Dime and Nickel Problem

A parking meter coin slot receives dimes and nickels. When the meter box is emptied, there were 148 coins and a total of $10.65. How many dimes and how many nickels were there?

Solution
Before we can derive a valid algebraic equations, we have to first define what values are represented with a dime and a nickel. In US currency, the following coins have the following monetary values:

        1 US dollar = 100 cents
        1 nickel = 5 cents
        1 dime = 10 cents

Therefore, based on this definition and the problem statement, we can derive the following representation:

Let
        d = dime
        n = nickel

Hence,
        $d + n = 148$                            Eq. 1    Total number of coins in the slot
        $0.1d + 0.05n = 10.65$           Eq. 2    Total amount of all coins

It will be easier to solve this system of equations in two unknowns by first eliminating the decimals in Eq. 2. Then using addition/elimination method we get

        $(0.1d + 0.05n = 10.65)(100)$
        $10d + 5n = 1065$
        $(d + n = 148)(-5)$            Multiply Eq. 1 with -5 to eliminate n during addition
       
        $10d + 5n = 1065$
        $-5d - 5n = -740$
        $5d + 0n = 325$                        Divide both sides by 5
        $d = \frac{325}{5}$
        $d = 65$

Substituting this value of d in Eq. 1, we will get the value of n as follows:
        $d + n = 148$
        $65 + n = 148$
        $n = 148 - 65$
        $n = 83$

Therefore, there were 65 dimes and 83 nickels in the meter box.

Check
There were 148 coins: 65 (dimes) + 83 (nickels) = 148
The total money is $10.65: 65(0.1) + 83(0.05) = 10.65
                                    6.5 + 4.15 = 10.65
                                    10.65 = 10.65