# Finding all the critical points of a system of equations

How can I find all the critical points of a system such as:

$x' = x(3-x-y)$

$y' = y(1-3x+y)$

or

$x' = x-4y+2xy$

$y' = x+3y-4xy$

I know I'm supposed to set them equation to 0, but I can't seem to always find them all. Is there a general method I can follow that will help me obtain all the critical points? If not, are there any tips that will lead me in the right direction?

For example, if we add the second set of equations together we get

$2x-y-2xy = 0$

$2x-y(1+2x) = 0$

$2x = y(1+2x)$

Doesn't this show that there are an infinite number of critical points, then?

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How can I find all solutions to the system of linear equations (1) $2x+y=2$, (2) $x-2y=1$? If we add the two equations together, we get $3x-y=3$. Doesn't this show that there are infinitely many solutions, then? – user1551 Nov 27 '12 at 12:28

For your systems, each system happens to have at least one equation which is linear in one of the two variables. In such a case you can solve the one equation for one of the two variables, and substitute the result in the other equation. Then you have two equations, one of which has one of the original variables, say $y$, on the left and does not have $y$ on the right, and the other equation is all in terms of the other variable $x$. Then if you can solve the one variable equation for $x$ these results may be put into the "solved for $y$" equation to get the corresponding $y$ values for each $x$ obtained.
Added note: when the expression in the "solved for" variable has a denominator, one has to separately consider the case in which that denominator is zero. For example solving the first equation of your first system for $y$ gives $y=(3x-x^2)/x$, so that the possibility $x=0$ has to be treated separately, and then the substitution may be used as above suggested, in this case (once $x=0$ already treated) the substitution is $y=3-x$.
How is the possibility that $x = 0$ treated separately? Doesn't $x = 0$ make the equation undefined? – user1038665 Nov 27 '12 at 20:59
In the first system, if $x=0$ then the first equation is true, and putting $x=0$ into the second equation makes it $y(1+y)=0$ which has solutions $y=0,y=-1$. So the $x=0$ possibility gives the solutions $(x,y)=(0,0)$ and $(x,y)=(0,-1)$. These are only two of the critical points of the first system. For the others, let $y=3-x$ in the second equation of the first system, and then solve for $x$ in that result. After that, these $x$ values go with corresponding $y=3-x$ to get the rest of the pairs $(x,y)$. – coffeemath Nov 27 '12 at 21:45
You should check this, but for your first system I got, along with the points $(0,0),(0,-1)$ mentioned in my last comment, also the two points $(1,2),(3,0)$ by letting $y=3-x$ as suggested. – coffeemath Nov 27 '12 at 21:49