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Course: praxis core math   >   unit 1.

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## SSAT Upper Level Math : Algebraic Word Problems

Study concepts, example questions & explanations for ssat upper level math, all ssat upper level math resources, example questions, example question #1 : algebraic word problems.

Michael scores a 95, 87, 85, 93, and a 94 on his first 5 math tests. If he wants a 90 average, what must he score on the final math test? To solve for the final score: If David wants to drive to his friend's house, which is 450 miles away, in 6 hours, what is the average speed David has to drive at? Plug in the the values for distance and time, and solve for rate. If the sum of the smallest and largest of three consecutive even numbers is 28, what is the value of the second largest number in the series? The three numbers would be Add the first and third value and you get Beth and Sam are 500 miles apart. If Beth travels at 60mph and leaves her house at 1pm, what time will she arrive at Sam's house? ## Example Question #1 : How To Solve Algebraic Word Problems

Greg is trying to fill a 16 oz. bottle with water. If Greg fills the bottle at 1 oz per second and the bottle leaks .2 oz per second, how long would it take for Greg to fill the bottle? You first find the rate at which the bottle is being filled at, which is Two inlet pipes lead into a large water tank. One pipe can fill the tank in 45 minutes; the other can fill it in 40 minutes. To the nearest tenth of a minute, how long would it take the two pipes together to fill the tank if both were opened at the same time? Look at the work rates as "tanks per minute", not "minutes per tank". The inlet pipe to a large water tank can fill the tank in 25 minutes; the drain of the tank can empty it in 55 minutes.

Once, the drain was left open by mistake when the tank was being filled. The mistake was not caught until the tank was full. To the nearest tenth of a minute, how long did it take to fill the tank? ## Example Question #9 : Algebraic Word Problems The total of the measures of the angles of a pentagon is ## Example Question #10 : Algebraic Word Problems

John and Julie represented West High in a math contest. John outscored Julie by 16 points; as a team, they scored 80 points.

David and Dana represented East High in the same contest. Dana outscored David by 10 points; as a team, they scored 60 points.

Arrange the four students from highest score to lowest score.

Dana, John, David, Julie

John, Dana, David, Julie

John, Dana, Julie, David

John, Julie, Dana, David

Dana, John, Julie, David Julie scored 32, and John scored 16 higher, or 48. David's score was 25, and Dana's score was 10 higher, or 35.

In descending order of score, the students were John, Dana, Julie, David. ## Report an issue with this question

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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.

/en/algebra-topics/solving-equations/content/

## 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:

• Read through the problem carefully, and figure out what it's about.
• Represent unknown numbers with variables.
• Translate the rest of the problem into a mathematical expression.
• Solve the problem.

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: • What question is the problem asking? • What information do you already have? 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:

• The van cost $30 per day. • In addition to paying a daily charge, Jada paid$0.50 per mile.
• The total cost was $360 . ## 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 . 60 + .5m = 360 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? Answer:$29

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:

• A single ticket costs $8 . • The family pass costs$25 more than half the price of the single ticket.

## 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.

• f is already alone on the left side of the equation, so all we have to do is calculate the right side.
• First, multiply 1/2 by 8 . 1/2 ⋅ 8 is 4 .
• Next, add 4 and 25. 4 + 25 equals 29 .

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. 1/2s = 29 - 25 • Let's work on the right side first. 29 - 25 is 4 . • To find the value of s , we have to get it alone on the left side of the equation. This means getting rid of 1/2 . To do this, we'll multiply each side by the inverse of 1/2: 2 . 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 .

Here's Problem 2:

Answer: $70 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 amount Flor donated is three times as much the amount Mo donated • Flor and Mo's donations add up to$280 total

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 .

m + 3m = 280

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

• To get the correct answer, we'll have to get m alone on one side of the equation.
• To start, let's add m and 3 m . That's 4 m .
• We can get rid of the 4 next to the m by dividing both sides by 4. 4 m / 4 is m , and 280 / 4 is 70 .

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. 70 + 210 = 280 • 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. /en/algebra-topics/distance-word-problems/content/

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## REAL WORLD PROBLEMS: How to Write Equations Based on Algebra Word Problems

I know that you often sit in class and wonder, "Why am I forced to learn about equations, Algebra and variables?"

But... trust me, there are real situations where you will use your knowledge of Algebra and solving equations to solve a problem that is not school related. And... if you can't, you're going to wish that you remembered how.

It might be a time when you are trying to figure out how much you should get paid for a job, or even more important, if you were paid enough for a job that you've done. It could also be a time when you are trying to figure out if you were over charged for a bill.

