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## Praxis Core Math

Course: praxis core math > unit 1.

- Algebraic properties | Lesson
- Algebraic properties | Worked example
- Solution procedures | Lesson
- Solution procedures | Worked example
- Equivalent expressions | Lesson
- Equivalent expressions | Worked example
- Creating expressions and equations | Lesson
- Creating expressions and equations | Worked example

## Algebraic word problems | Lesson

- Algebraic word problems | Worked example
- Linear equations | Lesson
- Linear equations | Worked example
- Quadratic equations | Lesson
- Quadratic equations | Worked example

## What are algebraic word problems?

What skills are needed.

- Translating sentences to equations
- Solving linear equations with one variable
- Evaluating algebraic expressions
- Solving problems using Venn diagrams

## How do we solve algebraic word problems?

- Define a variable.
- Write an equation using the variable.
- Solve the equation.
- If the variable is not the answer to the word problem, use the variable to calculate the answer.

## What's a Venn diagram?

- Your answer should be
- an integer, like 6
- a simplified proper fraction, like 3 / 5
- a simplified improper fraction, like 7 / 4
- a mixed number, like 1 3 / 4
- an exact decimal, like 0.75
- a multiple of pi, like 12 pi or 2 / 3 pi
- (Choice A) $ 4 A $ 4
- (Choice B) $ 5 B $ 5
- (Choice C) $ 9 C $ 9
- (Choice D) $ 14 D $ 14
- (Choice E) $ 20 E $ 20
- (Choice A) 10 A 10
- (Choice B) 12 B 12
- (Choice C) 24 C 24
- (Choice D) 30 D 30
- (Choice E) 32 E 32
- (Choice A) 4 A 4
- (Choice B) 10 B 10
- (Choice C) 14 C 14
- (Choice D) 18 D 18
- (Choice E) 22 E 22

## Things to remember

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## Algebraic Expressions and Word Problems

Related Topics: More Lessons for Grade 7 Math Worksheets

Examples, solutions, videos, worksheets, games and activities to help Algebra 1 or grade 7 students learn how to write algebraic expressions from word problems.

Beginning Algebra & Word Problem Steps

- Name what x is.
- Define everything in the problem in terms of x.
- Write the equation.
- Solve the equation.
- Kevin’s age is 3 years more than twice Jane’s age. The sum of their ages is 39. How old is Kevin and Jane?
- The difference between two numbers is 7. Find the two numbers if the larger number is three times the smaller.
- Mary and Jim collect baseball cards, Mary has 5 more than 3 times as many cards as Jim. The total number of cards they both have is 253. How many cards does Mary have?

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## How to Solve an Algebraic Expression

Last Updated: October 27, 2023 Fact Checked

This article was co-authored by David Jia . David Jia is an Academic Tutor and the Founder of LA Math Tutoring, a private tutoring company based in Los Angeles, California. With over 10 years of teaching experience, David works with students of all ages and grades in various subjects, as well as college admissions counseling and test preparation for the SAT, ACT, ISEE, and more. After attaining a perfect 800 math score and a 690 English score on the SAT, David was awarded the Dickinson Scholarship from the University of Miami, where he graduated with a Bachelor’s degree in Business Administration. Additionally, David has worked as an instructor for online videos for textbook companies such as Larson Texts, Big Ideas Learning, and Big Ideas Math. There are 10 references cited in this article, which can be found at the bottom of the page. This article has been fact-checked, ensuring the accuracy of any cited facts and confirming the authority of its sources. This article has been viewed 480,933 times.

An algebraic expression is a mathematical phrase that contains numbers and/or variables. Though it cannot be solved because it does not contain an equals sign (=), it can be simplified. You can, however, solve algebraic equations , which contain algebraic expressions separated by an equals sign. If you want to know how to master this mathematical concept, then see Step 1 to get started.

## Understanding the Basics

- Algebraic expression : 4x + 2
- Algebraic equation : 4x + 2 = 100

- 3x 2 + 5 + 4x 3 - x 2 + 2x 3 + 9 =
- 3x 2 - x 2 + 4x 3 + 2x 3 + 5 + 9 =
- 2x 2 + 6x 3 + 14

- You can see that each coefficient can be divisible by 3. Just "factor out" the number 3 by dividing each term by 3 to get your simplified equation.
- 3x/3 + 15/3 = 9x/3 + 30/3 =
- x + 5 = 3x + 10

- (3 + 5) 2 x 10 + 4
- First, follow P, the operation in the parentheses:
- = (8) 2 x 10 + 4
- Then, follow E, the operation of the exponent:
- = 64 x 10 + 4
- Next, do multiplication:
- And last, do addition:

- 5x + 15 = 65 =
- 5x/5 + 15/5 = 65/5 =
- x + 3 = 13 =

## Solve an Algebraic Equation

- 4x + 16 = 25 -3x =
- 4x = 25 -16 - 3x
- 4x + 3x = 25 -16 =
- 7x/7 = 9/7 =

- First, subtract 12 from both sides.
- 2x 2 + 12 -12 = 44 -12 =
- Next, divide both sides by 2.
- 2x 2 /2 = 32/2 =
- Solve by taking the square root of both sides, since that will turn x 2 into x.
- √x 2 = √16 =
- State both answers:x = 4, -4

- First, cross multiply to get rid of the fraction. You have to multiply the numerator of one fraction by the denominator of the other.
- (x + 3) x 3 = 2 x 6 =
- Now, combine like terms. Combine the constant terms, 9 and 12, by subtracting 9 from both sides.
- 3x + 9 - 9 = 12 - 9 =
- Isolate the variable, x, by dividing both sides by 3 and you've got your answer.
- 3x/3 = 3/3 =

- First, move everything that isn't under the radical sign to the other side of the equation:
- √(2x+9) = 5
- Then, square both sides to remove the radical:
- (√(2x+9)) 2 = 5 2 =
- Now, solve the equation as you normally would by combining the constants and isolating the variable:
- 2x = 25 - 9 =

- |4x +2| - 6 = 8 =
- |4x +2| = 8 + 6 =
- |4x +2| = 14 =
- 4x + 2 = 14 =
- Now, solve again by flipping the sign of the term on the other side of the equation after you've isolated the absolute value:
- 4x + 2 = -14
- 4x = -14 -2
- 4x/4 = -16/4 =
- Now, just state both answers: x = -4, 3

## Community Q&A

- The degree of a polynomial is the highest power within the terms. Thanks Helpful 9 Not Helpful 1
- Once you're done, replace the variable with the answer, and solve the sum to see if it makes sense. If it does, then, congratulations! You just solved an algebraic equation! Thanks Helpful 7 Not Helpful 3
- To cross-check your answer, visit wolfram-alpha.com. They give the answer and often the two steps. Thanks Helpful 8 Not Helpful 5

