Word problems using expressions

Variables and expressions

These word problems involve expressions with variables. Students write expressions (using addition, subtraction, multiplication, division & fractions) which include an unknown represented by a variable.

Grade 5 Expressions and Variables word problem worksheets

Rational Equations: Applications — Work

He’s not as efficient as Anya, so it takes them a little bit longer. This information right here. When both pipes are opened, they fill the pool in five hours. So we figured out Anya. Shalini can run 3 miles per hour faster than her sister Meena can walk. We are actually adding the Work they complete together and alone using formula , where the Time is 1 hour or whatever the unit is.

So the number of hours it takes Bill to paint the deck is twice as long as the number of hours it takes Anya to paint a deck. And then we know that it takes Bill twice as long as Anya. Anya and Bill stained a large porch deck in 8 hours. It takes Jill 3 hours to make sandwiches for a charity organization as many as they need. Find the average speed rate of the canoe in still water. How much part of the water melon will each get including Mary? We solve these problems directly by performing the operations.

Takes them 8 hours. Let me scroll down a little bit. In these kind of problems, we have to solve for x. The part of the house cleaned by jack is: Let t be the time taken to complete the job together. Now, let’s do another variable for Bill just like that. If the two hoses are working, and the drain is open by mistake , how long will it take to fill the swimming pool?

So we can rewrite this equation as– I’ll stick to the colors for now. Let the whole water melon be denoted by 1. So, Daniel reaches the house at The individual rates of each worker is , and when you multiply the rate by the time they work, you get.

How long would it take to fill the pool if only the slower pipe is used? But it makes sense that combined they’re going to take less time than individually. Solving Word Problems with Rational Expressions A rational expression is an expression which is expressed in the form of two polynomials.

Combined rates rational equations Video transcript Working together Anya and Bill stained a large porch deck in 8 hours. If the first painter can do the entire job in twelve hours and the second painter can do it in eight hours, then this here is the trick! So the number of hours it takes Anya to paint a deck, or A, is 12 hours. Plus– do that same color. The year before Bill painted it by himself, but took twice as long as Anya did. One pipe can fill a pool 1.

Here are some examples. For word problems with rational expressions, first we have to convert the word problem into a rational expression and then, we can easily solve the rational expression. So B is equal to 2A.

Published
August 30, 2017

An algebraic sentence when written in equation form involves algebraic expressions (which contain variables such as letters in the alphabet), constants, and an equal symbol. Each algebraic sentence may contain a combination of algebraic expressions and constants, or with just two or more algebraic expressions. Almost always, the word “is” in an algebraic sentence denotes the symbol of equality.

Five more than twice a number is 43 can be written as 2x+5=43

In our example above, the algebraic sentence, “Five more than twice a number is forty-three“, is translated and written into its equation form: 2x + 5 = 43.

But before we delve into solving word problems that involve algebraic sentences, it’s crucial that we become familiar with how to translate and write algebraic expressions.


Algebraic Expressions

Learning how to write and translate algebraic expressions is the foundation in writing algebraic sentences. Eventually, we will use this knowledge to write algebraic equations where we solve for the value of the unknown variable.

What is an algebraic expression?

Think of an algebraic expression as the equivalent of a phrase in the English language. It does not convey a complete thought but is an important building block in constructing a sentence. Algebraic expressions consist of numbers, variables, and arithmetic operations.

Let’s take a quick look at some math phrases that were translated and written into algebraic expressions.

  • the sum of a number and 8 large{,,, to ,,,, {color{red}{c + 8}}}
  • 3 less than a number large{,,, to ,,,, {color{red}{y — 3}}}
  • the product of 45 and m large{,,, to ,,,,{color{red}{45m}}}
  • a number divided by 2 large{,,, to ,,,,Large {color{red}{x over 2}}}

Translating and Writing Algebraic Sentences

Once you know how to translate math phrases into algebraic expressions, it’ll be easy for you to translate and write algebraic sentences in an equation form as well.

For instance, how do you write “the difference of 16 and k is seven” into an algebraic equation?

Let’s deconstruct this algebraic sentence.

The difference of 16 and k is 7 can be written as 16-k=7.

You may notice that we translated “the difference of 16 and k” into an algebraic expression, 16-k, then used the equal symbol (=) in place of the word “is”.

As I mentioned at the beginning of this lesson, the word “is” in an algebraic sentence, most of the time, signifies the equality symbol which is the case in this example.

