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Solving for a variable in terms of two other variables

  1. Aug 23, 2012 #1
    solve the following equation for z in terms of x and y
    3[itex]y^{2}[/itex]+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. jcsd
  3. Aug 23, 2012 #2

    Mute

    User Avatar
    Homework Helper

    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).
     
  4. Aug 23, 2012 #3

    Mark44

    Staff: Mentor

    Show us what you tried.
     
  5. Aug 23, 2012 #4

    Mark44

    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.
     
  6. Aug 23, 2012 #5
    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
     
  7. Aug 23, 2012 #6
    yes that is what i meant
     
  8. Aug 23, 2012 #7

    Mark44

    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.
     
  9. Aug 23, 2012 #8
    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
     
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