Of 29 students, 11 have cats, 14 have dogs, and 7 have no pets. How many
students have 2 pets?

There are three methods.  Here are all three:

1. Venn diagram method:  Draw this with the two sets C and D overlapping

_____________________
|   ________         |
| C|     ___|_____   |
|  |____|___|     |D |
|       |_________|  |
|____________________|

The rectangle C contains the 11 students with cats
The rectangle D contains the 14 students with dogs
The overlaping part contains the students with both cats and dogs

This is the required answer so we put x in the overlapping part

_____________________
|   ________         |
| C|     ___|_____   |
|  |    | x |     |D |
|   ¯¯¯¯|¯¯¯      |  |
|        ¯¯¯¯¯¯¯¯¯   |
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
Since rectangle C must contain 11 students, the part of C which does not
overlap with D contains 11-x students, so write 11-x in that part of C

_____________________
|   ________         |
| C|11-x ___|_____   |
|  |    | x |     |D |
|   ¯¯¯¯|¯¯¯      |  |
|        ¯¯¯¯¯¯¯¯¯   |
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

Since rectangle D must contain 14 students, the part of D which does not
overlap with C contains 14-x students, so write 14-x in that part of C

_____________________
|   ________         |
| C|11-x ___|_____   |
|  |    | x |     |D |
|   ¯¯¯¯|¯¯¯ 14-x |  |
|        ¯¯¯¯¯¯¯¯¯   |
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
The part of the large rectangle which is outside the smaller rectangles
   contains the number with neither cats nor dogs.  This is given to be 7,
   so we put 7 in that part:

_____________________
|   ________         |
| C|11-x ___|_____   |
|  |    | x |     |D |
|   ¯¯¯¯|¯¯¯ 14-x |  |
| 7      ¯¯¯¯¯¯¯¯¯   |
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
 
Now since there are 29 students, we add up all the expressions in the various
parts and equate this to 29:

(11-x) + x + (14-x) + 7 = 29

Solve that equation and get x = 3

---------------------------------------------------------------

2. Set formula method
                               
N(C OR D) = N(C) + N(D) - N(C AND D)

Your book may have "union" for "OR" and "intersection" for "AND"

We are looking for N(C and D) so this will be x

N(C OR D) = N(C) + N(D) - x

We are given N(C) = 11 and N(D) = 14

N(C or D) = 11 + 14 - x

All of the 29 have cats or dogs except for the 7 which have no pets,
so since 29-7 = 22 we substitute 22 for N(C or D)

22 = 11 + 14 - x

Solve this and get x = 3

--------------------------------------------------------

3. System of equations method.

Let x = the number of students with cats and no dogs.
Let y = the number of students with dogs and no cats.
Let z = the number of students with both dogs and cats. This will be
         the desired answer
Given 7 = the number of students with no dogs and no cats. 

There are 29 students, so

x + y + z + 7 = 29 or, subtracting 7 from both sides

                x + y + z = 22

x and z have cats, so

                    x + z = 11

y and z both have dogs, so

                    y + z = 14

So we have the system of three equations and three unknowns: 

                x + y + z = 22
                x     + z = 11
                    y + z = 14


x = 8, y = 11, z = 3

So z = 3 = the answer.

Edwin
AnlytcPhil@aol.com