A common algebra problem often goes something like this:

Tony has 11 more nickels than quarters. How many coins does he have if the total value of his coins is $2.65?

As with most algebra 'word problems', let's start by looking at what we know.

First, we are told that Tony has 11 more nickels than quarters. We can call the number of quarters Tony has q. Then Tony must also have q + 11 nickels since he has 11 more nickels than quarters.

We're also told that Tony has $2.65. To solve the problem, we need to know that a dollar is 100 cents so $2.65 = 265 cents, a quarter is worth 25 cents (a quarter of a dollar), and a nickel is worth 5 cents. We can now use the following equation:

    25 cents x no. of quarters + 5 cents x no. of nickels = 265 cents

Translating this into symbols:

25q + 5(q +11) = 265
 25q + 5q + 55 = 265  (distributing)
      30q + 55 = 265  (adding like terms)
           30q = 210  (subtracting 55 from each side)
             q = 7

Thus Tony has a total of 7 quarters + 18 nickels = 25 coins.

From the Dr. Math archives:

1 Dollar, 50 Coins

5 Coins = 55 Cents

21 Nickels, Dimes and Pennies in a Dollar

25 Coins in a Dollar

50 Coins = One Dollar

Coin Combinations to Make a Dollar

Dimes and Quarters Puzzle

Five Dollars in Change

Make $5 Using One of Each Coin

Making Change for a Dollar

Number of Quarters