# 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. Does the method for finding the intersection of 2 single variable functions work for multivariable functions?

$2$ multivariable functions $Q\left(x,y\right)$ and $P\left(x,y\right)$, I was wondering if finding the point of intersection between these $2$ functions is as easy as making $Q\left(x,y\right)=P\left(x,y\right)$ as you would do for most single variable functions.
Does the method for finding the intersection of $2$ single variable functions work for multivariable functions?
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Vicente Macias
If you have two functions of one variable, say $f\left(x\right)$ and $g\left(x\right)$, then you just write $f\left(x\right)=g\left(x\right)$ or, equivalently, $f\left(x\right)-g\left(x\right)=0$ 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: $P\left(x,y\right)=ax+by$, $Q\left(x,y\right)=cx+dy$. If $ax+by=0$, then $x=-\frac{b}{a}y$. Put it in $Q\left(x,y\right)=0$ 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 $P\left(x,y\right)={x}^{2}+{y}^{2}-1=0$ determines a circle in ${\mathbb{R}}^{2}$ and equation $Q\left(x,y\right)=y=0$ determines a line ($x$-axis). Just draw this two figures and you'll see that they intersect exactly in two points $\left(1,0\right)$ and $\left(-1,0\right)$, which are desireable solutions.

Liam Keller
Yes, finding the intersection between $P\left(x,y\right)$ and $Q\left(x,y\right)$ is done by simply setting $P\left(x,y\right)=Q\left(x,y\right)$. Note that it won't always be just a point (or a finite number of points) though!