Algebra leg problem!! HELP!!!?
Okay so this is the problem:
You are bored at a party and count 50 legs. There are 4 less dogs in the room than people. Since the room is filled with people and dogs, how many people are in the room?
Please help! I'm lost :O
Update:thanks everyone!
Answers & Comments
Verified answer
y = dogs, x = humans
y = x - 4
4y + 2x = 50
4(x - 4) + 2x = 50
4x - 16 + 2x = 50
6x - 16 = 50
6x = 66
x = 11
y = 11-4
x = 7
2 legs x 11 humans = 22 human legs
4 legs x 7 dogs = 28 dog legs
50 total legs
You counted 50 legs right?Every person has 2 legs.
50 divided by 2 = 25
There are 25 people.
25 - 4 = 21
There are 21 dogs.
So the answer is:
There are 25 people and 21 dogs.
Edit:Oh wait.NM.Ignore this post.
12 people = 24 legs
8 dogs = 32 legs
add em up to get 50 legs.....i just use the guess and check method cause it seems that im a good guesser but there are the answers
no. of people=x
no. of dogs=x-4
--> tot. no. of legs = 2x+4(x-4)=50
--> 2x+4x-16=50
6x=66
==>x=6