Question 1009623

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


                      Value      Value
Type       Number       of         of
 of          of        EACH       ALL
coin        coins      coin      coins
-------------------------------------------
quarters      x      $0.25     $0.25x
dimes         y      $0.10     $0.10y
-------------------------------------------
TOTALS       15      -----     $2.70

 The first equation comes from the second column.

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

                   x + y = 15

 The second equation comes from the last column.

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

           0.25x + 0.10y = 2.70

Get rid of decimals by multiplying every term by 100:

               25x + 10y = 270

 So we have the system of equations:
           {{{system(x + y = 15,25x + 10y = 270)}}}.

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

                   x + y = 15
                       y = 15 - x

Substitute (15 - x) for y in 25x + 10y = 270

        25x + 10(15 - x) = 270
         25x + 150 - 10x = 270
               15x + 150 = 270
                     15x = 120
                       x = 8 = the number of quarters.

Substitute in y = 15 - x
              y = 15 - (8)
              y = 7 dimes.


Checking:  8 quarters is $2.00 and 7 dimes is $0.70
            That's 15 coins.
            And indeed $2.00 + $0.70 = $2.70

Edwin</pre>