SOLUTION: The total number of coins in a box is 312. The number of dimes is three times the number of nickels and quarters together. If the box has 31 dollars and 50 cents find the number of

Originally this problem is in three unknowns and three equations.
But it can be reduced to the problem in two unknowns and two equations.
I will show you now how to make this reduction.

Let M be the number of nickels and quarters together.
Then the number of dimes is 3M, according to the condition.
Since the total number of coins in a box is 312, it gives you an equation

M + 3M = 312,  or  4M = 312.

Hence, M =  = 78.

Thus the number of dimes is 3M = 3*78 = 234. Correspondingly, the amount in dimes is $0.10*234 = $23.40.

Then the amount in other coins (in nickels and quarters together) is $31.50 - $23.40 = $8.10.

Thus the reduction is done. We reduced the original problem to this one:

   There are 78 coins in a box, nickels and quarters. If the box has $8.10, 
   find the number of nickels and quarters.

For this auxiliary problem we have two equations in two unknowns

 n +   q =  78,    (1)
5n + 25q = 810,    (2)

where n = # of nickels and q = # of quarters.

To solve it, express n = 78-q from (1) and substitute into (2). You will get a single equation for n

5(78-q) + 25q = 810.

Simplify and solve it:

390 - 5q + 25q = 810,
20q = 810 - 390,
20q = 420,

q =  = 21.

Thus the number of quarters is 21.
Hence, the number of nickels is n = 78-q = 78-21 = 57.

Check:  5*57 + 25*21 = 810.

Answer. n = 57, d = 234, q = 21.