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!

  • $\begingroup$ I'd make this a comment if I could and I'm not completely sure of your goal here, but consider what you get if you merely subtract $y$ from each side of your original equation $x = c_{0}y + c_{1}z$ $\endgroup$ – Keith Flower Jun 27 '12 at 23:55
  • $\begingroup$ Oh, sorry I forgot to mention that I want x and y only to appear on the left side. $\endgroup$ – MrNiceGuy Jun 27 '12 at 23:57

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



$$x=\pi y+az$$

and similar examples.

  • $\begingroup$ 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! $\endgroup$ – MrNiceGuy Jun 28 '12 at 0:35

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 $y=\frac {c_2-c_1}{c_0-1}$

  • $\begingroup$ Thanks for your answer! I have solved my problem now, see comment to Peter Tamaroffs answer. $\endgroup$ – MrNiceGuy Jun 28 '12 at 0:36

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