Solve for 3 variables with 2 equations

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In summary: You could try substituting a b c for u v w so that a = constant, a2 + b2 + c2 = 1. However, this won't solve the equation.
  • #1
staples
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Homework Statement


Solve the system of equations for x, y, and z.
-------------------------
[tex]4x^4+8y^4+z^4=9[/tex]
[tex]x^8+y^8+z^8=1 [/tex]
--------------------------

Homework Equations


The Attempt at a Solution


Both equations can be rearranged and factored (difference of squares). I've tried it and it doesn't seem to help solve the problem. Any hints? Thanks :)
 
Last edited:
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  • #2
welcome to pf!

hi staples! welcome to pf! :smile:

(i assume that's x8, not xy?)

have you tried the obvious substitution? :wink:
 
  • #3
staples said:

Homework Statement


Solve the system of equations for x, y, and z.
-------------------------
[tex]4x^4+8y^4+z^4=9[/tex]
[tex]x^y+y^8+z^8=1 [/tex]
--------------------------
Is the exponent on x in the 2nd equation a typo? Just checking.
staples said:

Homework Equations


The Attempt at a Solution


Both equations can be rearranged and factored (difference of squares). I've tried it and it doesn't seem to help solve the problem. Any hints? Thanks :)
 
  • #4
Sorry about that. That was a typo. It's corrected now.
tiny-tim: I don't see it :frown:
 
  • #5
u = x4 … ?
 
  • #7
So I substituted x^4 = u, y^4 = v, z^4 = w.
I still don't see how it would help. The only thing I could see to isolate one variable in the first equation and substitute that into the second equation. That still leaves us with 2 variables. I don't understand how we can solve for 3 variables with 2 equations. I always thought if you have n variables you need at least n equations. Please explain. Thanks
 
  • #8
staples said:
So I substituted x^4 = u, y^4 = v, z^4 = w.
I still don't see how it would help. The only thing I could see to isolate one variable in the first equation and substitute that into the second equation. That still leaves us with 2 variables. I don't understand how we can solve for 3 variables with 2 equations. I always thought if you have n variables you need at least n equations. Please explain. Thanks
What does each of the resulting equations describe?

[itex]4u+8v+w=9[/itex]

[itex]u^2+v^2+w^2=1[/itex]

How would you describe their intersection, if they intersect?
 
  • #9
If you have n variables you need at least n equations to get a unique solution (a single point). Since you have only two equations in three variables, there won't be a unique solution. This means that the solution set will be all the points along some curve, assuming the two equations represent surfaces that intersect.
 
  • #10
So like Marks says, the intersection will be infinite over a certain interval(s)? I don't understand what you mean by what they represent? 3 dimensional planes/surfaces? :uhh: Thanks.
 
  • #11
In (u, v, w) space, the two equations are surfaces that might or might not intersect. If they intersect, they will do so at an infinite number of points - all the points along the curve of intersection.
 
  • #12
I think I get it now. Divide the first equation by 9 and compare it to the second equation. This shows that u,v,w must be equal to their coefficients because of the square in the second equation. So u=4/9, v=8/9, w=1/9. Now we take the (positive and negative) fourth root of these values. So there are 8 solutions (WolframAlpha: http://www.wolframalpha.com/input/?i=%284x^4%2B8y^4%2Bz^4%3D9%29+%28x^8%2By^8%2Bz^8%3D1%29)

Is this correct? Can I just "compare" the two equations like I did? Thanks.

P.S. Mark: I think you're under the impression that I'm doing higher level math courses. I am only in high school and we've never worked with 3d coordinate systems...
 
  • #13
staples said:
I don't understand what you mean by what they represent? 3 dimensional planes/surfaces? :uhh: Thanks.

the first equation represents a plane, the second represents a sphere :smile:

(so what will be the shape of the points where they intersect?)
 
  • #14
Normally, you cannot solve two equations for specific values of three unknown numbers but here you can. It is crucial to this problem that [itex]8^2+ 4^2+ 1^2= 64+ 16+ 1= 81= 9^2[/itex]
 
  • #15
tiny-tim said:
the first equation represents a plane, the second represents a sphere :smile:

(so what will be the shape of the points where they intersect?)

I'm guessing the answer is a circle or ellipse depending on how they intersect. But I highly doubt it that the teacher wants an answer using the geometry of the equations. After all we've never learned the equation of a sphere. I'm pretty sure he's looking for an algebraic approach...

I'm going to look into what you said HallsofIvy. So I guess, my "solution" is wrong? A less subtle hint would be great :) I appreciate everyone's effort to help me. Thanks.
 
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  • #16
staples said:
I'm guessing the answer is a circle or ellipse depending on how they intersect.

a plane always intersects a sphere in a circle :smile:
But I highly doubt it that the teacher wants an answer using the geometry of the equations. I'm pretty sure he's looking for an algebraic approach...

yes, but knowledge of the geometry points you towards an easy algebraic approach :wink:

try substituting a b c for u v w so that a = constant, a2 + b2 + c2 = 1 :smile:
 
  • #17
tiny-tim said:
a plane always intersects a sphere in a circle :smile:


yes, but knowledge of the geometry points you towards an easy algebraic approach :wink:

try substituting a b c for u v w so that a = constant, a2 + b2 + c2 = 1 :smile:

|a|,|b|,|c| ≤ 1 ?
 
  • #18
Can someone please help me with this question? Thanks.
 
  • #19
staples said:
|a|,|b|,|c| ≤ 1 ?

on the intersection, yes

conversion from u v and w axes to a b and c axes is just a rotation of the axes

(eg if instead of the usual x and y axes you use p = (x+y)/√2 and q = (x-y)/√2, you've simply rotated the axes by 45°, and x2 + y2 = 1 becomes p2 + q2 = 1)

here's another geometrical snippet that may help …

4u + 8v + w = 9 is a plane whose normal is the direction (4,8,1) …

if we call the endpoints of the diameter of the sphere parallel to (4,8,1) the north and south poles, then the solutions (in u,v,w coordinates) are a circle of latitude on the sphere :wink:
 
  • #20
tiny-tim said:
a plane always intersects a sphere in a circle :smile:
...

... unless the plane is tangent to the sphere.
 
  • #21
SammyS said:
... unless the plane is tangent to the sphere.
In which case it's a teeny tiny sphere. :tongue:
 
  • #22
the plane(in question) will touch the sphere at only one point.
 

What is the concept of solving for 3 variables with 2 equations?

Solving for 3 variables with 2 equations involves using algebraic methods to find the values of three unknown variables when given two equations. This allows for the determination of a unique solution for the system of equations.

What are the requirements for solving for 3 variables with 2 equations?

In order to solve for 3 variables with 2 equations, the equations must be independent, meaning they are not multiples or rearrangements of each other. Additionally, the number of variables must equal the number of equations for a unique solution.

What is the process for solving for 3 variables with 2 equations?

The first step is to rearrange the equations to put them into standard form, with the variables on one side and the constants on the other. Then, use substitution or elimination methods to solve for one variable in terms of the other two. Finally, substitute the solved variable into either of the original equations to find the remaining variables.

What happens if there is no solution when solving for 3 variables with 2 equations?

If there is no solution, it means that the system of equations is inconsistent, meaning the equations are parallel or the same line. This indicates that there is no point of intersection and therefore no solution exists.

Can you solve for 3 variables with 2 equations using matrices?

Yes, it is possible to solve for 3 variables with 2 equations using matrices. This method involves setting up a matrix with the coefficients of each variable, and then using matrix operations to solve for the values of the variables. This method is particularly useful for larger systems of equations.

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