Question 832822
Michael's bank contains only nickels, dimes and quarters.
<pre>
N = number of nickels
D = number of dimes
Q = the number of quarters</pre>There are 63 coins in all<pre>
           N + D + Q = 63
</pre>valued at $5.50.<pre>The N nickels are worth $0.05N
The D dimes are worth $0.10D
The Q quarters are worth $0.25Q

0.05N + 0.10D + .25Q = 5.50

Multiply every term by 100 moves the decimals two places over:

      5N + 10D + 25Q = 550

Divide every term by 5

         N + 5D + 5Q = 110</pre>The number of nickels is 5 short of being 3 times the sum of
the number of dimes and quarters.<pre>the sum of number of dimes and quarters = D+Q
3 times the sum of number of dimes and quarters = 3(D+Q)
5 short of being 3 times the sum of the number of dimes and quarters = 3(D+Q)-5
So

                   N = 3(D+Q) - 5
                   N = 3D + 3Q - 5
         N - 3D - 3Q = -5

So we have the system of 3 equations in 3 unknowns:

         N +  D +  Q =  63
         N + 2D + 5Q = 110 
         N - 3D - 3Q =  -5

Solve that system by elimination.
</pre>How many dimes are in the bank?<pre>
Solve the system and see.  If you can't post again
asking how to solve it.

Edwin</pre>