Algebra Word Problems: Age Questions

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Problem-Solving to Determine Missing Variables

Using Algebra to calculate missing variable values. Rick Lewine/Tetra Images/Brand X Pictures/Getty Images

Many of the SATs, tests, quizzes, and textbooks that students come across throughout their high school mathematics education will have algebra word problems that involve the ages of multiple people where one or more of the participants' ages are missing.

When you think about it, it is a rare opportunity in life where you would be asked such a question. However, one of the reasons these types of questions are given to students is to ensure they can apply their knowledge in a problem-solving process.

There are a variety of strategies students can use to solve word problems like this, including using visual tools like charts and tables to contain the information and by remembering common algebraic formulas for solving missing variable equations.

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"Birthday:" An Algebra Age Problem

The Algebra Age Problem.

In the following word problem, students are asked to identify the ages of both of the people in question by giving them clues to solve the puzzle. Students should pay close attention to key words like double, half, sum, and twice, and apply the pieces to an algebraic equation in order to solve for the unknown variables of the two characters' ages.

Check out the problem presented to the left: Jan is twice as old as Jake and the sum of their ages is five times Jake's age minus 48. Students should be able to break this down into a simple algebraic equation based on the order of the steps, representing Jake's age as a and Jan's age as 2a: a + 2a = 5a - 48.

By parsing out information from the word problem, students are able to then simplify the equation in order to arrive at a solution. Read on to the next section to discover the steps to solving this "age-old" word problem.

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Steps to Solving the Algebraic Age Word Problem

First, students should combine like terms from the above equation, such as a + 2a (which equals 3a), to simplify the equation to read 3a = 5a - 48. Once they've simplified the equation on either side of the equals sign as much as possible, it's time to use the distributive property of formulas to get the variable a on one side of the equation.

In order to do this, students would subtract 5a from both sides resulting in -2a = - 48. If you then divide each side by -2 to separate the variable from all real number in the equation, the resulting answer is 24.

This means that Jake is 24 and Jan is 48, which adds up since Jan is twice Jake's age, and the sum of their ages (72) is equal five times Jake's age (24 X 5 = 120) minus 48 (72).

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An Alternate Method for the Age Word Problem

birthday problem 3 algebra
Alternate Method.

No matter what word problem you're presented in algebra, there's likely going to be more than one way and equation that's right to figure out the correct solution. Always remember that the variable needs to be isolated but it can be on either side of the equation, and as a result, you can also write your equation differently and consequently isolate the variable on a different side.

In the example on the left, instead of needing to divide a negative number by a negative number like in the solution above, the student is able to simplify the equation down to 2a = 48, and if he or she remembers, 2a is the age of Jan! Additionally, the student is able to determine Jake's age by simply dividing each side of the equation by 2 to isolate the variable a.