# Solving for a variable in terms of two other variables

1. Aug 23, 2012

### iancurtis

solve the following equation for z in terms of x and y
3$y^{2}$+2yz-5z-2x=0
i've spent a lot of time on this question, but i keep getting something along the lines of 0=0
i tried solving for x in temrs of y and z to replace the x but i get 0=0
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

Last edited: Aug 23, 2012
2. Aug 23, 2012

### Mute

You have one equation with three variables. The best you are going to be able to do is solve for z and write z = f(x,y). You could alternatively solve for x = g(y,z) or y = h(x,z), but all "three" of these solutions contain the exact same information, so if you tried to plug, say, x = g(y,z), into f(x,y), you would get z = f(g(y,z),y) = z, because all three equations came from the same equation.

Typically, if you have n variables, you need n independent equations to solve for each variable. If you have n variables but only m < n independent equations, you are only going to be able to solve for m of the variables in terms of the other n-m variables. (Independent equations means that no two equations can be rearranged into the same equation up to an overall constant).

3. Aug 23, 2012

### Staff: Mentor

Show us what you tried.

4. Aug 23, 2012

### Staff: Mentor

This is exactly what the OP is supposed to do, although the problem statement is garbled. You can't "solve for z in terms of z" other than to say that z = z, which is trivally true. Presumably the OP meant to say "solve for z in terms of x and y."
Under certain circumstances, including the one presented in this problem, you can solve algebraically for one variable in terms of the others. What you're talking about is solving a system of n equations in n variables for a possibly unique solution for the variables. That is NOT what this problem is about.

5. Aug 23, 2012

### iancurtis

well first i tried to solve for x and substitute, as stated in the original post, which gave me 3y^2+2yz-5z-2((3y2+2yz-5z-2x=0)/2)=0 which ends up being 3y^2+2yz-5z-(3y^2+2yz-5z)=0 which is 0=0. and after that i started over and tried setting it up as 3y^2-2x=5z-2yz and dividing both sides by yz which gave me (3y)/z -2x/yz = 5/y - 2 and after that i pretty much just tried anything that came to mind, but it all just kept taking me in circles

6. Aug 23, 2012

### iancurtis

yes that is what i meant

7. Aug 23, 2012

### Staff: Mentor

First off, you are not asked to solve for x -- you're asked to solve for z.
Second, it looks like instead of substituting for x, you replaced x by the entire equation. That makes no sense.
In the above, you replaced x by (3y2+2yz-5z-2x=0)/2. You can't do that. x is not "equal" to an equation.
For your equation, get the two variables with z in them on one side of the equation, and all other terms on the other sided.
Then factor out z from the terms that have z in them.

8. Aug 23, 2012

### iancurtis

wow i feel so stupid that never occured to me.
and as for the first part, i was trying to get the x out of the equation completely by solving for x and plugging it in so that i would get z in terms of just y. which makes less sense now that i think about it. (also i didnt meant to put the "=0" in there i was copy and pasting to save time.)
but anyways thanks man i appreciate it