SOLUTION: A collection of nickels, dimes, and quarters consist of 60 coins with a total of $6.50. If there are 3 times as many dimes as quarters, find the number of each type of coins.

From the condition, you have these tree equations for three unknowns

n + d + q = 60,              (1)
d = 3q,                      (2)
5n + 10d + 25q = 650.        (3)

Since you have 3 independent equations for three unknowns, it is enough to get the solution.

Substitute d = 3q from the equation (2) first into equation (1) and then into equation (3).
In this way you exclude the unknown d and will get the system of TWO equations in TWO unknowns

n + 3q + q = 60,             (1')
5n + 10*(3q) + 25q = 650.    (2')

or simplifying,

 n +  4q =  60,              (1'')
5n + 55q = 650.              (2'')

To solve this system, let's do one more step of the substitution method.
Express n = 60 - 4q from (1'') and substitute it into equation (2''). You will get

5*(60-4q) + 55q = 650.

Simplify and solve it for q:

300 - 20q + 55q = 650,

35q = 650 - 300,

35q = 350,

q = 10.

Congratulations! You just found the number of quarters.

Now you can find the number of nickels. 
Use the formula n = 60 - 4q that you obtained above. You will get

n = 60 - 4*10 = 60 - 40 = 20.   So, the number of nickels is 20.

Last step is to find the number of dimes. For it, use the formula d= 3q (equation (2)). You have d = 3*10 = 30.

Answer. n = 20, d = 30, q = 10.

Please check yourself that the solution is correct.

Let number of nickels and quarters, N, and Q, respectively

Then number of dimes = 3Q

We then get: N + Q + 3Q = 60_____N + 4Q = 60 ------- eq (i)

Also, .05N + .25Q + .1(3Q) = 6.5____.05N + .25Q + .3Q = 6.5____.05N + .55Q = 6.5 ---- eq (ii)

- .05N - .2Q = - 3 ------- Multiplying eq (i) by – .05 ------- eq (iii)

.35Q = 3.5 ------- Adding eqs (iii) & (ii)

Q, or number of quarters = , or 

N + 4(10) = 60 --------- Substituting 10 for Q in eq (i)

N + 40 = 60

N, or number of nickels = 60 – 40, or 

Number of dimes: 3(10), or