Venn diagram problem with three beverages liking 
In a group of people who like beverages tea, coffee and milk, 9 people like tea, 9 people like coffee and 15 people like milk. If 5 people like tea and milk, 7 people like coffee and milk and 3 people like all three beverages. How many people like tea and coffee but not milk? The total number of people is 20.
To my understanding, there are 15 people who like milk. These 15 includes 3 that likes all the three. Since 5 likes both tea and milk, this is included with the 3 that likes all the three beverages. So, 2 likes both tea and milk only as per the Venn diagram. Similarly, 4 likes only coffee and milk. Going by this, I tried to find those that like only tea and coffee (x) which is 9-5-x under tea side (refer attached diagram). The x should be the same as 9-7-x for coffee. The equation 9-5-x=9-7-x is not solvable. I know I am wrong somewhere. Help appreciated.
 A: Illustrating the ideas already indicated by the comments:
Consider the following chart:
\begin{array}{| r | r | r | r | r |}
  \hline                       
\text{Variable} & \text{Tea} & \text{Milk}  
& \text{Coffee} & \text{Value} \\
\hline 
v_1 & t & t & t & 3 \\ \hline
v_2 & t & t & f & 2 \\ \hline
v_3 & t & f & t &   \\ \hline
v_4 & t & f & f &   \\ \hline
v_5 & f & t & t & 4 \\ \hline
v_6 & f & t & f & 6 \\ \hline
v_7 & f & f & t &   \\ \hline
v_8 & f & f & f &   \\ \hline
\end{array}
From the constraints: 
$v_1 + v_2 + v_3 + v_4 = 9$. 
$v_1 + v_3 + v_5 + v_7 = 9$. 
$v_1 + v_2 + v_5 + v_6 = 15$. 
$v_1 + v_2 = 5$. 
$v_1 + v_5 = 7$. 
$v_1 = 3$.
It is required to solve for $v_3$.
From the above constraints:

*

*As shown in the chart, $v_1, v_2, v_5, v_6$ are now all known.

*$v_3 + v_4 = 4.$

*$v_3 + v_7 = 2.$

*Based on the added constraint that the total number of people is $20$, you have that $v_7 + v_8 = 1$ and $v_4 + v_8 = 3.$
The constraint that $v_7 + v_8 = 1$ clearly has two possibilities, that lead to two separate solutions for $v_3$.
Either $v_7 = 1 \implies v_3 = 1$ 
or $v_7 = 0 \implies v_3 = 2.$