## You Might Also Like

- ↑ https://www.math4texas.org/Page/527
- ↑ https://www.khanacademy.org/math/cc-sixth-grade-math/cc-6th-expressions-and-variables/cc-6th-combining-like-terms/v/combining-like-terms-2
- ↑ https://www.mathsisfun.com/algebra/factoring.html
- ↑ https://www.mathsisfun.com/operation-order-pemdas.html
- ↑ https://sciencing.com/tips-for-solving-algebraic-equations-13712207.html
- ↑ https://www.mathsisfun.com/algebra/equations-solving.html
- ↑ https://tutorial.math.lamar.edu/Classes/Alg/SolveExpEqns.aspx
- ↑ https://www.mathsisfun.com/algebra/fractions-algebra.html
- ↑ https://math.libretexts.org/Courses/Coastline_College/Math_C045%3A_Beginning_and_Intermediate_Algebra_(Chau_Duc_Tran)/10%3A_Roots_and_Radicals/10.07%3A_Solve_Radical_Equations
- ↑ https://www.mathplanet.com/education/algebra-1/linear-inequalitites/solving-absolute-value-equations-and-inequalities

## About This Article

If you want to solve an algebraic expression, first understand that expressions, unlike equations, are mathematical phrase that can contain numbers and/or variables but cannot be solved. For example, 4x + 2 is an expression. To reduce the expression, combine like terms, for example everything with the same variable. After you've done that, factor numbers by finding the lowest common denominator. Then, use the order of operations, which is known by the acronym PEMDAS, to reduce or solve the problem. To learn how to solve algebraic equations, keep scrolling! Did this summary help you? Yes No

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## Formulating Algebraic Equations from Word Problems

Content objectives.

If you have 4 pencils and I have 7 apples, how many pancakes will fit on the roof? Purple, because aliens don’t wear hats. 1

This is a meme that I have encountered in various forms on social media, and even from students joking around in person. Word problems seem to be a commonly cited justification for hating math, particularly algebra. Jokes aside, in my experience as a secondary math teacher, even students who are skilled in solving given algebraic equations struggle in creating their own equation to solve a word problem. This is not so surprising, because the demands of solving an equation are very different from the demands of formulating an equation from a verbal or written context. To formulate an algebraic equation that correctly represents a situation, students must interpret a verbal or written statement, and symbolically represent the relationship of its parts. Thus, to improve student success with formulating algebraic equations, we must provide instruction in both the process of decoding a problem, and using mathematical notation that maintains the meaning of the original problem.

The purpose of this curriculum unit is to help students translate word problems into first order equations that can be written in the form ax + b = c. For purposes of this unit, a, b, and c will be whole numbers. To do this, I will help students identify and organize the given information, the unknown value, and the relationship of the values of the problem through verbal discussions and visual models. By organizing this information, students will have an easier time representing these quantities and relationships symbolically. In the long run, I hope this curriculum unit also gives my students confidence to approach more complicated word problems as they progress further in math.

This unit is the first of four complementary curriculum units developed in the 2017 Yale National Initiative (YNI) math seminar, “From Arithmetic to Algebra,” about translating, simplifying, and solving algebraic expressions and equations. For the rest of the related curriculum units, please see the work of 2017 math Fellows Jeffrey Rossiter, Xiomara Pacheco, and Sally Yoo. (See Figure 1 below, which summarizes the connection between the concepts of our pieces.)

This order is surprising because, in most textbooks, translating and simplifying algebraic expressions comes first, followed by solving equations, and ending with word problems. However, starting with word problems provides the motivation for using algebraic notation to represent increasingly complex situations. While bar models are useful in representing, and even solving simple word problems, students will need more sophisticated skills in representing and manipulating equations with compound expressions. Therefore, the intention of this curriculum unit is to merely begin a broader investigation into algebraic expressions and equations, which may last for an entire semester.

I teach at Peter Burnett Middle School in San Jose, California. Currently, Burnett Middle School’s overall math performance on the California Assessment of Student Progress and Performance (CAASPP) falls well below the state average. 2 Last year, 13% of Burnett’s students met or exceeded standards, compared to a 37% statewide average. Since CAASPP results are one of the measures used in course placement in my district, I worry that these scores will limit students’ opportunities to take algebra 1, and serve as indicators of their likelihood to succeed in that course. Moreover, having taught or tutored every higher-level course from geometry to calculus, I know that algebra skills are necessary to take and succeed in each of these classes as well. So very few of Burnett’s students are on track to take any of these higher-level math classes, which will limit their opportunities in college. Colleges will want students to take remedial math, which may lead to falling behind in their other courses, and hindering their path to a diploma. Even those who go directly into the workforce, or training outside of higher education, may be affected. Among high school graduates who don’t attend college, “those with low levels of math are 50 percent more likely to be unemployed than those with higher levels of math, and they typically earn about $ 1.30 less per hour.” 3 So while the CAASPP itself is a single test, what the scores reveal about math competence among Burnett students can have much longer-lasting repercussions. The goal of this unit is to improve my students’ algebra skills, and consequently, their future opportunities.

This curriculum unit targets students in 7 th grade math, which covers pre-algebra concepts and skills. Similar to whole school performance stated above, 13% of Burnett 7th graders met or exceeded math standards on the CAASPP, compared to 36% statewide. This number covers all 7th graders, including those who are in an additional math support course, those who are “on grade level”, and those who are in advanced placement (7th grade accelerated, or Algebra 1). My curriculum unit will focus on students “on grade level” who, despite that classification, underperform on district benchmark exams. Fewer than 1 in 25 of my 55 “grade level” students were proficient on the 2016-2017 exam on my district’s Algebraic Expressions and Equations, which assessed using variables to represent real-world quantities, and formulate simple equations to solve those problems. (See APPENDIX A for full list of standards.) The primary objective of this unit will be to equip students with the tools to interpret a word problem, and to write an algebraic equation that accurately relates its quantities.

In the 2016-2017 school year, my district started using the Springboard Course 2 Textbook for 7th grade math; in this book, there is only one lesson on “Modeling and Writing Two-Step Equations”. After teaching this lesson, I have concluded that it has several deficiencies. First, the lesson over-scaffolds the translation process, by providing a numeric expression to represent the word problem, and by defining the variable for students. Then, the lesson over-simplifies the distinction between an equation and an expression. Specifically, equations “have an equal sign”, and expressions do not. Finally, the lesson introduces vocabulary, such as “coefficient” and “constant”, without connecting back to the word problem. As a result of these shortcomings, students do not get enough practice in creating the structures of the expression or equation themselves, including operations, or in creating appropriate notation for unknown values.

