Recent content by staples

Can someone please help me with this question? Thanks.

|a|,|b|,|c| ≤ 1 ?

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...

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...

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? :rolleyes: Thanks.

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...

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

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...

Thanks :)

Homework Statement Solve the equation (x & y are integers): (x^3+4)(xy^2-x^2y+3y^2-12)=x^6 Homework Equations The Attempt at a Solution xy^2-x^2y+3y^2-12=\frac{x^6}{x^3+4} \\ xy^2-x^2y+3y^2-12=x^3-4 + \frac{16}{x^3+4} \\ 16 \geq x^3+4 \\ x^3 \leq 12 That's all I can...