Question 1030311
<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       35      -----      $5.00

 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 = 35

 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 = 5.00

Get rid of decimals by multiplying every term by 100:

               10x + 25y = 500

 So we have the system of equations:
           {{{system(x + y = 35,10x + 25y = 500)}}}.

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

                   x + y = 35
                       y = 35 - x

Substitute (35 - x) for y in 10x + 25y = 500

        10x + 25(35 - x) = 500
         10x + 875 - 25x = 500
              -15x + 875 = 500
                    -15x = -375
                       x = 25 = the number of dimes.

         Substitute in y = 35 - x
                       y = 35 - (25)
                       y = 10 quarters.

Checking:  25 dimes is $2.50 and 10 quarters is $2.50

            That's 35 coins.

And indeed $2.50 + $2.50 = $5.00

Edwin</pre></b>