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Introduction to Algebra Word Problems Study Guide

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Updated on Oct 3, 2011

Introduction to Algebra Word Problems

There is no branch of mathematics, however abstract, which may not some day be applied to phenomena of the real world.

—Nikolai Lobatchevsky (1792–1856) Russian Mathematician

In this lesson, you'll learn how to use algebra to solve real–life problems.

You might not realize how often algebra can be used to answer everyday questions. Anytime a situation has an unknown value, that unknown can be found using algebra. If a gas tank holds 20 gallons of gas, and a car goes 16 miles for every gallon, how much gas will be left in the tank after a trip of 300 miles? That's an important question to answer if you do not want to run out of gas!

To solve a word problem, we follow these six steps:

  1. Read the entire word problem.
  2. Underline the keywords.
  3. List the possible operations.
  4. Represent the unknown with a variable.
  5. Write an equation or inequality to solve the problem.
  6. Solve the number sentence.

The first step might seem simple, but it is actually the most important. If you do not read a problem carefully, you might not answer the question that is being asked. Next, underline the keywords that might signal which operations must be performed. Remember the chart from Lesson 6?

Algebra Word Problems

We used this chart to help us turn algebraic phrases into algebraic expressions. We can also use it to help us turn algebraic word problems into algebraic equations or inequalities. Once we know what operation or operations are needed, we use a variable to hold the place of the unknown and write an equation or inequality to solve the problem. Finally, we solve that for the variable and check our work.

Example

Bethel has five less than four times the number of postcards that Loring has. If Bethel has 11 postcards, how many postcards does Loring have?

First, read the entire problem. We are told how many postcards Loring has and we are looking for how many postcards Bethel has. Next, underline the keywords. The words less than can signal subtraction and the word times can signal multiplication. We don't know how many postcards Loring has, so we can use x to represent that value.

Now, we must write an equation that we can use to solve this problem. Bethel has five less than four times the number of postcards that Loring has. The number of postcards Loring has is x, so Bethel has five less than four times x. Four times x means "four multiplied by x," which is 4x. Five less than 4x is 4x – 5. The expression 4x – 5 represents how many postcards Bethel has. We are told that Bethel has 11 postcards, so we can set these two values equal to each other: 4x – 5 = 11. Finally, solve the equation for x to find how many postcards Loring has:

4x – 5 = 11

4x = 16

x = 4

Loring has 4 postcards. We can check our work by rereading the word problem. Bethel has five less than four times the number of postcards that Loring has. Four times four is 16, and five less than that is 11, which is the number of postcards Bethel has. We have found the correct answer.

Tip:

If you are not sure which operation should be used to solve a problem, write more than one equation, using different operations in each equation. Solve each equation and decide which answers seem reasonable. Then, check each answer against the information given in the question and choose the one that correctly solves the problem.

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