Question 1007859
<pre>
Let the number of 44¢ stamps be x
Let the number of 20¢ stamps be y


                      Value      Value
Type       Number       of         of
 of          of        EACH       ALL
stamp      stamps     stamp     stamps
-------------------------------------------
44¢s         x        $0.44     $0.44x
20¢s         y        $0.20     $0.20y
-------------------------------------------
TOTALS      40       -----     $13.28

 The first equation comes from the second column.

  {{{(matrix(3,1,Number,of,"44c's"))}}}{{{""+""}}}{{{(matrix(3,1,Number,of,"20c's"))}}}{{{""=""}}}{{{(matrix(4,1,total,number,of,stamps))}}}

                 x + y = 40

 The second equation comes from the last column.
  {{{(matrix(4,1,Value,of,ALL,"44c's"))}}}{{{""+""}}}{{{(matrix(4,1,Value,of,ALL,"20c's"))}}}{{{""=""}}}{{{(matrix(5,1,Total,value,of,ALL,stamps))}}}

           0.44x + 0.20y = 13.28

Get rid of decimals by multiplying every term by 100:

          44x + 20y = 1328

 So we have the system of equations:
           {{{system(x + y = 40,44x + 20y = 1328)}}}.

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

           x + y = 40
               y = 40 - x

Substitute (40 - x) for y in 44x + 20y = 1328

   44x + 20(40 - x) = 1328
    44x + 800 - 20x = 1328
          24x + 800 = 1328
                24x = 528
                  x = 22 = the number of 44¢s.

Substitute in y = 40 - x
              y = 40 - (22)
              y = 18 = the number of 20¢s.



Checking:  22 44¢s is $9.68 and 18 20¢s is $3.60
            That's 40 stamps.
            And indeed $9.68 + $3.60 = $13.28
Edwin</pre>