6th grade standards also cover some of the skills needed in translating situations into expressions or equations. By 7 th grade, students should already know how to use variables to represent unknown numbers, and write expressions and equations to solve real world problems; specifically, in the 6 th grade, students practice problems that can be written as one-step addition or multiplication equations. (See APPENDIX A for full list of standards.)

When I taught 6th grade math in 2015-2016, I used my own worksheets to instruct students on identifying “key words” when translating verbal sentences into algebraic expressions and equations. For addition, I gave the words increase, more, sum, and deposit . For subtraction, the words/phrases were difference, decreased by, less than, and withdrawal. For multiplication, I gave times and product . I indicated division by quotient . And for equal, I gave the word is, and the phrase the same as.

However, both in my experience, and according to cognitive research, 4 this type of syntactic —or word-by-word—translation directly leads to many misconceptions and mistakes. Students often made reversal errors, as when translating “four less than a number” to “4-x,” rather than “x-4.” Moreover, as my lists were not exhaustive, students struggled when translating word problems with unfamiliar words indicating operation, such as “each” in, “There were 56 students in a room, arranged in 4 rows. How many students were in each row?” Even the key words that I have listed can have multiple meanings, depending on the context. If I say, “Andrew had three cookies, then bought two more, ” this indicates addition, whereas, “Andrew has three cookies, which is two more cookies than Robert has,” indicates subtraction. These examples illustrate Polya’s 5 comparison of translating a word problem into an equation, to translating between English and French, which requires understanding the English sentence, knowing the forms of expression specific to French, and being familiar with idiomatic expressions that cannot be translated word-for-word. Similarly, we must be familiar with rules and notation specific to mathematical representation, and focus on whole meaning rather than word-for-word translation from a word problem. I am confident that I have not been alone in using the key word approach. Thus, 7th graders in 2016-2017 may have come in with prior misconceptions about translating word problems to expressions or equations.

For the 2017-2018 school year, my rising 7th graders will be the first class that used the Springboard curriculum in 6th grade. Their Course 1 textbook has more material on formulating algebraic expressions and equations on “Writing Equations”, “Representing Situations with Equations”, and “Modeling and Solving Addition, Subtraction, and Multiplication Equations”.

“Writing Expressions” has the same syntactic format as my own lesson on identifying key words, thus enforcing the same misconceptions. However, the given examples focus on translating verbal phrases such as “12 less than twice a number,” without instruction on how to transition to meaningful word problems. So students lack guidance in considering how context and phrasing of a real world problem affect the structure of the equation. This lesson also asks students to “simplify” or show “equivalence” between expressions, prior to any lessons on how to simplify, or what equivalence means. It would be better to discuss these terms when students actually encounter a situation that can be represented by multiple expressions; this way, both the process of simplifying, and the concept of equivalence will have applied meaning to the students.

“Representing Situations with Equations”, like the lesson in Course 2, over-scaffolds by providing the structure of the equation, rather than having students derive it themselves. Again, there is also little instruction about transitioning from purely verbal phrases to word problems that represent an actual situation. Thus, students lack guidance in semantic , or conceptual, understanding of word problems. Similar issues repeat themselves in the remainder of the lessons named above, although they do provide actual contexts to translate.

In the 2017 YNI math seminar, we examined the Common Core State Standards (CCSS) taxonomies for one-step equations 6 by looking at a given context, examining which problem type is represented, and how the equation for that context changes, depending on what is given, and what is unknown. However, I worry that having students memorize the problem types will essentially be the same as having them memorize key words. Instead, I want the students to learn how to carefully read each problem, and critically think about the data, the unknown values, and the relationship between the values. To scaffold this process, I will use visual models, based on the bar models from the Singapore Bar Method, as described below. While some students will be able to translate a word problem directly into symbolic terms, visual models may assist struggling students, and may help all students with more complex problem types. Additionally, students who are able to write an equation directly from a word problem may use Singapore bar models to check consistency between the relationship represented in the equation and the relationship represented in the word problem.

## Singapore Bar Models to Categorize Problems

The Singapore Bar Method scaffolds the process of translating word problems by requiring students to identify and organize the unknown information, the given data, and the condition(s) relating these values in a single model. Thus, students are using a visual means of answering Polya’s guiding questions in understanding “Problems to Solve.” 7 Furthermore, the process of answering these questions pushes students to conduct multiple reads of a word problem. “The first reading provides the context of the story. In subsequent readings of the text, information is further processed to ascertain what the givens are and what is to be found.” 8 Additionally, a collaborative paper between a neuroscientist and math educators references multiple studies, anecdotal information from teachers, and neuroimaging to link math performance and use of visuals. 9 So along with the specific information bar models prompt and convey, using visual representations may help students, and raise math achievement, in general.

The way that students will set up a bar model will depend on the type of problem to be represented. Practice with providing the models for these one-step equations should prepare students to use them to then model multi-step equations. I can assign the various one-step problems to four types of bar models: Part-Whole, Comparison, Multiplication and Division, and Multiplicative Comparison. The first three models are those identified in resources I have encountered on the Singapore method. 10 The last is a combination of the Comparison and Multiplication and Division models. As I describe the problem types that apply to each bar model, I will refer to the CCSS taxonomies, which I have adapted in the table below. However, as I mentioned above, I do not plan to teach the students this language.

## Selected CCSS Problem Taxonomies

I have modified some of the terminology of the taxonomies about the problem types and their parts. Instead, I will use alternative language we used during the YNI seminar, as it better describes the situations that apply to a specific problem type. For example, the problems I label as “Part-Part-Whole” are called “Put Together/Take Apart” in the CCSS guide. However, in the example situation about the red and green apples, you really aren’t putting anything together, as both types of apples are already there. It is more accurate to think of red apples and green apples as being parts of a whole group of apples. Also, the CCSS guide identifies the components of “Compare (Multiplication/Division)” problems as Number of Groups • Group Size= Product. However, I find that the language of Factor • Smaller = Bigger better matches the “Compare (Addition/Subtraction”)” language. (For a more in-depth explanation, see Multiplication & Division Models section below.)

I am purposely skipping problems based on the Array category in the taxonomy, because I did not come across many problems that require arrangement into rows and columns. Also, since the Singapore Bar Method relies on a linear model, rather than an area model, it is not ideal when approaching these types of problems.

