# Easier solution for system of equations.

I was given this task and I solved it, but only after solving for $x^4+x^3$... $$x^2+y^2+4z^2=6y-4\quad \& \quad 2xy-4xz+4yz=y^2+5$$

How I tried: 1. I summed (and multiplying the second one with $-1$) the two equations, getting: $$(x-y+z)^2=-(y-3)^2 \implies y-3=0 \quad \& \quad x-y+z=0$$

I than put $y$ into the equations but in the end I got something long with $x^4$...

I know this task is not ment to solve polynomials, so I am asking for a simpler solution.

• If you know $y=3$ and $x=-z+3$ then isn't this a simple quadratic? – lulu Apr 19 '18 at 20:37
• @lulu it was actually $x = 3 - 2 z,$ the error might have made things complicated – Will Jagy Apr 19 '18 at 22:00

You need to be more careful. The two surfaces are $$x^2 + (y-3)^2 + 4 z^2 = 5 \; ,$$ $$x^2 - (x-y+2z)^2 + 4 z^2 = 5 \; .$$ The first is an ellipsoid centered at $(0,3,0),$ the second is a hyperboloid of one sheet centered at the origin, but not a surface of revolution.
You get $$(y-3)^2 + (x-y+2z)^2 = 0.$$ Then for real points, $y=3$ and $x = 3 - 2z.$ the ellipsoid gives $8 z^2 - 12 z + 4 = 0$ or $4 (2z-1)(z-1) = 0.$ The two points where the surfaces touch are $$(1,3,1)$$ $$\left(2,3,\frac{1}{2} \right)$$ The surfaces must be tangent at these points, so the gradients of the two defining functions should be parallel.
From the original way you wrote them the gradients are $$(2x,2y-6,8z) \; ,$$ $$(2y-4z,2x-2y+4z, -4x+4y)$$
At $(1,3,1)$ the gradients are $$(2,0,8), \; \; \; (2, 0, 8) \; \; .$$
At $\left(2,3,\frac{1}{2} \right)$ the gradients are $$(4,0,4), \; \; \; (4, 0, 4) \; \; .$$