan 21, 2010

tiny-tim

I repeat … I don't think there's any way of making y the subject, without knowing the solution to a general cubic equation.

12. Jan 21, 2010

story645

Sorry, I was slow on the uptake and totally missing your hint.

13. Jan 22, 2010

HallsofIvy

Hint? What hint? tiny-tim said exactly that in his first response and you said it was not true.

From $x= (y^3+ 1)/(y^2+ 1)$ we have $x(y^2+ 1)= xy^2+ x= y^3+ 1$ so $y^3- xy^2+ (1- x)= 0$. There no way to solve that for y, with general x, except by using the (very complicated!) cubic formula.

14. Jan 22, 2010

Thanks guys