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Algebra Topics - Introduction to Word Problems

Algebra topics -, introduction to word problems, algebra topics introduction to word problems.

Algebra Topics: Introduction to Word Problems

Lesson 9: introduction to word problems.

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What are word problems?

A word problem is a math problem written out as a short story or scenario. Basically, it describes a realistic problem and asks you to imagine how you would solve it using math. If you've ever taken a math class, you've probably solved a word problem. For instance, does this sound familiar?

Johnny has 12 apples. If he gives four to Susie, how many will he have left?

You could solve this problem by looking at the numbers and figuring out what the problem is asking you to do. In this case, you're supposed to find out how many apples Johnny has left at the end of the problem. By reading the problem, you know Johnny starts out with 12 apples. By the end, he has 4 less because he gave them away. You could write this as:

12 - 4 = 8 , so you know Johnny has 8 apples left.

Word problems in algebra

If you were able to solve this problem, you should also be able to solve algebra word problems. Yes, they involve more complicated math, but they use the same basic problem-solving skills as simpler word problems.

You can tackle any word problem by following these five steps:

We'll work through an algebra word problem using these steps. Here's a typical problem:

The rate to rent a small moving van is $30 per day, plus $0.50 per mile. Jada rented a van to drive to her new home. It took two days, and the van cost $360. How many miles did she drive?

It might seem complicated at first glance, but we already have all of the information we need to solve it. Let's go through it step by step.

Step 1: Read through the problem carefully.

With any problem, start by reading through the problem. As you're reading, consider:

Let's take a look at our problem again. What question is the problem asking? In other words, what are you trying to find out?

The rate to rent a small moving van is $30 per day, plus $0.50 per mile. Jada rented a van to drive to her new home. It took 2 days, and the van cost $360. How many miles did she drive?

There's only one question here. We're trying to find out how many miles Jada drove . Now we need to locate any information that will help us answer this question.

There are a few important things we know that will help us figure out the total mileage Jada drove:

Step 2: Represent unknown numbers with variables.

In algebra, you represent unknown numbers with letters called variables . (To learn more about variables, see our lesson on reading algebraic expressions .) You can use a variable in the place of any amount you don't know. Looking at our problem, do you see a quantity we should represent with a variable? It's often the number we're trying to find out.

Since we're trying to find the total number of miles Jada drove, we'll represent that amount with a variable—at least until we know it. We'll use the variable m for miles . Of course, we could use any variable, but m should be easy to remember.

Step 3: Translate the rest of the problem.

Let's take another look at the problem, with the facts we'll use to solve it highlighted.

The rate to rent a small moving van is $30 per day , plus $0.50 per mile . Jada rented a van to drive to her new home. It took 2 days , and the van cost $360 . How many miles did she drive?

We know the total cost of the van, and we know that it includes a fee for the number of days, plus another fee for the number of miles. It's $30 per day, and $0.50 per mile. A simpler way to say this would be:

$30 per day plus $0.50 per mile is $360.

If you look at this sentence and the original problem, you can see that they basically say the same thing: It cost Jada $30 per day and $0.50 per mile, and her total cost was $360 . The shorter version will be easier to translate into a mathematical expression.

Let's start by translating $30 per day . To calculate the cost of something that costs a certain amount per day, you'd multiply the per-day cost by the number of days—in other words, 30 per day could be written as 30 ⋅ days, or 30 times the number of days . (Not sure why you'd translate it this way? Check out our lesson on writing algebraic expressions .)

$30 per day and $.50 per mile is $360

$30 ⋅ day + $.50 ⋅ mile = $360

As you can see, there were a few other words we could translate into operators, so and $.50 became + $.50 , $.50 per mile became $.50 ⋅ mile , and is became = .

Next, we'll add in the numbers and variables we already know. We already know the number of days Jada drove, 2 , so we can replace that. We've also already said we'll use m to represent the number of miles, so we can replace that too. We should also take the dollar signs off of the money amounts to make them consistent with the other numbers.

30 ⋅ 2 + .5 ⋅ m = 360

Now we have our expression. All that's left to do is solve it.

Step 4: Solve the problem.

This problem will take a few steps to solve. (If you're not sure how to do the math in this section, you might want to review our lesson on simplifying expressions .) First, let's simplify the expression as much as possible. We can multiply 30 and 2, so let's go ahead and do that. We can also write .5 ⋅ m as 0.5 m .

Next, we need to do what we can to get the m alone on the left side of the equals sign. Once we do that, we'll know what m is equal to—in other words, it will let us know the number of miles in our word problem.

We can start by getting rid of the 60 on the left side by subtracting it from both sides .

The only thing left to get rid of is .5 . Since it's being multiplied with m , we'll do the reverse and divide both sides of the equation with it.

.5 m / .5 is m and 300 / 0.50 is 600 , so m = 600 . In other words, the answer to our problem is 600 —we now know Jada drove 600 miles.

