a little help in this maths problem?

Question

a group of dog owners met to parade their dogs. some owners had more than one dog. at the meeting there wasa total of 400 heads and 1300 legs. how many dogs were there?

Answer 1

System of equations
Heads: h + d = 400
Legs: 2h + 4d = 1300

h = 400 - d
Substitute
2(400-d) + 4d = 1300
800 + 2d = 1300
2d = 500
d = 250
250 dogs

Just for the heck of it substitute:
h + 250 = 400
h = 150 humans

Answer 2

There are 250 dogs.
250*4=1000 legs; this leaves
150 people*2 legs=300
250+150=400 heads
1000+300=1300 legs
To figure it out:
I multiplied 400*2, because people have 2 legs. That makes 800 legs. But you need 1300, which is 500 more, or 250 more pairs of legs. So 250 dogs is 1000 legs, and 400 (total) minus the 250 dogs is 150, which is another 300 legs. :)

Answer 3

150 people = 300 legs
250 dogs = 1000 legs
total of 400 heads and 1300 legs

Answer 4

hope this is what you are looking for...

human heads= x
dog heads= 400-x

4(400-x) +2x=1300
the four is for how many dog legs the two is how many human legs, 1300 is how many total legs.

1600-4x+2x=1300
300=2x
150=x

conclusion:
150 human heads
250 dog heads

check:
150 times 2= 300 human legs
250 times 4= 1000 dog legs
300+1000=1300 legs

Answer 5

let x- heads of owners,
y- heads of dogs,
therefore,
x+y = 400,
x< or = y;
there are 1300 legs, hence,
2x+4y=1300;

Then we result it : x+y=400, and x+2y=650;
resukt: x=150, y=250