(For the example situations with various unknown and given quantities, please see the CCSS guide for math, pp.88- 89.)

## Addition/Subtraction Models

Part-Whole Models show a bracket as equal in length to a pair of bars, which indicates that the large one has the same value as the two small ones combined. This is how addition is realized in the domain of length measurement. These can, of course, be applied to “Part-Part-Whole” problems, which can be written as Part + Part = Whole . With this first model, I will discourage my students from making equal parts, unless this is specifically stated in the problem, as they may be tempted to simply divide a quantity by two if they draw the small bars the same length. Alternatively, introducing obviously different quantities may help students see this themselves. (See Figure 2 below.)

For example, “Sandy's high school played 14 hockey games this year. The team won most of their games. They were defeated only in 4 games. How many games did they win?” (The model for this is in Figure 3 below.)

The bracket above indicates the total number of hockey games, the “?” represents the number of games that her team won, and the 4 represents the number of games that the team lost. (See APPENDIX B and Teacher Resources for more example problems.) Although this model follows the order of the parts listed in the word problem—that is, “games won” in the leftmost bar, and “games lost” in the rightmost—this type of problem may be useful in introducing the commutative property of addition. This discussion would come naturally from students who draw the bars in the opposite order.

The diagram for the Part-Whole Model can also be applied to “Add to” and “Take from” problems, labeled appropriately for each problem type. Although these are distinguished in the taxonomy from “Put Together/Take Apart” problems, this model captures well any one-step problem with an additive relationship. It can also be applied to comparison problems.

For “Add to” problems, we will use the convention of labeling the leftmost bar as the Start , the rightmost bar as the Change , and the bracket as the Result. For example, “There were 8 roses in the vase. Melanie cut some more roses from her flower garden and added then to the vase. There are now 17 roses in the vase. How many roses did she cut?” (See Figure 4 below.)

The bracket above indicates how many roses are in the vase now, the 8 represents the roses that were in the vase before Melanie cut roses, and the “?” represents how many roses Melanie cut from her garden.

For “Subtract from” problems, we will label the bracket as the Start , the leftmost bar as the Change , and the rightmost bar as the Result. For example, “Tom received 80 dollars for his birthday. He went to a sporting goods store and bought a baseball glove, baseball, and bat. He had 39 dollars left over, how much did he spend on the baseball gear?” (See Figure 5 below.)

The bracket above indicates the amount of money Tom received for his birthday, the “?” represents how much he spent on baseball gear, and the 39 represents how much money he had left. (See APPENDIX B and Teacher Resources for more example problems.)

Comparison Models apply to “Compare (Addition/Subtraction)” problems, which can be written as Smaller + Difference = Bigger. (See Figure 6 below.)

The bars in this model are stacked vertically in order to show the size of one value in relation to the other values. For example, “Connie has 15 red marbles and 28 blue marbles. How many more blue marbles than red marbles does Connie have?” (See Figure 7 below.)

Although 15 red marbles are mentioned first in the word problem, students should figure out that the bigger amount, 28 blue marbles, should be the top bar, with the “?” symbol labeling the bracket, and representing how many more blue marbles Connie has, compared to the number of red marbles she has. Deciding on the correct relationships is the core of understanding the problem. The construction of bar models provides a structure to help students work through this process. (See APPENDIX B and Teacher Resources for more example problems.)

## Multiplication/Division Models

Just as multiplication can be viewed as repeated addition of equal parts, the Multiplication and Division Model can be viewed as an extension of the Part-Whole Model. This model applies to “Equal Group” problems that can be written in the form Number of Groups • Group Size = Product. (See Figure 8 below.)

As you can see, the generalization of this model is a bit more abstract than the Part-Whole Model. This is because, without knowing how many groups there are, we are unable to draw the correct number of equal parts. Instead, we use ellipses to represent a continuation of bars of equal size between the first and last bars that are drawn in. We can also use ellipses to represent a large number of groups that would be too tedious to draw in our model. A “large number” is subjective, and will be different according to the person using the model, but one book recommends to use “this approach when multiplying by a factor greater than 10.” 12

Here is an example of the same situation, with different unknown values. Notice how the model changes, depending on whether number of groups is unknown or group size is unknown.

“At a restaurant, Mike and his three friends decided to divide the bill evenly. If the total bill was $ 52, how much did each person pay?” (See Figure 9 below.)

In this example, the unknown value is the group size , as indicated by the “?” inside each of the equal groups. This is fairly straightforward to illustrate using a bar model.

In contrast, consider the situation written as, “At a restaurant, Mike and some of his friends decided to divide the bill evenly. If the total bill was $ 52, and each person paid $ 13, how many people divided the bill?” (See Figure 10 below.)

Here, we know that the group size is 13, representing the $ 13 that each person paid, but we don’t know the number of groups, which represents the number of people who split the bill. As mentioned in the general model, since we don’t now how many equal groups to draw, we use ellipses to indicate the continuation of the same sized group between the two that are illustrated. (See APPENDIX B and Teacher Resources for more example problems).

Finally, the Multiplicative Comparison Model combines aspects of the Comparison Model the Multiplication and Division Model. This model applies to “Compare (Multiplication/Division)” problems, which can be written in the form

Factor • Smaller= Bigger , for whole number factors. However, if a factor is between 0 and 1, then Factor • Bigger = Smaller would be an accurate statement. For example, “Tanya had half as many red toy trucks as Al” could be represented in the equation, ½ a = t, where a is the number of red toy trucks Al has (the Bigger amount), and t is the number of red toy trucks Tanya has (the Smaller amount). However, this curriculum unit will focus on cases wherein the multiplicative factor is a whole number. So, to attack this problem, I would ask my students to convert it into a comparison of Al to Tanya, rather than the other direction, and to write the equation a = 2t . This can then be handled nicely with a bar model similar to the one illustrated in Figure 9 above.

Although the CCSS guide uses the same language as the “Equal Group” problems, I am using the more descriptive names, “smaller” and “bigger” to mirror the language in the “Compare (Addition/Subtraction)” problems. (See Figure 11 below.)

As with the Comparison Model, the bars in this model are stacked vertically in order to show the size of one value in relation to the other value. Further, as with the Multiplication and Division Model, in order to represent the number of groups, we use ellipses in the general model, when the number of groups is unknown, or when the number of groups is too large.

For example, “A bus is 45 feet long. It is 3 times as long as a car. How long is the car?” (See Figure 12 below.)

The 45 represents the length of the bigger vehicle, and as you can see, the bar is three times the length of the smaller bar, the length of which is represented by “?”. The quantities throughout the model are in feet. (See APPENDIX B and Teacher Resources for more example problems.)