Therefore, we can write this algebraic sentence in equation form as 16 — k = 7.

Now that we are confident in translating algebraic sentences and writing them in an equation format, it’s time for us to take one step further. We will not only translate and write algebraic sentences into algebraic equations but also proceed to solve them.


Examples of Algebraic Sentences Word Problems

The main key when solving word problems with algebraic sentences is to accurately translate the algebraic expressions then set up and write each algebraic equation correctly. In doing so, we can ensure that we are solving the right equation and as a result, will get the correct answer for each word problem.

Example 1: Six more than seven times a number is thirty-four. Find the number.

First, let’s deconstruct the algebraic sentence. It’s important that we identify and separate the algebraic expressions from the constants as well as determine if the problem suggests equality between the terms.

For the unknown value, we’ll use large{x} as our variable.

algebraic equation: 6+7x=34 or 7x+6=34

Note: Since addition is commutative, changing the order of the addends on the left side of the equation does not change the sum. Therefore, we may also write the algebraic expression as 7x+6 or the algebraic equation as 7x + 6 = 34 instead.

However, for our discussion, we will use 6 + 7x = 34 as our equation.

The original problem asks us to find the number which in this case is large{x}. So our next step is to solve for large{x} in our equation, 6 + 7x = 34.

Solution:

x=4

Now, we’ll evaluate the value of large{x = 4} to verify if the number we found satisfies the original algebraic sentence.

Remember that we are asked to find the identity of “the number” in our algebraic sentence and NOT the value of the variable large{x}. Therefore, it will be incorrect to say that large{x=4} is the answer. This is a common mistake that we always need to pay attention to.

We’ll replace the variable large{x} with the number 4 to see if the algebraic expression on the left side of the equation also results to 34.

6+7(4)=34

Yes, it does! This means that number 4 is the answer.


Example 2: The difference between three times a number and five is sixteen. Find the number.

Right away, the word “difference” in our algebraic sentence gives us a clue that we will be using the subtraction operation. But, unlike, in addition, the order of the terms within an expression matters in subtraction. So we have to make sure that the terms in our algebraic expression are set up in the correct order.

Another keyword is “times”, which suggests that 3 is being multiplied to a number whose value is currently unknown.

This time, let’s use large{g} as our variable. Start by translating the algebraic expression correctly then continue to write the equation.

algebraic equation: 3g-5=16

Now that we have our equation, let’s find out what the unknown value is by solving for the variable, large{g}.

Solution:

g=7

As discussed in our first example, it is important that we verify if the result of our solution makes the algebraic equation true. For this one, I will leave it up to you to evaluate the value of large{g = 7}. Remember to replace the unknown value of the variable large{g} with the number 7 to see if the left side of the equation is also equal to 16.

Since the original algebraic sentence requires us to find the number, then the answer is number 7.


Example 3: A number decreased by half of the number is four. Find the number.

This algebraic sentence is quite interesting because we don’t just have one unknown value, but two! First, we have the “number” (the unknown) then the “half of the number”, which means one-half of the unknown.

The keyword “half of” also indicates multiplying something by one-half while the “decreased by” tells us that we will be subtracting one term from the other.

To organize our thoughts, let’s deconstruct this algebraic sentence into its meat and potatoes.

algebraic equation: m-[(1/2)m]=4

Being able to break apart our algebraic sentence into its basic parts allows us to read the sentence thoroughly and understand the relationship between its quantities.

After setting up our algebraic equation, our next step is to solve it for large{m}.

Solution:

m=8

We need to evaluate the value of large{m = 8} to check if the number we got makes our algebraic equation true.

four equals four

Great! The values on both sides of the equation are equal to each other. Therefore, the number 8 is the correct answer to our original algebraic sentence.


Example 4: Four times the sum of twice a number and six is thirty-two. Find the number.

Looking closely, we see a few keywords that would help us in translating this algebraic sentence into an equation.

  • times – means that we’ll be multiplying 4 by the quantity, “sum of twice a number and six”
  • sum – means that we need to add the terms, “twice a number” and six
  • twice – means that a number (the unknown value) is multiplied by 2

With these in mind, let’s write our algebraic equation.

algebraic equation: 4(2d+6)=32

Notice that we used the variable large{d} in our equation to stand for our unknown value. Let’s now proceed and solve for large{d} and afterward, check if the value we get indeed makes the equation true.

