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How to Do Intermediate Algebra Story Problems

Algebra students always want to know how algebra applies to real life, but seem to have the most trouble with story problems -- the application of algebra to real life situations. At the intermediate algebra level, there are only a few types of story problems and each type has its characteristic formula or technique that is central to solving the problem. When an expert reads one of these problems, the key formula of technique immediately suggests itself. The key to solving these problems is to recognize the key in the wording of the problem.

Instructions

    • 1

      Look for the key formula or technique -- there is one in every story problem. If the problem talks about cars leaving the same city at the same time, the formula distance = rate X time is going to be central to solving the problem. If the problem is about consecutive numbers there is going to be something like X + (X + 1) + (X + 2) in the problem. Find the central idea and you are on the way to the solution. Without the central idea, you are lost.

    • 2

      Learn how language is translated into mathematical symbols. For example, "In 10 years, A will be as old as B is now" translates into A + 10 = B. "Less than" implies subtraction and "three less than double" implies 2X - 3. "John is four times as old as Benny" is J = 4B. "X has more than twice as much money as Y" is X is-greater-than 2Y. Look at worked examples from the book or the class and pay attention to how the language is turned into mathematics.

    • 3

      Categorize the samples you see by the central formulas. For example, age problems will have phrases like "In 10 years, Jack will be twice as old as Bill" becomes J + 10 = 2 (B + 10). For coin problems, there is going to be an equation like W + 5X + 10Y + 25Z = K, because 1, 5, 10 and 25 are the values of coins. For people working together there will be an equation like 1/T = 1/A + 1/B when A and B working together can accomplish a job in T amount of time.

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