Algebraic Expressions and Word Problems Study Guide

Introduction to Algebraic Expressions and Word Problems

"What's one and one and one and one and one and one and one and one and one and one?" "I don't know," said Alice. "I lost count."

—Lewis Carroll (1832–1898)

In this lesson, you'll see that translating word problems into algebra is important for both math tests and issues that arise every day. Specifically, you'll learn to tackle distance, mixture, and work problems.

When you are translating sentences and word problems into algebraic expressions and equations, it can seem like you are translating between two different languages.

There are some strategies, however, that will help you become fluent in both:

First, read the problem to determine what you are looking for.
Then, write the amount you are looking for in terms of x (or whatever letter you want to use). You can do this by writing "Let x =… " Write any other unknown amounts in terms of x, too.
Last, set up the algebraic expressions in an equation with an equals sign and solve for the variable.

Tip:

When you are translating key words in an algebraic expression, the phrases less than and greater than do not translate in the same order as they are written in the sentence. For example, when you are translating the expression eight less than 15, the correct expression is 15 – 8, not 8 – 15.

Study the following to see how to translate word problems into mathematical statements and equations.

Equals Key Words is are has was were had

Addition Key Words Sum Together Total More Greater or Older than

Subtraction Key Words Difference Less Fewer or Younger than Remain Left Over

Multiplication Key Words Product Times Of

Division Key Words Per Evenly

Look at an example where knowing the key words is necessary:

Twenty less than five times a number is equal to the product of ten and the number. What is the number?

Let's let x equal the number we are trying to find. Now, translate the sentence piece by piece, and then solve the equation.

Twenty less than five times a number equals the product of 10 and x.

The equation is 5x– 20 = 10x. Subtract 5x from both sides:

Now, divide both sides by 5:

In this example, the key words less than tell you to subtract from the number and the key word product reminds you to multiply.

Problem Solving with Word Problems

There are a variety of different types of word problems you will encounter on tests or in your daily life. To help with these types of problems, always begin first by figuring out what you need to solve for and defining your variable(s) as what is unknown. Then write and solve an equation that matches the question asked.

Distance Word Problems

Distance word problems use the distance formula rate × time = distance. Distance formula problems include key words such as speed, plane, train, boat, car, walk, run, climb, and swim. They provide two of the three elements—rate, time, and distance—in the distance formula. Plug those two elements into the formula and solve for the third. Don't forget that rate and time have to be measured in common units. If the rate is measured in miles per hour, the time has to be measured in hours, not minutes or days.

Let's practice:

How far did the plane travel in 4 hours if it traveled at an average speed of 300 miles per hour?

The plane traveled 1,200 miles.

Mixture Word Problems

Mixture questions will present you with two or more different types of "objects" to be mixed together. Some common types of mixture scenarios are combining different amounts of money at different interest rates, different amounts of solutions at different concentrations, and different amounts of food (candy, soda, etc.) that have different prices per pound. Let's put this into play:

How many pounds of hot chocolate that costs $4 per pound need to be mixed with 10 pounds of hot chocolate that costs $6.40 per pound to create a mixture of hot chocolate that costs $5.50 per pound?

For this type of question, remember that the total amount spent in each case will be the price per pound times how many pounds in the mixture. Therefore, if you let x = the number of pounds of $4 hot chocolate, then $4(x) is the amount of money spent on $4 hot chocolate, $6.40(10) is the amount spent on $6.40 hot chocolate, and $5.50(x + 10) is the total amount spent. Write an equation that adds the first two amounts and sets it equal to the total amount. Then, multiply through the equation:

Subtract 4x from both sides:

Subtract 55 from both sides:

Divide both sides by 1.5:

You need 6 pounds of the $4 per pound hot chocolate.

Work Word Problems

Work problems often present the scenario of two people working to complete the same job. To solve this particular type of problem, think about how much of the job will be completed in one hour. For example, if someone can complete a job in 5 hours, then of the job is completed in 1 hour. If a person can complete a job in x hours, then of the job is completed in 1 hour. Take a look at the next example to see how this is used to solve work problems:

Jason can mow a lawn in 2 hours. Shantelle can mow the same lawn in 4 hours. If they work together, how many hours will it take them to mow the same lawn?

Think about how much of the lawn each person completes individually. Jason can finish in 2 hours, so in 1 hour he completes of the lawn. Because Shantelle can finish in 4 hours, then in 1 hour she completes of the lawn. If we let x = the time it takes both Jason and Shantelle working together, then is the amount of the lawn they finish in 1 hour working together. Then use the equation and solve for x. Multiply each term by the least common denominator (LCD) of 4x:

The equation becomes 2x + x = 4. Combine like terms and divide each side by 3.

So, hours. Because of an hour is of 60 minutes, which is 20 minutes, the correct answer is 1 hour and 20 minutes.

Find practice problems and solutions for these concepts at Algebraic Expressions and Word Problems Practice Questions.

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Algebraic Expressions and Word Problems Study Guide

Introduction to Algebraic Expressions and Word Problems

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