This is important stuff - when it comes time to spend YOUR money - you are going to want to make sure that you are getting paid enough and not spending more than you have to.

Ok... let's put all this newly learned knowledge to work.

There are a few rules to remember when writing Algebra equations:

## Writing Equations For Word Problems

• First, you want to identify the unknown, which is your variable. What are you trying to solve for? Identify the variable: Use the statement, Let x = _____. You can replace the x with whatever variable you are using.
• Look for key words that will help you write the equation. Highlight the key words and write an equation to match the problem.
• The following key words will help you write equations for Algebra word problems:

## subtraction

Multiplication.

double (2x)

triple (3x)

divided into

Let's look at an example of an algebra word problem.

## Example 1: Algebra Word Problems

Linda was selling tickets for the school play.  She sold 10 more adult tickets than children tickets and she sold twice as many senior tickets as children tickets.

• Let x represent the number of children's tickets sold.
• Write an expression to represent the number of adult tickets sold.
• Write an expression to represent the number of senior tickets sold.
• Adult tickets cost $5, children's tickets cost$2, and senior tickets cost $3. Linda made$700. Write an equation to represent the total ticket sales.
• How many children's tickets were sold for the play? How many adult tickets were sold? How many senior tickets were sold?

As you can see, this problem is massive!  There are 5 questions to answer with many expressions to write. 1. In this problem, the variable was defined for you.  Let x represent the number of children’s tickets sold tells what x stands for in this problem.  If this had not been done for you, you might have written it like this:

Let x = the number of children’s tickets sold

2. For the first expression, I knew that 10 more adult tickets were sold.  Since more means add, my expression was x +10 .  Since the direction asked for an expression, I don’t need an equal sign.  An equation is written with an equal sign and an expression is without an equal sign.  At this point we don’t know the total number of tickets.

3. For the second expression, I knew that my key words, twice as many meant two times as many.  So my expression was 2x .

4.  We know that to find the total price we have to multiply the price of each ticket by the number of tickets.  Take note that since x + 10 is the quantity of adult tickets, you must put it in parentheses!  So, when you multiply by the price of $5 you have to distribute the 5. 5. Once I solve for x, I know the number of children’s tickets and I can take my expressions that I wrote for #1 and substitute 50 for x to figure out how many adult and senior tickets were sold. ## Where Can You Find More Algebra Word Problems to Practice? Word problems are the most difficult type of problem to solve in math. So, where can you find quality word problems WITH a detailed solution? The Algebra Class E-course provides a lot of practice with solving word problems for every unit! The best part is.... if you have trouble with these types of problems, you can always find a step-by-step solution to guide you through the process! Click here for more information. The next example shows how to identify a constant within a word problem. ## Example 2 - Identifying a Constant A cell phone company charges a monthly rate of$12.95 and $0.25 a minute per call. The bill for m minutes is$21.20.

1. Write an equation that models this situation.

2. How many minutes were charged on this bill? ## Notes For Example 2

• $12.95 is a monthly rate. Since this is a set fee for each month, I know that this is a constant. The rate does not change; therefore, it is not associated with a variable. •$0.25 per minute per call requires a variable because the total amount will change based on the number of minutes. Therefore, we use the expression 0.25m
• You must solve the equation to determine the value for m, which is the number of minutes charged.

The last example is a word problem that requires an equation with variables on both sides.

## Example 3 - Equations with Variables on Both Sides

You have $60 and your sister has$120. You are saving $7 per week and your sister is saving$5 per week. How long will it be before you and your sister have the same amount of money? Write an equation and solve. ## Notes for Example 3

• $60 and$120 are constants because this is the amount of money that they each have to begin with. This amount does not change.
• $7 per week and$5 per week are rates. They key word "per" in this situation means to multiply.
• The key word "same" in this problem means that I am going to set my two expressions equal to each other.
• When we set the two expressions equal, we now have an equation with variables on both sides.
• After solving the equation, you find that x = 30, which means that after 30 weeks, you and your sister will have the same amount of money.

I'm hoping that these three examples will help you as you solve real world problems in Algebra!

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## 3.7 Numeric Word Problems

Number-based word problems can be very confusing, and it takes practice to convert a word-based sentence into a mathematical equation. The best strategy to solve these problems is to identify keywords that can be pulled out of a sentence and use them to set up an algebraic equation.

Variables that are to be solved for are often written as “a number,” “an unknown,” or “a value.”

“Equal” is generally represented by the words “is,” “was,” “will be,” or “are.”

