Given a plane: $$x - y + z = 0$$
And a cylinder $$x^2 + y^2 = 2$$
Why can't I get the intersection of the two by equaling both equations? i.e. $$x^2 + y^2 - 2 = x - y + z \implies x^2+y^2 -x + y - z -2=0$$
Instead, I have to isolate $x$ for example in eq(1) and substitute in eq(2).
On
The intersection is an ellipse in the space and the equation is defined by the two curves, that is
As an alternative representation, we can use
with $t\in[0,2\pi)$.
On
To elaborate on why the technique you used in general doesn't work, it is because you introduce a new degree of freedom.
More precisely, given two equations $$f(x,y,z)=0,$$ and $$g(x,y,z)=0,$$ if you consider the equation $$f(x,y,z)=g(x,y,z),$$ the solutions to this new equation should contain the entire intersection of the two original surfaces, but it will also contain the imtersection of the surfaces $$f(x,y,z)=1,$$ and $$g(x,y,z)=1.$$ Or, in other words, the equation $$f(x,y,z)=g(x,y,z)$$ can have solutions where $$f(x,y,z)\neq 0,$$ but these solutions violate the original equations.
On
I think that you have been badly misled by sloppy teaching at some level, probably high-school. You should never expect that you can find the common solution(s) of two equations by setting them equal, except in some rather special situations. Let me go into some detail:
One case where your strategy will work is when you have two equations, $y=f(x)$ and $y=g(x)$, and want to know where the two graphs intersect. Then you may (and should) write $f(x)=g(x)$ and solve the single equation $f(x)-g(x)=0$ for $x$. The reason this works is that by setting $f=g$, you have in effect eliminated the letter $y$ from the picture, and can now use one-variable methods to come to the desired answer.
Your particular problem presented you with a three-variable situation, and, as @B.Goddard and @gimusi have said in their answers, the simple methods applicable to two-variable problems have to be handled rather differently.
But let me come back to the two-variable case, where you expect finitely many common points to two planar curves. There, setting $f=g$ in the case that you were intersecting two graphs was just a neat and easy trick for eliminating $y$ from the situation, and in the two-variable situation, you should always base your strategy on eliminating one of the variables. Take an interesting case like intersecting a parabola with a circle, like $y=x^2+x-2$ and $x^2+y^2=4$. Here you easily eliminate $y$ by substituting its expression in the first equation for the appearance(s) of $y$ in the second, to get \begin{align} x^2+(x^2+x-2)^2&=4\\ x^2+x^4+2x^3-3x^2-4x+4&=4\\ x^4+2x^3-2x^2-4x&=0\\ x(x+2)(x^2-2)&=0\,, \end{align} to give the four(*) points $(-2,0)$, $(-\sqrt2,-\sqrt2)$, $(0,-2)$, and $(-\sqrt2,-\sqrt2)$ of the intersection.
In cases where neither equation is already solved for $x$ nor $y$, you may have to try to solve one by your own means. If the equations are of higher degree than $2$, this can get tricky.
(*) According to a wonderful theorem of Bézout, any two different conic sections will have exactly four points in their intersection, as long as you count properly.
On
Eliminating $x$ we get $(y-z)$ orthogonal plane projection ( and eliminating $y$ gives $(x-y)$ plane projection) etc.
A parameter can be chosen to represent the three coordinates of the curve of intersection by eliminating or isolating one variable.
But by directly adding the two equations together in linear combinations as in an example
$$ \lambda(x^2+y^2) + (\lambda-1) (x - y + z +2) =0 $$
we get a set or family of second degree conicoids/quadrics passing through their common intersection curve ( ellipse in this case). For equal weightage as in the case $\lambda=\frac12$ you mention, it is a paraboloid.
Generally, a curve in 3-space can't be expressed with one equation. Try at 2-D example. Take the two equations
$$x^2+y^2 =1$$
and
$$x=0$$.
The intersection is just the two points $(0,\pm 1)$. But if we do what you did above, we have
$$x^2+y^2=1 =x$$
which has infinitely many solutions.