# Multiple variables with multiple solutions

I'm not sure how best to explain this so please ask questions to help clarify. I'm trying to find a solution to the following formula, where I know there are multiple "correct" solutions.

Essentially I have three materials, each having a different density. They need to be mixed in some proportion to create the mean density provided.

Formula for density is: $d = m/v$, or $density = {mass \over volume}$

$x$ has a mass of $8g/cm^3$
$y$ has a mass of $3g/cm^3$
$z$ has a mass of $1g/cm^3$, therefore:

$m = 8x + 3y + 1z$
$v = x + y + z$
$d = 5.515g/cm^3$

$5.515 = {8x + 3y + 1z \over x + y + z}$

How do I go about solving this? I can approximate it by plugging into values randomly ($x=3, y=1, z=1$), but is there a more exacting process?

You can parametrize the solution as:

$$\left(x,y,\frac{1}{5.515-1}[(8-5.515)x+(3-5.515)y]\right)$$

where all I've done is just rearrange your equation and solve for $z$ in terms of $x,y$. In other words, any choice of $x,y$ above will give you a solution. Presumably since $x,y,z$ refer to volume, they must be non-negative for any of this to make sense.

• Thanks for that Alex. That makes a little sense, rearranging it like that. The part I'm unclear of is what I do with the $x, y,$ at the start? Are they just multiplied in? Plugging in values of $x=3, y=1$, how might it work? (sorry if this is a really basic question) – Carl Sep 29 '12 at 19:32
• @Carl: You have one equation for 3 unknowns, which effectively means that you can solve for at most one of the unknowns in terms of the other two. In other words, by solving for $z$ in terms of $x,y$, we are guaranteeing that for any choice of $x,y$, the triple I've written above shall satisfy your density equation. So, by plugging in $x=3,y=1$, you will get a value for $z$ as above, and then the triple $(x,y,z)$ shall be a solution to your equation. Again, it may be worth considering that $x,y,z\geq 0$ should be a restriction, so for some $x,y\geq 0$ you might get $z<0$. – Alex R. Sep 29 '12 at 20:22
• Thanks for the further explanation. That makes sense now. Yes, $x, y, z \ge 0$. I guess rearranging the equation to solve for $x$ or $y$ would be tricky? – Carl Sep 29 '12 at 20:35

You can add another condition, maybe something like minimizing the price of the materials, then it restricts the solutions. If nothing else matters, pick randomly.

• Hi Berci. Thanks for that idea too. I would love to add in another condition, but in this particular instance, I can't. Otherwise it might certainly make it easier to restrict the values of $x, y, z$. – Carl Sep 29 '12 at 20:49