Solution:

d=1

Evaluate the value of large{d = 1}:

32=32

Looks like everything checks out, so the answer to our algebraic sentence is the number 1.


Example 5: Two-thirds of the sum of three times a number and six is ten. What is the number?

This problem involves a fraction and an algebraic expression as you can tell. The keyword “of” indicates multiplication, so it tells us that we need to multiply the fraction large{2 over 3} by the sum of the two given quantities (“three times a number” and six).

Let’s move on and write our algebraic equation. We’ll use the variable large{y} as the placeholder for our “number” whose value is currently unknown.

algebraic equation: (2/3)(3y+6)=10

Awesome! We are able to translate our algebraic sentence into an equation. It’s time for us to solve for large{y} then I’ll leave it up to you to do a check if the number we find as the value of large{y} makes the algebraic equation true.

Solution:

y=3

Once you evaluate large{y = 3}, you’ll see that the answer to our word problem is indeed the number 3.


Example 6: Twice the difference of a number and three is four more than the number. Find the number.

We have an example here that is a little different from the word problems we’ve covered so far. In this algebraic sentence, both sides of the equation contain an algebraic expression. However, the keywords present will help us translate it correctly into an equation so there’s nothing to worry about.

Keywords:

  • twice – means that we will multiply the quantity (difference of a number and three) by 2
  • difference – tells us to use the subtraction operation between our unknown value (“number”) and three
  • more than – means that we need to add four to our unknown value
twice the quantity a minus 3 equals the sum of a and 4

Let’s solve for our unknown value, large{a}, then verify that our result makes the algebraic equation true.

Solution:

a equals 10

Evaluate the value of large{a = 10}:

fourteen equals fourteen

Perfect! The value we got for large{a} which is 10, made our equation equal on both sides. Thus, the answer is the number 10.


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Algebraic Expressions

Algebraic Expressions Worked Examples

Lesson Intro: Expressions with Variables

In this lesson, Juni Mathematics instructor Kadyn talks about variables and variable expressions — foundational concepts in pre-algebra. Variables are important to know for higher levels of math, and are even used similarly in other subjects like computer science!

Read Kadyn’s Intro to Variable Expressions lesson first to understand what variables are and how to use them in expressions. Then, use what you’ve learned with Kadyn’s word problems below to translate sentences and real-life problems into expressions with unknown variables. Learn to solve for unknowns, and tackle real-world applications.

Once you’ve checked your answers, you can also keep practicing using variables with Variable Expressions Drills and Warmup Problems.

Problem 1:

Write out the following mathematical expression in full and then in simplified form: negative three times the difference of five times x and the absolute value of negative two minus three.

Problem 2:

Ted is hosting a birthday party. If you take the number of friends he invited and multiply it by 4 and subtract it by 7, you get 53. How many people did Ted invite to his party?

Problem 3:

Carmen wants to collect stamps and the store charges $3 for a pack of 12 stamps. How many stamps will Carmen have if she spent $36 on stamps? (disregard taxes)

Problem 4:

Peter wants to build a rectangular fence around his yard. One side of the fence is 3m shorter than 4 times the other side. If the shorter side of the fence is 15m, what is the perimeter of the fence in meters?

Problem 5:

Elliott and Gretchen are renting their own apartments in the city. Elliott has paid $10,000 in rent so far and Gretchen has paid $8,000. If both of them have been renting their own place for the same number of months and Gretchen’s monthly rate is $1,000, how much is Elliott’s monthly rent?


Find Solutions Below


Answer Key

Problem 1

-3 ⋅ (5x — |-2 -3|) = -3 ⋅ (5x — 5) = -15x + 15

Full form: -3 ⋅ (5x — |-2 -3|)

Simplified form: -15x + 15

Problem 2

Ted invited 15 people to his birthday party.

Let the number of people at Ted’s birthday party be n. Then n ⋅ 4 — 7 = 53 → n ⋅ 4 = 60. Then we know that n = 15.

Problem 3

Carmen has 144 stamps.

Let the number of stamps that Carmen has be s. Since she spent $36 on stamps and each pack of stamps cost $3, she then bought 36 3 = 12 packs. Since each pack comes in with 12 stamps, she has 12 ⋅ 12 = 144 stamps.

Problem 4

The perimeter of the fence is 144m.

