Question 1029975
<pre><b>
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       41      -----       $6.80

 The first equation comes from the "Number of coins" column.

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

                 x + y = 41

 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))}}}

         0.10x + 0.25y = 6.80

Get rid of decimals by multiplying every term by 100:

             10x + 25y = 680

 So we have the system of equations:
           {{{system(x + y = 41,10x + 25y = 680)}}}.

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

                 x + y = 41
                     y = 41 - x

Substitute (41 - x) for y in 10x + 25y = 680

      10x + 25(41 - x) = 680
      10x + 1025 - 25x = 680
           -15x + 1025 = 680
                  -15x = -345
                     x = 23 = the number of dimes.

Substitute in y = 41 - x
              y = 41 - (23
              y = 18 quarters.

Checking:  23 dimes is $2.30 and 18 quarters is $4.50
            That's 41 coins.
            And indeed $2.30 + $4.50 = $6.8

Edwin</pre></b>