Step 5: Check the problem.

To make sure we solved the problem correctly, we should check our work. To do this, we can use the answer we just got— 600 —and calculate backward to find another of the quantities in our problem. In other words, if our answer for Jada's distance is correct, we should be able to use it to work backward and find another value, like the total cost. Let's take another look at the problem.

According to the problem, the van costs $30 per day and $0.50 per mile. If Jada really did drive 600 miles in 2 days, she could calculate the cost like this:

$30 per day and $0.50 per mile

30 ⋅ day + .5 ⋅ mile

30 ⋅ 2 + .5 ⋅ 600

According to our math, the van would cost $360, which is exactly what the problem says. This means our solution was correct. We're done!

While some word problems will be more complicated than others, you can use these basic steps to approach any word problem. On the next page, you can try it for yourself.

Let's practice with a couple more problems. You can solve these problems the same way we solved the first one—just follow the problem-solving steps we covered earlier. For your reference, these steps are:

If you get stuck, you might want to review the problem on page 1. You can also take a look at our lesson on writing algebraic expressions for some tips on translating written words into math.

Try completing this problem on your own. When you're done, move on to the next page to check your answer and see an explanation of the steps.

A single ticket to the fair costs $8. A family pass costs $25 more than half of that. How much does a family pass cost?

Here's another problem to do on your own. As with the last problem, you can find the answer and explanation to this one on the next page.

Flor and Mo both donated money to the same charity. Flor gave three times as much as Mo. Between the two of them, they donated $280. How much money did Mo give?

Problem 1 Answer

Here's Problem 1:

A single ticket to the fair costs $8. A family pass costs $25 more than half that. How much does a family pass cost?

Let's solve this problem step by step. We'll solve it the same way we solved the problem on page 1.

Step 1: Read through the problem carefully

The first in solving any word problem is to find out what question the problem is asking you to solve and identify the information that will help you solve it . Let's look at the problem again. The question is right there in plain sight:

So is the information we'll need to answer the question:

Step 2: Represent the unknown numbers with variables

The unknown number in this problem is the cost of the family pass . We'll represent it with the variable f .

Step 3: Translate the rest of the problem

Let's look at the problem again. This time, the important facts are highlighted.

A single ticket to the fair costs $8 . A family pass costs $25 more than half that . How much does a family pass cost?

In other words, we could say that the cost of a family pass equals half of $8, plus $25 . To turn this into a problem we can solve, we'll have to translate it into math. Here's how:

First, replace the cost of a family pass with our variable f .

f equals half of $8 plus $25

Next, take out the dollar signs and replace words like plus and equals with operators.

f = half of 8 + 25

Finally, translate the rest of the problem. Half of can be written as 1/2 times , or 1/2 ⋅ :

f = 1/2 ⋅ 8 + 25

Step 4: Solve the problem

Now all we have to do is solve our problem. Like with any problem, we can solve this one by following the order of operations.

That's it! f is equal to 29. In other words, the cost of a family pass is $29 .

Step 5: Check your work

Finally, let's check our work by working backward from our answer. In this case, we should be able to correctly calculate the cost of a single ticket by using the cost we calculated for the family pass. Let's look at the original problem again.

We calculated that a family pass costs $29. Our problem says the pass costs $25 more than half the cost of a single ticket. In other words, half the cost of a single ticket will be $25 less than $29.

We could translate this into this equation, with s standing for the cost of a single ticket.

According to our math, s = 8 . In other words, if the family pass costs $29, the single ticket will cost $8. Looking at our original problem, that's correct!

So now we're sure about the answer to our problem: The cost of a family pass is $29 .

Problem 2 Answer

Here's Problem 2:

Let's go through this problem one step at a time.

Start by asking what question the problem is asking you to solve and identifying the information that will help you solve it . What's the question here?

To solve the problem, you'll have to find out how much money Mo gave to charity. All the important information you need is in the problem:

The unknown number we're trying to identify in this problem is Mo's donation . We'll represent it with the variable m .

Here's the problem again. This time, the important facts are highlighted.

Flor and Mo both donated money to the same charity. Flor gave three times as much as Mo . Between the two of them, they donated $280 . How much money did Mo give?

The important facts of the problem could also be expressed this way:

Mo's donation plus Flor's donation equals $280

Because we know that Flor's donation is three times as much as Mo's donation, we could go even further and say:

Mo's donation plus three times Mo's donation equals $280

We can translate this into a math problem in only a few steps. Here's how:

Because we've already said we'll represent the amount of Mo's donation with the variable m , let's start by replacing Mo's donation with m .

m plus three times m equals $280

Next, we can put in mathematical operators in place of certain words. We'll also take out the dollar sign.

m + three times m = 280

Finally, let's write three times mathematically. Three times m can also be written as 3 ⋅ m , or just 3 m .