Since the shorter side of the fence is 15m long, the longer side of the fence is 4 ⋅ 15 — 3 = 57m long. Then the perimeter of the fence is 15 + 15 + 57 + 57 = 144m.

Problem 5

Elliott’s monthly rent is $1,250.

Since Gretchen has paid $8,000 in rent and her monthly rent is $1,000, she has lived at her place for 8,000 1,000 = 8 months. Since Gretchen has lived at her place as long as Elliott has lived at his, we know that Elliott has lived at his place for 8 months as well. Then since Elliott has paid $10,000 in rent, his monthly rate is $10,000 8 = $1,250.

More Exercises on Variables

We hope you enjoyed Kadyn’s Word Problems with Variable Expressions! This lesson falls under our Pre-Algebra A course curriculum.

Continue practicing variable expressions with warmup problems and practice drills below. Or, review key terms and concepts with Kadyn’s Intro to Variables lesson.

  • Intro to Variable Expressions: Definitions and Approaches
  • Warmup Questions with Variable Expressions
  • Variable Expressions Drills

Need help or want to keep learning?

To keep practicing or learning, please check out all of our math and coding tutorials on our Home Learning Resources page.

Need help? Looking up your questions is one of the best ways to learn! Another great way to learn is from an experienced math instructor. Read more about our online math courses or speak with a Juni Advisor by calling __(650) 263-4306__ or emailing __advisors@learnwithjuni.com__.

When a problem contains a word for an unknown number or quantity, we use a variable to represent that unknown number or quantity. Just replace those unknowns with x, or y, or z, or just about any other letter. We could also replace them with some other variable, but these are the three that get the most play. They must have great agents.

Here are some English phrases translated into mathematical symbols.

  • «A number plus four» translates to x + 4.
     
  • «A quantity doubled» translates to 2x.
     
  • «One third of an amount» is .
     
  • «Two less than a value» is x – 2.

Sometimes a problem already contains a variable and asks for some other quantity in terms of that variable. In that case, we don’t need to plug in and solve anything. Instead, we only need to describe the quantity using an expression that has the variable in it. It comes in handy when your mother storms into your room demanding to know how many chocolate chip cookies you ate, and you can tell her, «x – 3.» She can’t possibly deduce from that remark how many you ate, and you’ll be off the hook. We promise.

Sample Problem

Alice has x dollars. Tanya has 3 dollars more than Alice. How many dollars does Tanya have?

Tanya has (the amount Alice has) + 3, which is x + 3 dollars.

This may actually be good news for Alice. She’ll be in a lower tax bracket.

Sometimes a problem doesn’t come right out and say there’s a variable, but there’s still one hiding in there somewhere. When all else fails, check under the bed and in the closet.

Sample Problem

Dwight has some brownies, but Bill has three times as many brownies. Lucky Bill. Express the number of brownies Bill has in terms of the number of brownies Dwight has. Not to make Dwight feel bad about it or anything, just so’s we know.

Bill has 3(the number of brownies Dwight has). That’s kinda obnoxious to write out, though.

We need a variable to express the number of brownies that Dwight has. Since the problem didn’t tell us what letter to use, we can use whatever letter we want. Let’s use D for «Dwight.» Bet you didn’t see that coming. Then we can answer the question like this:

«Let D be the number of brownies Dwight has. Then Bill has 3D brownies.»

Wow. They’re so real-looking you could almost reach out and touch them.

When a problem doesn’t tell us what letters to use for variables, we can pick our own. This situation is great for us, because we have control issues. When we do this, it’s very, very, very important to write down—somewhere, anywhere—the meanings of the letters that we’re using. Otherwise, we might arrive at the end of the problem and not remember if x is height, speed, or the number of peanuts per mile. One would hope that you’d never forget something as important as the variable that stands for peanuts per mile, but stranger things have happened.

If a student answered the previous problem by saying «Bill has 3D brownies» but never bothered to say what D was, someone could be confused. Either the teacher trying to read and grade the homework, or the student trying to use the homework to study for a test later. With enough time (and brownies) you can probably figure it out, but why leave it to chance?

This is one of those little details that can make math way easier than it would be otherwise. You’ll be on your way to the next problem, while your friends are busy trying to figure out what x means. Man, there’s nothing you love more than leaving your friends in the dust.

Be Careful: Whenever you introduce a new variable, write down what that variable means so you can remember when you come to the end of the problem. Don’t trust your brain to remember. It has failed you before, and it will fail you again.