It is worth noting that this type of word problem is another common source of reversal errors. For example, for the situation “6 times as many students as professors,” students are likely to write the equation “6S=P,” 13 with S representing the number of students, 6 as the Factor relating the number of students and professors, and P representing the number of professors.

However, we could construct a bar model. (See Figure 13 below.)

Here, the number of students, represented by S , is equivalent to 6 times the number of professors, represented by P . Also note here that, rather than using a “?” to represent an unknown value, I used the variables themselves. As students become more comfortable with the use and meaning of variables, I will encourage them to make that transition as well.

However, this model can be a source of confusion in and of itself. There is only one S , and 6 P s; does this mean, therefore, that there are more professors than students? Before I answer this question, it may be useful to review some common proportional relationships—or multiplicative comparisons—through the lens of unit conversions.

## Proportions as Multiplicative Comparison s

There are 3 feet in 1 yard, which can be written as the ratio, 3 feet/1 yard =1. However, you want to know the number of yards you have, Y , if given F number of feet, it is better to write this relationship as a multiplicative comparison. It would be tempting to write 3F=Y , as this equation matches almost word-for-word with the sentence, “there are 3 feet in 1 yard.”

Yet, if we add units into the equation, we can see why it is incorrect.

3feet/yard∙F feet=Y yards simplifies to 3 F feet 2 /yard=Y yards.

Therefore, the units are not compatible.

Instead, if we make 3 the coefficient for Y , we get 3feet/yard∙Y yards =F feet , which simplifies to 3Y feet=F feet , which will make sense in standard numerical form, 3Y=F .

To clarify this point further, it may be helpful to repeat the reasoning above using other commonly known unit conversions, such as “12 inches in 1 foot”, “60 minutes in 1 hour”, and “7 days in 1 week”. All of these example illustrates how objects in multiplicative comparison problems have an inverse relationship between the number of units and the size of the unit. That is, because there are 3 feet in a yard, you must multiply 3 by the number of yards (the larger unit) to get the number of feet (the smaller unit); this means that there is a larger number of the smaller unit , and a smaller number of the larger unit .

Thus, revisiting the student and professor example in the previous section, “6 times as many students as professors” can be also thought of as the ratio 6 students/1 professor . The correct equation, therefore would look like:

(6 students/professor)∙ (P professors)=S students

This simplifies to 6P students = S students . Since the units are the same, the equation will still make sense in standard numerical form, or 6P=S . Saying that there is one professor for 6 students is parallel to saying that there is one yard for 3 feet. “Professor” is functioning as the larger unit, and “student” as the smaller. Hence the number of students must be 6 times the number of professors.

## Transitioning from Models to Algebraic Equations

The next step for students using a visual model to transition into writing an equation is to introduce suitable notation. 14 The two parts that students often struggle with are where to place the “=” symbol, and how to use a variable.

Another benefit of using visual models prior to writing equations is that it emphasizes the relational meaning of the “=” symbol, as opposed to the operational meaning. “ Operational meaning stands for the...asymmetric use: ‘operation equals answer,’ which is often described for elementary arithmetic… Relational meaning focuses on a symmetric use of the equal sign.” 15

Since the sizes of bars and brackets show the relationship between the values in the models, it is easier to understand why a given situation represents an equation rather than an expression . Looking at the bus example for the Multiplicative Comparison Model, each bar is its own expression. The top bar represents one quantity (the length of the bus, in feet), and the bottom bar represents another quantity (the length of the car, in feet). Since the top bar is comprised of three copies of the bottom bar , the length of the bus is the same as three times the length of the car, or the length of the bus “=” three times the length of the car. Therefore, the relationship of the quantities in the bar models helps you can see an equation relating two expressions , or the lengths of the different vehicles

Regarding variables, there is a common misconception of “symbols as labels that can shift within an activity; for example…b = books, which can be interpreted later as the number of books, the value of the book, the value of the total number of books, etc.” 16 To avoid this, I will model and reinforce the need to identify the unknown value as a number with appropriate units. 17 I will make sure that my students develop the habit of specifying the units of their variables.

It may be useful for struggling students to identify the units of all of the parts of a problem. Additive problem types should have the exact same unit for each of its parts, in order for the addition to make sense in standard numerical form. Multiplicative problem types are a bit more challenging, as each factor may have different units, but, once multiplied, should result in the same unit on both sides of the equation. The models themselves are one means of checking that units correspond. In the additive case, in the Part-Whole Model example on Sandy’s hockey team, although one part is “number of games lost,” and another part is “number of games won,” these parts are both a certain number of the unit “game,” which is also the unit of the “total number of games.” Meanwhile, in the Multiplication and Division example on plums divided among bags (see Classroom Activities), if you multiply the number of groups "number of bags of plum," by the group size, "number of plums per bag", the product will be the total number of plums.

I expect that even students who can directly translate word problems into equations without the help of bar model(s), by explicitly identifying units for variables and known values alike, will have an easier time verifying that their equation makes sense when they solve a problem arithmetically, using the corresponding model. Can you combine the units in the equation in a way that is compatible with the situation? Does the solution match the magnitude and the units of the context? This last check also reinforces students’ conceptual understanding of the word problem as a whole, and the connection between the word problem, visual model, and equation.

Once students understand how to categorize one-step word problems by bar model type, represent the problem using the appropriate bar model, and translate those models into equations, they should be ready to put those pieces together to categorize and symbolically represent multi-step situations.

## Introducing Inverse Operations

Along with providing a scaffold for translating word problems into equations, Singapore Bar Models can help with introducing the concept of inverse operations. “By a process of undoing, the value of the unknown [quantity] is found. Hence, children [using bar models] circumvent the cognitive demands that are inherent in solving algebraic equations, which normally require the construction of a system of equivalent equations.” 18 In the Part-Whole Model examples in the previous section, the models represent additive relationships. However, to figure out the missing value in each of these examples, students must actually subtract, which is the inverse operation of adding. The diagram helps them understand what to do. Similarly, the Multiplicative Comparison Model example represents a multiplicative relationship. But in order to solve this problem, students should divide , which is the inverse operation of multiplying. By providing a visual wherein students can literally take away part of an amount, or split up an amount, use of bar models makes students less likely to perform the wrong operation, even when it is suggested by a “key word” (for example, getting r = 25 for 8 + r = 17 by adding 8+17).

