Does the method for finding the intersection of 2 single variable functions work for multivariable functions? I have $2$ multivariable functions $Q(x,y)$ and $P(x,y)$, I was wondering if finding the point of intersection between these 2 functions is as easy as making $Q(x,y) = P(x,y)$ as you would do for most single variable functions.
I know you would have more steps to do after that as there would still be $x$'s and $y$'s, I have everything on one side equal to $0$ but don't really know where to go after that.
Edit: Probably worth mentioning that both the $Q$ and $P$ functions are paraboloids(?) I thought about it and realised that I would have found the parabolic/hyperbolic curve of intersection(?) and I'm asked to find the maximum of the two functions $P$ and $Q$ using the same $x and $y$.
Another edit:
$$P(x,y) = 2x - \frac{2x^2+y^2}{10^6}\quad\text{and}\quad Q(x,y) = 2y - \frac{4y^2+x^2}{2\times 10^6}.$$
 A: Yes, finding the intersection between $P(x,y)$ and $Q(x,y)$ is done by simply setting $P(x,y) = Q(x,y)$. Note that it won't always be just a point (or a finite number of points) though!
A: This is correct, but the difference is that typically you will have level curves. That is, if you define a function $$f:\mathbb R\times\mathbb R \rightarrow \mathbb R$$
by $$f(x,y) = P(x,y)-Q(x,y)$$ then the sets $$S_c\equiv\left\{(x,y) : f(x,y) = c\right\}$$ are the level curves of $f$ (of course, adjust the domain for your particular functions as necessary).
Your special case is the level curve $S_0$ (the set of points $(x,y)$ in the domain for which $f(x,y)=0$, i.e., $P(x,y)=Q(x,y)$).
Think of the graph of $f$ as a surface. The points of the surface are of the form $(x,y,f(x,y))$, so an equation of the surface is $z=f(x,y)$. The horizontal plane $z=c$ makes a slice through the surface, typically in some curve lying in the surface. The projection or "shadow" of this curve down in the domain is the level curve $S_c$.
Comment: The analogue in the single-variable case is to solve $P(x)=Q(x)$ by setting $f(x) = P(x)-Q(x)$, and you again have "level sets" $S_c=\left\{x : f(x)=c\right\}$; the points in $S_0$ are exactly the points for which $f(x)=0$, i.e., $P(x)=Q(x)$.
A: If you have two functions of one variable, say $f(x)$ and $g(x)$, then you just write $f(x)=g(x)$ or, equivalently, $f(x)-g(x)=0$ and solve this equation. If $f$ and $g$ are linear functions then it's really easy. If they are quadratic polynomials, you know the formulas for its solutions. In fact there are also some formulas for the case when $f$ and $g$ are cubic and quartic polynomials (Cardano and Ferrari formulas). 
Finally, you have the Abel–Ruffini theorem which states that there is no general solution in radicals to polynomial equations of degree five or higher with arbitrary coefficients.
Of course, if $f$ and $g$ are not polynomials, you can use some clever tricks to find a solution, but there is no general recipe.
Now, when you have two variables $x$ and $y$, it's still easy to find a solution for linear equations: $P(x,y)=ax+by$, $Q(x,y)=cx+dy$. If $ax+by=0$, then $x=-\frac{b}{a} y$. Put it in $Q(x,y)=0$ and you get a single variable equation, which you can solve, of course.
If $P$ and $Q$ are not linear functions, then life is not so easy and you may want to use numerical methods to solve this system of equations. 
However, again, if your functions $P$ and $Q$ are not two difficult (polynomials, for example) then you can try to find solutions geometrically. To give you an idea: equation $P(x,y)=x^2+y^2-1=0$ determines a circle in $\mathbb{R}^2$ and equation $Q(x,y)=y=0$ determines a line ($x$-axis). Just draw this two figures and you'll see that they intersect exactly in two points $(1,0)$ and $(-1,0)$, which are desireable solutions.
