Let no. of each animal be, d, c, m
:
Total animal equation
d + c + m = 100
Total cost equation
15d + 1c + .25m = 100
:
We got three unknowns and two equations, but we know the solutions are integers
Eliminate the cats
15d + 1c + .25m = 100
1d + 1c + 1m = 100
---------------------- Subtraction eliminates c
14d - .75m = 0
or
.75m = 14d
m = 14%2F.75d
substitute values for d, until you get an integer value for m
when you get to d = 3
m = 14%2F.75*3
m = 42%2F.75
m = 56 mice
:
We have 3 dogs and 56 mice, then 100 - 59 = 41 cats
:
:
Check that in the cost equation
15(3) + 1(41) .25(56) =
45 + 41 + 14 = 100
So 3 dogs 41 cats and 56 mice