# 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.

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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