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Answer to Question #2685 in Algebra for Bonnie Morales

Question #2685
Kendra photographed the Dogs-n-Suds annual car wash. She photographed 20 dogs and owners in all. If there was a total of 64 legs, how many dogs and owners are there?
Expert's answer
Assume that every owner has 2 legs, each dog has 4 legs.
Let X be the total number of owners and Y be the total number of dogs.
Then
& X+ Y = 20
On the other hand the total number of legs is equal to
4 X + 2 Y = 64

We have to resolve this system of equations:
X+ Y = 20
4 X + 2 Y = 64

From the first equation:
& X=20-Y
Substituting to the second one we get:
4(20-Y)+2Y = 64
80 – 4Y + 2Y = 64
2Y = 24
Y=12,
Whence X=20-12=8

Thus there were 8 owners and 12 dogs.

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