Similarly, a bar model representing a two-step equation can help justify the convention of removing the constant prior to the coefficient. Below, I am using a combination of a Part-Whole Model and a Multiplication and Division Model to represent the general form of a first order equation in one variable, ax + b=c . Using the structure we have established for the Part-Whole Model, ax will be the first Part and b will be the second Part of the model. Then, using the structure of the Multiplication and Division Model, a will represent the number of groups, and x will be the group size. (See Figure 14 below.)

Using the model, you would first subtract b from c to be left with a groups with equal, unknown size, x. To figure out the value of x , divide the new amount (c-b) by a .

Symbolically, this looks like

ax + b – b = c – b

ax + (b – b) = (c – b)

ax + 0 = (c – b)

ax = (c – b)

ax/a=(c-b)/a

Thus, bar models can help students both translate word problems into equations, and introduce them to aspects of solving these equations algebraically.

## Teaching Strategies

Assessing through an international baccalaureate lens.

As an International Baccalaureate (IB) school, 19 Burnett grades students in each subject on a scale from 1-8 across four different criteria in a subject. Therefore, rather than having a single grade that summarizes a students’ mastery of math as a whole, I can assign four scores on their knowledge of math skills and concepts, investigation of patterns, communication, and application of math in real-life contexts. In addition to breaking down the subject into four criteria, each score comes with a descriptor that provides qualitative feedback for the student and parents. For this curriculum unit, I will assess students using Criterion C and Criterion D. (See Teacher Resources for criteria rubrics.)

Criterion C, Communicating: Students use appropriate mathematical language and different forms of representation when communicating mathematical ideas, reasoning and findings, both orally and in writing.

For Criterion C, students will represent the information in a word problem using models, symbols, and words. Within their equations, I will check that students clearly define their variables as numbers, with appropriate units. I will also give students opportunities to create a bar model, an algebraic equation, or a word problem when given one of the other representations, as well as the opportunity to create their own original word problem, using all of the mentioned representations. Throughout the curriculum unit, I will require students to write and/or discuss how the model, equation, and word problem all show the same values and relationships.

Criterion D, Applying Mathematics in real-life contexts: Students transfer theoretical mathematical knowledge into real-world situations and apply appropriate problem-solving strategies, draw valid conclusions and reflect upon their results. For Criterion D, students will be engaging with real-life contexts presented in word problems. Applying appropriate problem-solving strategies will require writing a correct equation, which may include choosing the correct bar model to represent a situation. Then, after solving either arithmetically or algebraically, students will reflect whether their solution is reasonable for the given context. This includes checking that the units of each part of the equation correspond, and that they make sense after calculation.

Thus, assessing students using Criteria C and D will push me to frame each class activity around using multiple mathematical representations to understand and solve word problems from a variety of contexts. Additionally, having students use the criteria rubrics to self-assess and assess and their peers, they will gain practice on critiquing their own representations—models and/or equations—and the reasonability of their solutions in context of a given word problem.

## Explicit Direct Instruction (EDI) with Graphic Organizers

When introducing each new model, and the idea of combining models, I plan to use EDI, consisting of orientation, gradual release (teacher-led, collaborative, then independent problem-solving), and closure. As students will be responsible for multiple representations of a problem, I will provide graphic organizers for students to record each word problem’s model type, the bar model itself, and the resulting algebraic equation with a defined variable and unit. This graphic organizer will mirror the questions that students should ask themselves when translating between representations. (See Figure 15 below.)

(See APPENDIX C for a blank template of the graphic organizer.)

Just as the bar method itself is merely a scaffold as opposed to an end in and of itself, the graphic organizer is merely a helpful tool that students should not rely on. Instead, I hope that students can find their own way to identify important aspects of a word problem, and to organize their multiple representations. The graphic organizer is just to get them started.

## Think-Write-Pair-Share

A strategy I will use within the EDI framework is giving students individual time to work on a translation problem, record their thoughts, and then discuss with a partner. One thing that I do, in particular, is create different student pairings so that students can hear from multiple perspectives. There is the usual pairing of speaking to the student sitting next to you (“elbow partners”) or across from you (“face partners”), but I have also had students go to a corner of the room, corresponding with the colored sticker on their desks, and other pairings that will also require students to get up and move. To keep each student accountable, I will then call on a non-volunteer to share his or her solution to the whole class, and the rest of the class would verify this solution. Peers may ask specific clarifying questions, or they may help the student sharing if they believe that the solution was incorrect. This would also be an opportunity for broader whole-class discussion on the translation process.

## Structured Student Groups

In addition to collaborating during direct instruction lessons through Think-Write-Pair-Share, students will have opportunities to collaborate for more sustained periods through structured group work. To facilitate team interdependence, I will assign group roles 20 of Facilitator/Timekeeper, Recorder, Errand Monitor/Presenter, and Summarizer (see Teacher Resources for link to role descriptors.) To balance teamwork and individual accountability, I will assess students as a group, as well as individually; students will have an opportunity to grade themselves, as well as their group mates.

I will use structured group work for activities that require a final product that will be shared either as a presentation, or through a gallery walk 21 . This may include activities where students must create their own word problem, given either a model or an algebraic equation, or showing how changing what is unknown in a context will change the model and algebraic equation.

## Classroom Activities

Preparing for the unit.

Prior to teaching the different models to students, I plan to give students a short pre-assessment on simple word problems. A sample task on this pre-assessment is this matching activity.

Students should correctly identify:

I will then ask students to verbally explain how they matched each word problem with its respective equation. In this way, I can see what intuition students already have about translating real world situations to symbolic representations.

## Sequence of Content

I will instruct students in the different bar models and their corresponding problem types, according to complexity of the model.

- Part-Whole Model
- Comparison Model
- Multiplication and Division Model
- Multiplicative Comparison Model
- Combining Model Types

I will teach the Part-Whole Model first, as it is needed to later understand the Multiplication and Division model. Similarly, both the Multiplication and Division Model, and the Comparison Model are needed to represent “Compare (Multiplication/Division)” problem types. Of course, it is necessary to learn all of the models for the one-step problem types, before representing multi-step situations.

For each new bar model, and when we first combine models, I will introduce a word problem that will produce the desired model. For each problem type applied to the model, I will ask: How do you set up the model? What is the unit of each piece of the model? What is the unknown value, and its unit, and what variable will you use? How can you represent each part of the model in an equation? What is the solution to the situation (either using the model or the equation)? Check that the solution makes sense for the units and the magnitude of the situation If you change what is unknown in the word problem, how does that change the model and the equation?

Below, find a one-step example, a two-step example, and a multi-step example. (See APPENDICES B-F for more examples.) Note that I will not have students change the unknown in the multi-step word problem.

