SOLUTION: In a dog park, there are 84 total legs. If there are 3 more dogs in the park than people, how many dogs are in the park?

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Question 1141747: In a dog park, there are 84 total legs. If there are 3 more dogs in the park than people, how many dogs are in the park?
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52784)   (Show Source): You can put this solution on YOUR website!
.
    4d + 2p = 84     (1)     (counting legs)

     d -  p =  3     (2)     (counting heads)


where "d" is the number of dogs and "p" is the number of people.


From eq(2) express p = d - 3  and substitute it into eq(1). You will get


    4d + 2*(d-3) = 84

    6d - 6 = 84

    6d = 84 + 6 = 90

    d = 90/6 = 15.


ANSWER.  15 dogs.


Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!


(1) Formally, using algebra....

The thing we are to find is the number of dogs, so let that be our variable:

let x = # of dogs
then x-3 = # of people

The total number of legs -- 4 per dog and 2 per person -- is 84:






The number of dogs is 15.

(2) Informally, using logical reasoning and mental arithmetic....

(a) Add 3 more people, making the numbers of people and dogs the same. That brings the total number of legs to 84+6 = 90.
(b) Each dog-person pair has a total of 6 legs.
(c) Therefore, the number of dogs in the park is 90/6 = 15.
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