How should I solve this kind of set theory problem? The exercise is the following:

In a group of $30$ people, 20 people speak French, 22 people speak German, and 25 people speak English. We assume everyone speaks at least one language. What could the number of people speaking all three languages be?

So using set theory notation, we could say:
$$ |U| = 30 \quad |F| = 20 \quad |G| = 22 \quad |E| = 25 \\ |F\cap G\cap E|=?$$
All I can think of is start guessing; For example, I would say that let us see what happens if we say that $20$ people learn all three languages. This way, there must be $2$ more people distributed somehow in $G$ and $5$ more in $E$. No matter where I put these remaining people (drawing a Venn-diagram helped me), the total number of people can't be $30$ this way. So I could now start going downwards, what if $19$, $18$,$\dots$ people speak all three...
I don't think that this is a very elegant solution. How would you do it?
 A: To solve for at most you take the smallest set which is the set of people how speak French which is $20$.
To solve for at least you add the three set together and subtract twice the amount of the set of all people $20+22+25-2*30 = 7$ , so to conclude.
note : if you have say $k$ set, then you add them together and subtract $k-1$ the cardinal of the united set and if the result is less or equal to $0$ then we say that at least $0$ people speak all language
... and if its a number bigger than zero then at least $x$ people speak all languages.   
there is at least 7 people and at most 20 how speak all languages. 
A: Denote the number of people speaking solely English by $E$, the number of people speaking French and German, but not English, by $E'$, and the number of people speaking all three languages by $x$. Looking at a Venn diagram we obtain the four equations
$$\eqalign{
x+G'+F'\quad &+E=25\cr
x+F'+E'\quad&+G=22\cr
x+G'+E'\quad&+F=20\cr
x+G'+F'+E'\quad&+E+F+G=30\ .\cr}$$
Solving for $x$ and the $\>'$-variables gives
$$x=7+E+F+G\tag{1}$$ and  $$G'=8-E-F,\quad F'=10-E-G,\quad E'=5-F-G\ .$$
It is therefore necessary that
$$E+F\leq8,\quad E+G\leq10,\quad F+G\leq5\ .\tag{2}$$
Adding these inequalities gives the condition $2(E+F+G)\leq23$, or $E+F+G\leq11$. The values $E=6$, $F=2$, and $G=3$ satisfy $(2)$ and give $E+F+G=11$; furthermore any smaller nonnegative $E$, $F$, $G$ satisfy $(2)$ as well. Looking at $(1)$ we see that the possible values for $x$ are $7\leq x\leq18$.
