How do I solve this algebra problem with unknown constant k?

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Homework Help Overview

The discussion revolves around an algebra problem involving an unknown constant k. The original poster presents an equation relating y and k, expressing difficulty in transitioning between steps in the problem-solving process.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore rewriting the equation to isolate y, with some questioning the implications of manipulating the equation, such as concerns about maintaining the integrity of the expression while simplifying.

Discussion Status

There is ongoing dialogue with participants offering different perspectives on how to approach the problem. Some guidance has been provided regarding rewriting the equation, but no consensus has been reached on the best method to proceed.

Contextual Notes

The original poster expresses a need for clarity and step-by-step guidance, indicating a potential lack of familiarity with the algebraic manipulation involved. There are also hints of external pressures, such as upcoming resits, influencing the urgency of the discussion.

franky2727
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first of all forgive me for the writing of this as i don't have any programs to properly write it up, shouldn't be some hard for some of you wizz kids out there just don't see how to get from one line to the next to help would be much apprichiated (y - 1)/(y+2) = +or- ke3/4x4 then the next line is y=(2Ke3/4x4 +1)/1-Ke3/4x4

where k is an unknown constant, thanks again in advance for a step by step guide through
 
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If you write
[tex]\frac{y - 1}{y + 2} = \frac{y + 2 - 3}{y + 2} = \frac{y + 2}{y + 2} - \frac{3}{y + 2}[/tex]
there's only one y left.

And don't worry, it's perfectly readable.
 
dont i then get similar problems getting the y back with that 1 hanging arround, like when i multiply through? thanks for the help so far tho i can see where its going now just need another little push lol
 
You don't multiply through, you invert.
If you have
[tex]\frac{a}{b} = c = \frac{c}{1}[/tex]
then also
[tex]\frac{b}{a} = \frac{1}{c}[/tex]
(provided neither a nor c is zero).

Of course, c needn't be just a number, it could also be, say, [itex]c = 1 + k e^{-3/4 x^4}[/itex], if you get what I mean :-p
 
chears bud i owe u one dam resits doing my head in making enough at poker to earn like £100k a year but want a dam degree incase the industry goes tits up
 

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