Question 1052972
Harry has $6.10 in his pocket.
<pre>
$0.25Q + $0.10D + $0.05N = $6.10

or multiplying through by 100,

25Q + 10D + 5N = 610

dividing through by 5

5Q + 2D + N = 122
</pre>
He has twice as many quarters as nickels
<pre>
Q = 2N
</pre>
 and 

one more dime than quarters. 
<pre>
D = Q + 1

So we have this system of equations:

1.     5Q + 2D + N = 122
2.               Q = 2N
3.               D = Q + 1

Use eq. 2 to substitute 2N for Q in eq. 3

3.               D = Q + 1
4.               D = 2N + 1

Use eq. 2 to substitute 2N for Q in eq. 1 and
use eq. 4 to substitute 2N + 1 for D in eq 1:

1.               5Q + 2D + N = 122
       5(2N) + 2(2N + 1) + N = 122 
            10N + 4N + 2 + N = 122
                     15N + 2 = 122
                         15N = 120
                           N = 8

Substitute 8 for N in eq. 4:

4.               D = 2N + 1
                 D = 2(8) + 1
                 D = 16 + 1
                 D = 17


Substitute N=8 in eq. 2

2.               Q = 2N
                 Q = 2(8)
                 Q = 16

Edwin</pre>