## One Step Equations

1) Word Problem: If 18 plums are shared equally into 3 bags, then how many plums will be in each bag?

Model Type(s): Multiplication and Division Model

Bar Model: (See Figure 16 below.)

Algebraic Equation:

? = the number of plums in each bag = p

State & Check your Solution:

There will be 6 plums in each bag, because 3bags of plums•(6 plums per bag) =18 plums.

2) Word Problem:

If some plums are shared equally into 3 bags, and there are 6 plums in each bag, how many plums are there in total?

Bar Model: (See Figure 17 below.)

? = the total number of plums = t

3•6 = t

There will be 18 plums total, because 3 bags of plums•6 plums per bag =18 plums.

3) Word Problem:

If 18 plums are shared equally into some bags, and there are 6 plums in each bag, how many bags are there?

Bar Model: (See Figure 18 below.)

? = the number of bags of plums = b

b•6 = 18

There will be 3 bags of plums, because (3 bags)•6 plums per bag =18 plums.

## Two Step Equations

1) Word Problem: Oceanside Bike Rental Shop charges a 12 dollar fixed fee plus 7 dollars an hour for renting a bike. Fred paid 54 dollars to rent a bike. How many hours did he pay to have the bike checked out?

Model Type(s): Part-Whole Model, Multiplication and Division Model

Bar Model: (See Figure 19 below.)

? = the number of hours he had the bike checked out=h

Fred paid to rent the bike for 6 hours, because $ 12+$ 7 per hour•(6 hours) = $ 12+$ 42 = $ 54.

2) Word Problem: Oceanside Bike Rental Shop charges a fixed fee plus 7 dollars an hour for renting a bike. Fred rode for 6 hours, and paid 54 dollars to rent a bike. How much did Fred pay for the fixed fee?

Bar Model: (See Figure 20 below.)

? = the amount Fred paid for the fixed fee, in dollars=f

f+7•6 = 54

Fred paid a fixed fee of $ 12 because ($ 12)+$ 7 per hour•6hours = $ 12+$ 42 = $ 54.

3) Word Problem: Oceanside Bike Rental Shop charges a 12 dollar fixed fee plus some dollars an hour for renting a bike. Fred rode for 6 hours, and paid 54 dollars to rent a bike. How much did Fred pay per hour of renting the bike?

Bar Model: (See Figure 21 below.)

?=amount Fred paid per hour of renting the bike, in dollars=p

12+p•6 = 54

Fred paid $ 7 per hour to rent the bike, because $ 12+($ 7 per hour)•6hours = $ 12+$ 42 = $ 54. (See APPENDIX B and Teacher Resources for more two-step word problems)

## Multistep Equations

1) Word Problem: Divide $ 79 among three people so that the second will have three times the amount of the first, and the third will have two dollars more than the second

Model Type(s): Part-Whole Model, Multiplicative Comparison Model, Comparison Model

Bar Model: (See Figure 22 below.)

?=the amount of money the first person gets, in dollars=f

f+(3f)+(3f+2) = 79

The first person will get $ 11, because ($ 11)+(3•$ 11)+(3•$ 11+$ 2)= $ 11+$ 33+($ 33+$ 2)= $ 44+$ 35= $ 79.

It is worth noting that this last problem is quite complex, as it requires adding three unknown values. Using bar models makes this problem more manageable, and should help students see that they need 7 copies of the unknown, and then $ 2 more, even if they are not comfortable with “adding like terms”. However, as mentioned in the overview, students will later encounter problems that require more sophisticated methods of translating and solving. (See APPENDIX B and Teacher Resources for more multi-step word problems).

Once students have shown proficiency in problems of at least two model types, I will provide a mix of word problems. Students will then determine:

- What is the structure of this problem, and what is it asking?
- What model do you need?

Then they will then continue the process outlined in the examples above to represent each word problem as algebraic equations.

Student-Created Word Problems

Up to this point, I have given students a word problem, and they have been creating a model—or combination of models for multi-step problems—and an algebraic equation that represent the situation. Next, I would give students either an algebraic equation or a visual model, and ask them to create a word problem that matches the given representation. Then I will have them explain why the equation or model accurately represent the word problem they have created.

A given equation: Create a word problem that can be represented as 2x+3=9

A given model: Create a word problem that can be represented by the model (See Figure 23 below.)

Once students have practiced creating a word problem that matches a specific model or equation, I will have them come up with an original word problem. Then, they will

- Represent the word problem using an appropriate bar model (or combination of bar models);
- Define the variable, with appropriate units, and write a corresponding algebraic equation;
- Explain how the word problem, model, and algebraic equation match; and
- Rewrite the word problem with the same situation and amounts, but with a different quantity for the unknown value;
- Determine how the new unknown value effects the other representations of the situation.

This will be one of their summative assessments, completed in groups, in addition to a more traditional individual paper-and-pencil test.

## Annotated Bibliography

Ciobanu, Mirela, “In the Middle-Misconceptions About the Equal Sign in Middle School,” OAME/AOEM Gazette (2014), 14-16.

This article defines the “operational” meaning of the “=” sign ( calculation equals answer), and describes how this interpretation causes errors, especially when solving algebraic equations.

Fuchs, Lynn S. et al., “The Effects of Schema-Broadening Instruction on Second Graders’ Word-Problem Performance and Their Ability to Represent Word Problems with Algebraic Equations: A Randomized Control Study,” Elem Sch J. (2010), 446-463.

This is a study on teaching students to representing the structural, defining features of word problems with the schema of “overarching equations.” Essentially, this unit is doing something analogous, but using the “problem types” and their corresponding bar models rather than “overarching equations.”

Howe, Roger, “From Arithmetic to Algebra,” Mathematics Bulletin—A Journal for Educators , 49 (2010), 13–21.

This is a paper on teaching algebra through analyzing and discussing both arithmetic and algebraic word problems. In particular, I found his discussion on units of variables, and of terms in an addition equation, to be helpful in my unit.

James, Jonathan “The Surprising Impact of High School Math on Job Market Outcomes,” Economic Commentary , (2013).

This is a study on the effect of high school math courses on unemployment rates and income, both for high school graduates who attend college, and those who go straight into the workforce. I was surprised at the effect for those who don’t attend college!

Kinzel, Margaret Tatem, “Understanding Algebraic Notation from the Students’ Perspective”, The Mathematics Teacher, Vol. 92, No. 5 (1999).

This paper examines students’ various struggles with representing and interpreting word problems using algebraic symbols (variables).

MacGregor, Mollie and Stacey, Kaye “Cognitive Models Underlying Students’ Formulation of Simple Linear Equations”, Journal for Research in Mathematics Education , Vol. 24, No.3 (1993), 217-232.