It will only take a few steps to solve this problem.

We've got our answer: m = 70 . In other words, Mo donated $70 .

The answer to our problem is $70 , but we should check just to be sure. Let's look at our problem again.

If our answer is correct, $70 and three times $70 should add up to $280 .

We can write our new equation like this:

70 + 3 ⋅ 70 = 280

The order of operations calls for us to multiply first. 3 ⋅ 70 is 210.

The last step is to add 70 and 210. 70 plus 210 equals 280 .

280 is the combined cost of the tickets in our original problem. Our answer is correct : Mo gave $70 to charity.

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5 Easy Steps to Solving Word Problems

child learning how to solve a word problem

Word problems strike fear into the hearts of many students, and the trauma can even carry into adulthood. This is why word problems are the topic of many education jokes.

“If two trains start at the same station and travel in opposite directions at the same speed, when will the bacon be ready for breakfast?”

This is obviously a silly scenario, but it shows how word problems are regarded by many: a mangle of confusion that doesn’t make sense and can’t be solved!

Why Are Word Problems Difficult for Children?

Why can word problems be so confusing and scary? There are a few possible reasons.

A quick tip before we get started…

Explain to students that the word “problem” really means “question.” A word problem is just asking a question to which the students must find an answer. Show them that you need to identify the question before you even worry about which math operations are going to be used. Word problems can be treated like mysteries: the students are the detectives that are going to use the clues in the question to find the answer!

So what are the five easy steps to solving word problems? Let’s take a look!

Five Easy Steps to Solving Word Problems (WASSP)

Let’s look at an example word problem to demonstrate these steps.

Matt has twelve cookies he can give to his friends during lunchtime. If Matt has three friends sitting at his table, how many cookies can Matt give to each of his friends?

1. Write (or draw) what you know.

It is important to convince students that they do not have to immediately know what math operation is required to solve the problem. They first need only understand the scenario itself. In this example, the student could simply write down “12 cookies” and “3 friends,” or draw Matt with 12 cookies sitting at a table with three other children.

2. Ask the question.

Again, we don’t need to know what the math operation is yet! We just need to identify what is actually being asked. What do we NOT know?

The student could write, “How many cookies can each of Matt’s friends have?”

Alternatively, the student could draw a question mark over each friend’s head, maybe with a thought bubble of a cookie!

3. Set up a math problem that could answer the question.

In this example, the “clue” word (if you are using that method of reasoning) would be “each,” which indicates division. Or, the student could understand that Matt has to split, or divide, the cookies among his friends. Thus a division problem is needed!

Dividing 12 cookies among 3 friends means 12 is divided by 3.

4. Solve the problem.

It is important to note that using units can be a good idea . Otherwise, the answer could be misunderstood. Is it 4 cookies, or 4 friends, or something else?

12 cookies ÷ 3 friends = 4 cookies per friend

5. Put the answer in a sentence to see if the answer makes sense.

“Each of Matt’s friends can have four cookies.”

Does this answer make sense? It seems reasonable. How could this step help identify an incorrect answer?

What if the student had decided this was a multiplication problem?

12 cookies × 3 friends = 36 cookies per friend

If the student then writes a sentence using the answer, he may realize the answer can’t be right.

“Each of Matt’s friends can have 36 cookies.”

How would that be possible if Matt only had 12 cookies to start with? This must not be a multiplication problem. Let’s try again!

Practice the Five Easy Steps for Word-Problem Success!

Steps 1 and 2 ( Write what you know and Ask the question) help the student gain an understanding of the scenario.

Steps 3 and 4 ( Set up the math problem and Solve the problem) can be more easily navigated with critical thinking once the scenario is understood.

Step 5 ( Put the answer in a sentence) can help the student decide whether the answer makes sense.

Now your student is ready for word-problem success!

Make sure to start at the student’s level of understanding so he can experience success and build confidence, moving on to more challenging problems as appropriate. Customized curriculum is always best, which again makes masterygenius.com a great option if tutoring is needed. Students are assessed and then matched with a curriculum that strikes balance between building confidence and tackling challenges, leading to topic mastery.

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COMMENTS

Algebra Topics: Introduction to Word Problems
Step 1: Read through the problem carefully. The first in solving any word problem is to find out what question the problem is asking you to solve and identify the information that will help you solve it. Let's look at the problem again. The question is right there in plain sight: A single ticket to the fair costs $8.
5 Easy Steps to Solving Word Problems
Step 1: Visualize the Problem. The first step is to visualize the problem. See if you can picture what is going on. Draw pictures if that will help you. Pinpoint or highlight the important parts ...
Key Words for Solving Word Problems
Key Words for Solving Word Problems The hardest part of solving a word problem is actually understanding the problem and determining the operation (or operations) that needs to be performed. Listed below are a few of the most commonly used key words in word problems and the operations that they signal.