The Heads and Legs Problem Using Counting Boards

Children and dogs are playing in the park. There are 7 heads and 20 legs in the park. How many children and how many dogs?

We will solve this problem using four boards. The first two boards will represent the first equality, and the second two will represent the second equality.


C + D = 7

We call the number of children C and represent it on the board by a hexagon counter. We call the number of dogs D and represent it by the diamond counter. We call these counters “variables”, because we will find their values when we solve the problem. The value on the first board above is C + D, and the value of the second board is 7. We read the two boards together as the equation C + D = 7.


2C + 4D = 20

Children have 2 legs and dogs have 4 legs, and the total number of legs is 20. So the two boards above represent 2C + 4D = 20.

Now we have to get rid of one of the two variables on either the A boards or the B boards. First let’s double the values on the A boards:


2C + 2D = 14

Now we can subtract the values on the A boards from the values on the B boards, and we can get rid of 2C and have only 2D remaining on the left B board (and 20 – 14 = 6 remaining on the right B board):


2D = 6

We have two times the number of dogs equals 6. We divide by 2 by moving the counters one row below on the B boards.


D = 3
C + 3 = 7

So the number of dogs is 3. We return to our first A board and substitute in 3 for the number of dogs, and then we regroup:


C + 3 = 7


C = 4

We subtract 3 from both boards. There are 4 children! Let’s check:

3 dogs plus 4 children = 7 playing in the park
4 children times 2 legs per child = 8 legs
3 dogs times 4 legs per dog = 12 legs
8 legs + 12 legs = 20 legs.

We got it!

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