Question 984189
<pre>
The above solution uses only 1 unknown.  It can also be solved
using 2 unknowns.  The following uses 2 unknowns.


Let the number of nickels be x
Let the number of quarters be y
                      Value      Value
Type       Number       of         of
 of          of        EACH       ALL
coin        coins      coin      coins
-------------------------------------------
nickels       x       $0.05     $0.05x
quarters      y       $0.25     $0.25y
-------------------------------------------
TOTALS       29       -----     $4.45

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

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

               x + y = 29

 The second equation comes from the "VALUE of all coins" column

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

       0.05x + 0.25y = 4.45

Get rid of decimals by multiplying every term by 100:

            5x + 25y = 445

 So we have the system of equations:
{{{system(x + y = 29,5x + 25y = 445)}}}.

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

  x + y = 29
      y = 29 - x

Substitute (29 - x) for y in 5x + 25y = 445

     5x + 25(29 - x) = 445
      5x + 725 - 25x = 445
          -20x + 725 = 445
                -20x = -280
                   x = 14 = the number of nickels.

     Substitute in y = 29 - x
                   y = 29 - 14
                   y = 15 quarters.

The number of quarters is 29-x or 29-14 or 15 quarters.

Checking:  14 nickels is $0.70 and 15 quarters is $3.75
            That's 29 coins.
            And indeed $0.70 + $3.70 = $4.45
Edwin</pre>