
The boys plus the cats have a total of 22 heads.
Or B + C = 22. Subtracting C from each side of the
preceding results in a value for B of 22  C. The boys' legs equal two times the number of
boys, the cats' legs equal four times the number
of cats, and the total number of legs equals 68.
Or 2B + 4C = 68. Using the value for B already
calculated, the last equation can be rewritten
as 2 (22  C) + 4C = 68, or 44  2C + 4C = 68,
which, by addition and subtraction, becomes 2C
= 24, or C = 12. There are 12 cats and 10 boys in the back yard.
From a puzzle inPuzzles,
Patterns, and Pastimes: From the World of Mathematicsby Charles F. Linn. Garden City: Doubleday,
1969.

