The animal rescue shelter placed 134 cats and dogs for adoption. The adoption fee is $15 for a cat and $25 for
The animal rescue shelter placed 134 cats and dogs for adoption. The adoption fee is $15 for a cat and $25 for a dog.
If the shelter collected $2650 in adoption fees, how may cats and how many dogs were placed?
70 cats and 64 dogs
64 cats and 70 dogs
72 cats and 62 dogs
68 cats and 66 dogs
74 cats and 60 dogs
Public Comments
1. 70 cats and 64 dogs.2. x+y=134
15x+25y=2650
-15x-15y=-2010
15x+25y=2650
10y=640
y=64
64+x=134
x=70
so 70 cats and 64 dogs.
3. Let c be the number of cats placed, and
let d be the number of dogs placed.
Because there were 134 cats and dogs placed, the first equation is
c + d = 134
Because the adoption fee is $15 for a cat and $25 for a dog and the shelter collected $2650, the second equation becomes
15c + 25d = 2650
We can now solve these two equations by any convenient metnod, the substitution method of which seems easier but the addition (or elimination) method is not so hard either. Let's eliminate the cats from this set of equations by solving the first equation for the number of cats, namely,
c = 134 - d
and plugging in the expression on the right everywhere c appears in the second equation. Thus, we have one equation in one unknown:
15(134 - d) + 25d = 2650
After simplifying the left side and some simple subtracts and divides, we get
d = 64 dogs
Now that we know the number of dogs, we can substitute this back into c = 134 - d to get 70 cats.
