Question 941651
A purse contains 58 coins consisting of dimes and nickels. If the total amount of these coins amounted to $4.80, how many dimes and nickels are in the purse?
<pre>
Let the number of nickels be x
Then the number of dimes, using 
ONE PART = TOTAL MINUS OTHER PART,
is 58-x.  
                      Value      Value
Type       Number       of         of
 of          of        EACH       ALL
coin        coins      coin      coins
-------------------------------------------
NICKELS      x        $0.05    $0.05x 
DIMES      58-x       $0.10    $0.10(56-x)
-------------------------------------------
TOTALS      58        -----    $4.80

 The equation comes from the column on the right

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

0.05x + 0.10(58-x) = 4.80

Get rid of decimals by multiplying every term by 100

     5x + 10(58-x) = 480

    5x + 580 - 10x = 480

         -5x + 580 = 480

               -5x = -100

                 x = 20 = the number of nickels

The number of dimes is 58-x or 58-20 or 38 dimes.

Checking:  20 nickels is $1.00 and 38 dimes is $3.80

And indeed $1.00+$3.80 = $4.80.

Edwin</pre>