There are twice as many cats than dogs and the total of both animals is 21

Question

Need help with this simple equation, not so simple to me tonight ;(

Answer 1

Let 'c' represent the # of cats and let 'd' represent the # of dogs.

We are given the following:

    1.) There are twice as many cats as there are dogs  ==>   c = 2d

    2.) The total of cats and dogs is 21  ==>   c + d = 21

Since we know that c=2d, we can replace c in c+d=21 with 2d and combine like terms. That is,

      c + d = 21   ==>   2d + d = 21   ==>   3d = 21

Divide both sides by 3 to solve for d:

      (3d) / 3 = (21) / 3   ==>   d = 7

Now that we know d, we can solve for c by plugging the value for d in the original equation:

      c + d = 21   ==>   c + 7 = 21

Subtract 7 from both sides to solve for c:

      c + 7 = 21

         - 7     - 7

  ________________

      c + 0 = 14     ==>     c = 14

Thus, there are 14 cats and 7 dogs.

Answer 2

Well, let's say the number of dogs is called D. There are two times as many cats as dogs, so we can call the number of cats 2D. We know that the total of both animals is 21, so D+2D=21. Therefore, 3D = 21, and D=7. So there are 7 dogs and 14 cats. Hope this helps!

 

Eytan