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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
    
    
Since q = the number of quarters Tony has, this means that Tony has 7 quarters. Since we also know that Tony has 11 more nickels than quarters, he must have 7 + 11 = 18 nickels.

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



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