SOLUTION: Rob has 40 coins, all dimes and quarters, worth $7.60. How many dimes and how many quarters does he have?

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       40      -----     $7.60

 The first equation comes from the second column.

  
             
                  x + y = 40

 The second equation comes from the last column.
  

          0.10x + 0.25y = 7.60

Get rid of decimals by multiplying every term by 100:

              10x + 25y = 760

 So we have the system of equations:
           .

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

                  x + y = 40
                      y = 40 - x

Substitute (40 - x) for y in 10x + 25y = 760

       10x + 25(40 - x) = 760
       10x + 1000 - 25x = 760
            -15x + 1000 = 760
                   -15x = -240
                      x = 16 = the number of dimes.

        Substitute in y = 40 - x
                      y = 40 - (16)
                      y = 24 quarters.


Checking:  16 dimes is $1.60 and 24 quarters is $6.00

            That's 40 coins.

And indeed $1.60 + $6.00 = $7.60

Edwin