Hi, there--

The Problem:
Sarah has a collection of dimes, nickels, and quarters worth $15.75.  She has 10 more dimes than nickels and twice as many quarters as dimes.  How many coins of each type does she have?

A Solution:
Let d be the number of dimes.
Let n be the number of nickels.
Let q be the number of quarters.

Now we need three equations using the information from the problem.
Sarah has ten more dimes that nickels.
d = n + 10

She has twice as many quarters as dimes.
q = 2d

The collection of coins is worth $15.75.
The value of the dimes is .10 times the number of dimes, or 0.10d.
The value of the nickels is .05 times the number of nickels, or 0.05n.
The value of the quarters is 0.25q.
The values of the three types of coins equals $15.75, so
0.10d + 0.05n + 0.25q = 15.75

Now we have three equations in three variables. We can solve the system for d, n, and q.
d = n + 10
q = 2d
0.10d + 0.05n + 0.25q = 15.75

Rewrite the first equation in a "n=..." form. (Subtract 10 from both sides.)
d = n + 10
n = d - 10

Substitute d - 10 for n in the third equation
0.10d + 0.05n + 0.25q = 15.75
0.10d + 0.05(d-10) + 0.25q = 15.75

We see in the second equation that q and 2d are equivalent. Substitute 2 * d for q in the 
third equation.
0.10d + 0.05(d-10) + 0.25(2d) = 15.75

Solve this equation for d.
0.10d + 0.05d - 0.50 + 0.50d = 15.75
0.65d - 0.50 = 15.75
0.65d = 15.75 + 0.50
0.65d = 16.25
d = 25

There are 25 dimes. 
There are twice as many quarters as dimes, so there must be 50 quarters. 
There are ten more dimes than nickels, so there must be 15 nickels.

We need to make sure tis combination of coins equals $15.75.
25 dimes are worth 25*0.10= $2.50.
50 quarters are worth 50*0.25=$12.50.
15 nickels are worth 15*0.05 = $0.75.

$2.50 + 12.50 + 0.75 = $15.75

That's it! Feel free to email me if you have questions about the solution.

Ms.Figgy
math.in.the.vortex@gmail.com