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(52782) (Show Source):
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.
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.