Addition is often stated as “more than,” “the sum of,” “added to,” “increased by,” “plus,” “all,” or “total.” Addition statements are quite often written backwards. An example of this is “three more than an unknown number,” which is written as $x + 3.$

Subtraction is often written as “less than,” “minus,” “decreased by,” “reduced by,” “subtracted from,” or “the difference of.” Subtraction statements are quite often written backwards. An example of this is “three less than an unknown number,” which is written as $x - 3.$

Multiplication can be seen in written problems with the words “times,” “the product of,” or “multiplied by.”

Division is generally found by a statement such as “divided by,” “the quotient of,” or “per.”

Example 3.7.1

28 less than five times a certain number is 232. What is the number?

• 28 less means that it is subtracted from the unknown number (write this as −28)
• five times an unknown number is written as $5x$
• is 232 means it equals 232 (write this as = 232)

Putting these pieces together and solving gives:

$\begin{array}{rrrrrr} 5x&-&28&=&232& \\ &+&28&&+28& \\ \hline &&5x&=&260& \\ \\ &&x&=&\dfrac{260}{5}&\text{or }52 \end{array}$

Example 3.7.2

Fifteen more than three times a number is the same as nine less than six times the number. What is the number?

• Fifteen more than three times a number is $3x + 15$ or $15 + 3x$
• nine less than six times the number is $6x-9$

Putting these parts together gives:

$\begin{array}{rrrrrrr} 3x&+&15&=&6x&-&9 \\ -6x&-&15&=&-6x&-&15 \\ \hline &&-3x&=&-24&& \\ \\ &&x&=&\dfrac{-24}{-3}&\text{or }8& \\ \end{array}$

Another type of number problem involves consecutive integers, consecutive odd integers, or consecutive even integers. Consecutive integers are numbers that come one after the other, such as 3, 4, 5, 6, 7. The equation that relates consecutive integers is:

$x, x + 1, x + 2, x + 3, x + 4$

Consecutive odd integers and consecutive even integers both share the same equation, since every second number must be skipped to remain either odd (such as 3, 5, 7, 9) or even (2, 4, 6, 8). The equation that is used to represent consecutive odd or even integers is:

$x, x + 2, x + 4, x + 6, x + 8$

Example 3.7.3

The sum of three consecutive integers is 93. What are the integers?

The relationships described in equation form are as follows:

$x + x + 1 + x + 2 = 93$

Which reduces to:

$\begin{array}{rrrrrr} 3x&+&3&=&93& \\ &-&3&&-3& \\ \hline &&3x&=&90& \\ \\ &&x&=&\dfrac{90}{3}&\text{or }30 \\ \end{array}$

This means that the three consecutive integers are 30, 31, and 32.

Example 3.7.4

The sum of three consecutive even integers is 246. What are the integers?

$x + x + 2 + x + 4 = 246$

$\begin{array}{rrrrrr} 3x&+&6&=&246& \\ &-&6&&-6& \\ \hline &&3x&=&240& \\ \\ &&x&=&\dfrac{240}{3}&\text{ or }80 \\ \end{array}$

This means that the three consecutive even integers are 80, 82, and 84.

For questions 1 to 8, write the formula defining each relationship. Do not solve.

• Five more than twice an unknown number is 25.
• Twelve more than 4 times an unknown number is 36.
• Three times an unknown number decreased by 8 is 22.
• Six times an unknown number less 8 is 22.
• When an unknown number is decreased by 8, the difference is half the unknown number.
• When an unknown number is decreased by 4, the difference is half the unknown number.
• The sum of three consecutive integers is 21.
• The sum of the first two of three odd consecutive integers, less the third, is 5.

For questions 9 to 16, write and solve the equation describing each relationship.

• When five is added to three times a certain number, the result is 17. What is the number?
• If five is subtracted from three times a certain number, the result is 10. What is the number?
• Sixty more than nine times a number is the same as two less than ten times the number. What is the number?
• Eleven less than seven times a number is five more than six times the number. Find the number.
• The sum of three consecutive integers is 108. What are the integers?
• The sum of three consecutive integers is −126. What are the integers?
• Find three consecutive integers such that the sum of the first, twice the second, and three times the third is −76.
• Find three consecutive odd integers such that the sum of the first, two times the second, and three times the third is 70.  • Kindergarten
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## Equation Word Problems Worksheets

This compilation of a meticulously drafted equation word problems worksheets is designed to get students to write and solve a variety of one-step, two-step and multi-step equations that involve integers, fractions, and decimals. These worksheets are best suited for students in grade 6 through high school. Click on the 'Free' icons to sample our handouts. One Step Equation Word Problem Worksheets

Read and solve this series of word problems that involve one-step equations. Apply basic operations to find the value of unknowns.

