Solve for 3 variables with 2 equations

[staples]

Homework Statement

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

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 :)

 

[tiny-tim]

welcome to pf!

hi staples! welcome to pf!

(i assume that's x⁸, not x^y?)

have you tried the obvious substitution? ​

 

[Mark44]

Homework Statement

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

Is the exponent on x in the 2nd equation a typo? Just checking.

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 :)

 

[staples]

Sorry about that. That was a typo. It's corrected now.
tiny-tim: I don't see it

 

[tiny-tim]

u = x⁴ … ?

 

[Mark44]

And similar substitutions for y⁴ and z⁴.

 

[staples]

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

 

[SammyS]

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?

$4u+8v+w=9$

$u^2+v^2+w^2=1$

How would you describe their intersection, if they intersect?

 

[Mark44]

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.

 

[staples]

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? Thanks.

 

[Mark44]

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.

 

[staples]

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...

 

[tiny-tim]

I don't understand what you mean by what they represent? 3 dimensional planes/surfaces? Thanks.

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

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

 

[HallsofIvy]

Normally, you cannot solve two equations for specific values of three unknown numbers but here you can. It is crucial to this problem that $8^2+ 4^2+ 1^2= 64+ 16+ 1= 81= 9^2$

 

[staples]

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

(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.

 

[tiny-tim]

I'm guessing the answer is a circle or ellipse depending on how they intersect.

a plane always intersects a sphere in a circle

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 …

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

 

[staples]

a plane always intersects a sphere in a circle


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

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

|a|,|b|,|c| ≤ 1 ?

 

[staples]

Can someone please help me with this question? Thanks.

 

[tiny-tim]

|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 ​

 

[SammyS]

a plane always intersects a sphere in a circle
...

... unless the plane is tangent to the sphere.

 

[Mark44]

... unless the plane is tangent to the sphere.

In which case it's a teeny tiny sphere.

 

[pcm]

the plane(in question) will touch the sphere at only one point.