A vending machine only accepts Quarters and Dimes. At the end of a day there are 200 coins in the machine that total $48.05. How many quarters are in the machine and how many dimes?

Well first off...I am not good in math so I used my calculator to determine that there are 187 Quarters and 13 Dimes, totaling $48.05.

I am unsure how to set up the equation. Is this anywhere close?

.25x + .10y = 48.05?

Your equation is correct. While several people have pointed out the second equation that would enable you to solve in the traditional manner, you can proceed to a solution starting with your equation

1--.25Q + .10D = 48.05 or 5Q + 2D = 961

2--Dividing through by the lowest coefficient yields Q + Q/2 = 480 + 1/2

3--(Q - 1)/2 must be an integer k making Q = 2k + 1

4--Substituting back into (1) yields D = 478 - 5k

5--Now, recognizing the fact that the total number of quarters and dimes is 200, you can derive the other equation that people have been trying to convey to you, Q + D = 200. By substitution, we derive 2k + 1 + 478 - 5k = 200 from which k = 93.

6--Therefore, Q = 187 and D = 13

7--.25(187) + .10(13) = 46.75 + 1.30 = $48.05.

While this is a path to the solution, by making use of the interim factor of k, you can reach the same solution by simply combining the two basic equations of 25Q + 10D = 4805 and Q + D = 200.

I'll leave it for you to solve directly using the basic equations of 25Q + 10D = 4805 and Q + D = 200.

I'll leave it to you to follow that path.