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Writing the phrase as an algebraic expression

Example

Write the phrase as an algebraic expression.

“Four less than twice ???x???”

The phrase “twice ???x???” means “???2??? times ???x???” which we know means to multiply, and so we can write it as ???2x???. Now we have

“Four less than ???2x???”

Less means subtraction, so we’ll subtract ???4??? from ???2x???.

???2x-4???

It may be tempting to write the ???4??? first and subtract ???2x???. Let’s use numbers to help us visualize this. When we say ???4??? less than ???10??? you know that you’d have to subtract ???4??? from ???10???, written out as ???10-4???. So ???4??? less than ???2x??? will be ???2x-4???.

Example

Find the value of the expression.

???frac{1}{4}??? of ???120???

In math, the word “of” (immediately after a proper or improper fraction) tells us to multiply. Therefore, the mathematical expression of the phrase will be

???frac{1}{4} cdot 120???

Because we were asked to actually find the value of the expression, we’ll perform the multiplication to get the simplified value.

???30???

Not only can we translate phrases into expressions, but we can write equations from some phrases as well.

For instance, suppose you wanted to use algebra to solve the following word problem:

John’s age is four less than twice Mary’s age. If Mary is ???18???, how old is John?

The first step in solving a word problem like this is to define the variables. What that means is to state the particular quantity that each variable stands for.

In this problem, we have two quantities: Mary’s age and John’s age. So we’ll define the variables by saying “Let ???x??? be Mary’s age, and let ???y??? be John’s age.” (We could use any letters of the alphabet for the variables, but people often use ???x??? for one of the variables, and if there are one or two additional variables, they tend to use ???y??? and ???z???, in that order.)

The next step in solving a word problem is to “translate” each word or phrase into mathematical symbols. Here, “John’s age” is translated as “???2x-4???.”

How about the word “is” (in “John’s age is four less than twice Mary’s age”)? Well, “is” is translated as an equals sign. To see this, it may help to think of the word “is” as having the same meaning (in math) as “is equal to.”

Combining all of these, we get the equation

???y=2x-4???

The third step in solving a word problem is to use the given data and solve the equation. Here, we’re given Mary’s age as ???18???, so we substitute ???18??? for ???x??? and then solve for ???y???.

???y=2(18)-4???

???y=36-4???

???y=32???

The final step is to answer the question that was asked. Here, we’re asked for John’s age. Since we defined ???y??? as John’s age, the answer is ???32???.

Suppose we’d been given the following word problem instead:

Currently, John’s age is four less than twice Mary’s age. If Mary is now ???18???, how old will John be seven years from now?

To solve this problem, it would be convenient to define ???x??? as Mary’s age now, and ???y??? as John’s age now, because we’re given a relationship between Mary’s age (now) and John’s age (now). Then in the last step (answering the question that was asked), we’d have to evaluate ???y+7??? (to get John’s age seven years from now), and our answer would be ???32+7=39???.

Get access to the complete Algebra 1 course

Need extra help for EL students? Try the Explaining Expressions pre-lesson.

Learning Objectives

Students will be able to solve multi-step word problems by writing and solving expressions with the order of operations.

The adjustment to the whole group lesson is a modification to differentiate for children who are English learners.

EL adjustmentsOnOff

Expressions & Word Problems

  • Review PEMDAS (or GEMDAS). Hand out the Expressions & Word Problems worksheet and display its top section using a document camera. With student input, put the operations in order.
  • Remind students that this specific order is called the order of operations and it is important because it is a universal set of rules for how to solve multi-step problems.
  • Point out that multiplication and division are part of the same step. This means that these two operations are done during the same step, from left to right. The same is true for addition and subtraction.
  • Write the acronym (PEMDAS or GEMDAS) on the board in large letters and tell students that they can use it as a reminder throughout the lesson.
  • Explain that today students will practice using the order of operations and written statements to solve multi-step word problems.

Beginning

  • Allow students to use their home language (L1) or new language (L2) in their discussions throughout the lesson.
  • Have them tell you the meaning of PEMDAS.
  • Tell students to restate the meanings of the keywords, such as «multiplication» and «division.»

Intermediate

  • Write the academic language students can use throughout the lesson on the board as you introduce the language.
  • Have students define each of the components of PEMDAS (e.g., exponents, parentheses, etc.).
  • Ask students to identify the components of PEMDAS in the displayed worksheet.

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