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Im really stuck here, and feel quite dumb, because it looks so simple. However, I'm sure I'm overlooking something here.

I have given this equation:

$x=c_0y + c_1z$ where $x,y,z$ are some variables and $c_0,c_1$ are constants.

My goal is to get an equivalent equation in the form of $x-y = ?$

I went ahead and tried moving the components around, but whatever I do, I dont get it in the desired form:

$x - c_0y = c_1z$

$\frac{x}{c_0} - y = \frac{c_1z}{c_0}$

...

The constants just wont get to the other side... I just cant get my head around this. Any help or hint is very appreciated! Thanks in advance!

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    Oh, sorry I forgot to mention that I want x and$y$only to appear on the left side.2012-06-27

2 Answers 2

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Starting with

$x=c_0 y +c_1 z \; , \; c_0 \neq 1$

the best you can do is

$x-y=(C_0-1)y+c_1 z$

There is no way to obtain $x-y$ isolated in one side only (unless obviously $c_0=1$)

Think about it more concretely, with

$x=2y+az$

$x=-4y+az$

$x=\pi y+az$

and similar examples.

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    Thanks a lot. After your answer, I went back to the problem (a linear regression model) and realized that indeed $c_0$ **is** 1. $x$ is growing *exactly* the same amount as $y$ does... this is why I wanted to bring it into the form in the first place (I wanted $x-y$ to be some $\Delta x$). It was there the whole time, and I just didnt see it. Thanks a lot again!2012-06-28
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There is no equivalent equation of the form $x-y=$ something that doesn't involve either $x$ or $y$ (unless $c_0=1)$. If there were, without $z$ you would have $x=c_0y+c_1$, could find $x-y=c_2$ and could solve the one equation for two variables to get $y=\frac {c_2-c_1}{c_0-1}$

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    Thanks for your answer! I have solved my problem now, see comment to Peter Tamaroffs answer.2012-06-28