Question 1027510
<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       17      -----        $3.50

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

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

                 x + y = 17

 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 = 3.50

Get rid of decimals by multiplying every term by 100:

               25x + 10y = 350

 So we have the system of equations:
           {{{system(x + y = 17,25x + 10y = 350)}}}.

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

                   x + y = 17
                       y = 17 - x

Substitute (17 - x) for y in 25x + 10y = 350

        25x + 10(17 - x) = 350
         25x + 170 - 10x = 350
               15x + 170 = 350
                     15x = 180
                       x = 12 = the number of quarters.

Substitute in y = 17 - x
              y = 17 - (12)
              y = 5 dimes.

Checking:  12 quarters is $3.00 and 5 dimes is $0.50
            That's 17 coins.
            And indeed $3.00 + $0.50 = $3.50
Edwin</pre>