Question 1003305
<pre>
Let the number of dimes be x
Let the number of quarters be y


                      Value      Value
Type       Number       of         of
 of          of        EACH       ALL
coin        coins      coin      coins
-------------------------------------------
dimes        x        $0.10     $0.10x
quarters     y        $0.25     $0.25y
-------------------------------------------
TOTALS       23        -----    $4.55

 The first equation comes from the second column.

  {{{(matrix(3,1,Number,of,dimes))}}}{{{""+""}}}{{{(matrix(3,1,Number,of,quarters))}}}{{{""=""}}}{{{(matrix(4,1,total,number,of,coins))}}}
                 x + y = 23

 The second equation comes from the last column.

  {{{(matrix(4,1,Value,of,ALL,dimes))}}}{{{""+""}}}{{{(matrix(4,1,Value,of,ALL,quarters))}}}{{{""=""}}}{{{(matrix(5,1,Total,value,of,ALL,coins))}}}

           0.1x + 0.25y = 4.55

Get rid of decimals by multiplying every term by 100:

          10x + 25y = 455

 So we have the system of equations:
           {{{system(x + y = 23,10x + 25y = 455)}}}.

We solve by substitution.  Solve the first equation for y:

           x + y = 23
               y = 23 - x

Substitute (23 - x) for y in 10x + 25y = 455

     10x + 25(23 - x) = 455
      10x + 575 - 25x = 455
           -15x + 575 = 455
                 -15x = -120
                    x = 8 = the number of dimes.

Substitute in y = 23 - x
              y = 23 - (8
              y = 15 quarters.

The number of quarters is 23-x or 23-8 or 15 quarters.

Checking:  8 dimes is $0.80 and 15 quarters is $3.75
            That's 23 coins.
            And indeed $0.80 + $3.75 = $4.55


 So it checks.  8 dimes and 15 quarters
Edwin</pre>