(15 Worksheets) Two-Step Equation Word Problems: Integers

Interpret this set of word problems that require two-step operations to solve the equations. Each printable worksheet has five word problems ideal for 6th grade, 7th grade, and 8th grade students. Two-Step Equation Word Problems: Fractions and Decimals

Read each word problem and set up the two-step equation. Solve the equation and find the solution. This selection of worksheets includes both fractions and decimals. MCQ - Two-Step Equation Word Problems

Pick the correct two-step equation that best matches word problems presented here. Evaluate the ability of students to solve two-step equations with this array of MCQ worksheets. Multi-Step Equation Word Problems: Integers

Read each multi-step word problem in these high school pdf worksheets and set up the equation. Solve and find the value of the unknown. More than two steps are required to solve the problems. Multi-step equation Word Problems: Fractions and Decimals

Write multi-step equations that involve both fractions and decimals based on the word problems provided here. Validate your responses with our answer keys. Related Worksheets

» One-step Equation

» Two-step Equation

» Multi-step Equation

» Algebraic Identities

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Testimonial Members have exclusive facilities to download an individual worksheet, or an entire level.  #### IMAGES

1. [PDF] Learning to Automatically Solve Algebra Word Problems 2. Writing Equations From Word Problems Worksheet 2 3. Algebraic Sentences Word Problems 4. PPT 5. Pin on Kindergarten Worksheets 6. Algebra Word Problems #### VIDEO

1. STATEMENTS & ALGEBRAIC EXPRESSIONS

2. A Nice Algebra Equation

3. MIP 1502| word problems

4. Solving an algebraic equation 2

5. Solving Algebra Equations

6. Algebra Question

1. What Are Some Challenging Algebra I Questions and Answers?

The more challenging Algebra 1 problems are quadratic equations of the form ax^2 +bx +c =0, where the general solution is given by the quadratic formula: x = (-b +/- sqrt(b^2-4ac))/2a (where sqrt means a square root of the term in parenthes...

2. What Is a Demand Equation?

A demand equation is an algebraic representation of product price and quantity. Because demand can be represented graphically as a straight line with price on the y-axis and quantity on the x-axis, a demand equation can be as basic as a lin...

3. What Is a Mathematical Verbal Expression?

A mathematical verbal expression is a translation into words of an algebraic expression that can consist of different operations, numbers and variables. In math, students also may have to translate a verbal expression into an algebraic expr...

4. Algebraic word problems

and a single variable. Usually, the

5. Algebraic word problems

There was a depth of 18 inches of snow on the lawn when the storm ended and then it started melting at a rate of 1.5 inches per hour. Write an equation for S,S

6. SSAT Upper Level Math : Algebraic Word Problems

Explanation: To solve for the final score: Add the five past test scores and you get 454. Then set up an algebraic equation where you add 454 to

7. Algebra Topics: Introduction to Word Problems

Word problems in algebra · Step 1: Read through the problem carefully. · Step 2: Represent unknown numbers with variables. · Step 3: Translate the rest of the

8. Algebraic Sentences Word Problems

An algebraic sentence when written in equation form involves algebraic expressions (which contain variables such as letters in the alphabet), constants, and an

9. Algebra Word Problems Packet #1 On a separate sheet of paper

Solve each problem using an algebraic equation. 1. Twice a number is 500 more than six times the number. What is the number? 2. Three-sevenths of a

10. How to Write Equations Based on Algebra Word Problems

Example 1: Algebra Word Problems · Let x represent the number of children's tickets sold. · Write an expression to represent the number of adult tickets sold.

11. 3.7 Numeric Word Problems

The best strategy to solve these problems is to identify keywords that can be pulled out of a sentence and use them to set up an algebraic equation. Variables

12. 5 Tricks for Solving Algebra Word Problems

What Are the 4 Steps for Solving Algebra Word Problems? · Define and identify your variable. · Write an equation that uses the variable. · Solve

13. Algebra Word Problems Lesson 1 Worksheet 1 Algebra Word ...

Problem 1) Five times a number increased by seven is equal to forty-seven. What is the number? Solution: Let equal the number. Write and solve an equation: 5

14. Equation Word Problems Worksheets

Grab these equation word problem worksheets tailor-made for students of grade 6 through 8 to hone one-step, two-step and multi-step equation word problems.