Rearranging Equations: Finding y in x = (y^3+1)/(y^2+1)

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Homework Help Overview

The discussion revolves around rearranging the equation x = (y^3+1)/(y^2+1) to isolate y. Participants are exploring the complexities involved in solving for y, particularly in the context of cubic equations.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants express uncertainty about the feasibility of isolating y and question whether it is possible to express y as a function of x. Some suggest that the problem may involve complex algebraic manipulation, while others mention the need for knowledge of cubic equations.

Discussion Status

The conversation is ongoing, with various participants contributing their thoughts on the problem. Some have offered hints regarding factoring and rearranging terms, while others remain confused about the process and the potential for a solution.

Contextual Notes

There is a recurring theme of uncertainty regarding the ability to solve for y without resorting to the cubic formula, and some participants mention the complexity of the algebra involved. The original poster indicates a desire for clarity on whether a solution can be achieved.

Toby_Obie
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Homework Statement



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

Homework Equations



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

The Attempt at a Solution



Not sure , not been doing algebra for long
 
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Welcome to PF!

Hi Toby_Obie! Welcome to PF! :smile:

(try using the X2 tag just above the Reply box :wink:)
Toby_Obie said:
Rearrange x = (y^3+1)/(y^2+1) to find y

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


tiny-tim said:
I don't think there's any way of making y the subject, without knowing the solution to a general cubic equation.
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
the answer has a fraction
 
Last edited:
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
 
Toby_Obie said:
Can it be done
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:
Sorry I don't get it

Could someone help me out ?
 
sorry, not me … i don't get it either :redface:
 
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
tiny-tim said:
sorry, not me … i don't get it either :redface:

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
story645 said:
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.

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
tiny-tim said:
I repeat … I don't think there's any way of making y the subject, without knowing the solution to a general cubic equation.
Sorry, I was slow on the uptake and totally missing your hint.
 
  • #13
tiny-tim said:
I repeat … I don't think there's any way of making y the subject, without knowing the solution to a general cubic equation.

story645 said:
Sorry, I was slow on the uptake and totally missing your hint.
Hint? What hint? tiny-tim said exactly that in his first response and you said it was not true.

From [itex]x= (y^3+ 1)/(y^2+ 1)[/itex] we have [itex]x(y^2+ 1)= xy^2+ x= y^3+ 1[/itex] so [itex]y^3- xy^2+ (1- x)= 0[/itex]. There no way to solve that for y, with general x, except by using the (very complicated!) cubic formula.
 
  • #14
Thanks guys
 

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