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Solving for a variable in terms of two other variables

  • Thread starter iancurtis
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  • #1
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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

Homework Statement





Homework Equations





The Attempt at a Solution

 
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Answers and Replies

  • #2
Mute
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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
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solve the following equation for z in terms of z
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
Show us what you tried.
 
  • #4
33,075
4,779
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).
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."
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).
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
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Show us what you tried.
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
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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."
yes that is what i meant
 
  • #7
33,075
4,779
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
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.
3y^2+2yz-5z-2((3y2+2yz-5z-2x=0)/2)=0
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
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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.
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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