This paper shows students’ struggles in representing compared unequal quantities. Most notably, reversal errors can occur even if the student is not translating word-for-word (syntactically).

National Governors Association Center for Best Practices, Council of Chief State School Officers, Common Core State Standards for Mathematics , (Washington D.C.: National Governors Association Center for Best Practices, Council of Chief State School Officers, 2010), 88-89.

This is the list of the one-step problem taxonomies. In my unit, I changed the labels used in the table to ones we used in seminar, because I found them to be more descriptive than the ones in the CCSS guide. For example, the unknowns in the “Compare (Multiplication/Division)” problems changed from “number of groups” to “factor”, “group size” to “smaller”, and “product” to “bigger.” I also took some of my example problems in Appendices B-F from this document.

Ng, Swee Fong and Lee, Kerry, “The Model Method: Singapore Children’s Tool for Representing and Solving Algebraic Word Problems,” Journal for Research in Mathematics Education , Vol 40, No. 3 (2009), 282-313.

This paper studies the effect of using models to represent key information in a problem, in order to solve both arithmetic and algebraic problems. What I found most helpful were both the general, and applied examples of the Part-Whole model, the Comparison model, and the Multiplication and Division model.

Polya, G. How to Solve It: A New Aspect of Mathematical Method (Princeton and Oxford: Princeton University Press, 2004).

This is a book on methods of problem solving, broken up into four phases. In my curricular unit, I focus on the first phase, which is “understanding the problem.” The rest of Polya’s phases are devising a plan, carrying out the plan, and reviewing/extending the plan.

Powell, Sarah, "The Influence of Symbols and Equations on Understanding Mathematical Equivalence," Interventions in School and Clinic (ISC) , Vol. 50(5) (2015), 266-272.

This article focuses on the “relational” meaning of the “=” sign, where one side of the equation is “the same as” the other side of the equation. The article goes on to explain how instructing on this relational meaning of equality improved student outcomes in setting up and solving word problems.

## Teacher Resources

http://www.mathplayground.com/ThinkingBlocks/thinking_blocks_modeling%20_tool.html

This is a tool for creating bar models. Requires Adobe Flash. My diagrams in the “Key Ideas” section for this unit plan were screen-grabs from this tool.

Forsten, Char, Step-by-Step Model Drawing: Solving Word Problems the Singapore Way (Peterborough, NH: Crystal Springs Books, 2010), 30-32. I only used this book to see how she treated an unknown number of groups in the multiplication/division model. However, she is very step-by-step in explaining the process for beginners of the Singapore Bar Method. While this may be helpful as an entry, it is a bit prescriptive and too procedural.

The following are places where I got most of my word problems featured in the appendices.

http://www.math-aids.com/Algebra/Algebra_1/Word_Problems

http://cdn.kutasoftware.com/Worksheets/PreAlg/One-Step%20Word%20Problems.pdf

http://www.onlinemathlearning.com/comparison-word-problems.html

https://learnzillion.com/resources/11807?card_id=68694

https://www.ixl.com/math/algebra-1/solve-linear-equations-word-problems

https://www.khanacademy.org/math/algebra/one-variable-linear-equations/alg1-linear-eq-word-probs/e/linear-equation-world-problems-2

http://www.algebralab.org/lessons/lesson.aspx?file=Algebra_OneVariableWritingEquations.xml

https://www.cliffsnotes.com/study-guides/algebra/algebra-i/word-problems/age-problems

https://www.algebra.com/algebra/homework/word/age/Solving-Age-Problems.lesson

http://www.purplemath.com/modules/ageprobs.htm

Here is one version of group role cards you can give to students, including sentence starters for each group member.

http://www.readwritethink.org/files/resources/lesson_images/lesson277/cooperative.pdf

The CCSS Standards I am addressing in this unit are:

## 7 th Grade Standards

- MA.7.EE.B.A: Use variables to represent quantities in a real-world or mathematical problem, and construct simple equations and inequalities to solve problems by reasoning about the quantities.
- MA.7.7.EE.B.4.A: Solve word problems leading to equations of the form px+q=r and p (x + q)=r, where p, q, and r are specific rational numbers. Solve equations of these forms fluently. Compare an algebraic solution to an arithmetic solution, identifying the sequence of the operations used in each approach.

## 6 th Grade Standards

- MA.6.EE.B.6: Use variables to represent numbers and write expressions when solving a real-world or mathematical problem; understand that a variable can represent an unknown number.
- MA.6.EE.B.7: Solve real-world and mathematical problems by writing and solving equations in the form x + p = q and px = q.

## Part-Part Whole Model Problems

“Put Together, Take Apart” Problem

Alyssa is baking a cake. The recipe calls for 8 cups of flour. She already put in 2 cups. How many more cups does she need to add?

“Add To/Take From” Problem

Tom received 80 dollars for his birthday. He went to a sporting goods store and bought a baseball glove, baseball, and bat. He had 39 dollars left over, how much did he spend on the baseball gear?

## Comparison Model Problem

Bob ate14 crackers. Steve ate five fewer than Bob. How many did Steve eat?

## Multiplication and Division Model Problem

At a restaurant, Mike and his three friends decided to divide the bill evenly. If each person paid $ 13 then what was the total bill?

## Multiplicative Comparison Model Problems

The giraffe is 20 feet tall. The kangaroo is 5 feet tall. The giraffe is how many times taller than the kangaroo?

## Combining Bar Models -Multistep Word Problems

Part-Whole Combination Problem

When ringing up a customer, a cashier needs 17 seconds to process payment as well as 4 seconds to scan each item being purchased. How long does it take to ring up a customer with 6 items?

Comparison Combination Problem

Jose has a board that is 44 inches long. He wishes to cut it into two pieces so that one piece will be 6 inches longer than the other. How long should the shorter piece be?

## Blank Graphic Organizer

Note: Fill in the word problems yourself to save time for the students, OR have students only write down the important information (data, unknown values, relationship) in the top row of the organizer.

- “How I see math problems,” someecards , July 16, 2016 https://www.someecards.com/usercards/viewcard/MjAxMi1mOThmZjNhOGJiYmE5YzMy/
- “Peter Burnett Middle School,” Great Schools , May 30, 2017 http://www.greatschools.org/california/san-jose/5654-Peter-Burnett-Middle-School/
- Jonathan James, “The Surprising Impact of High School Math on Job Market Outcomes,” Economic Commentary , (2013)
- Mollie MacGregor and Kaye Stacey, “Cognitive Models Underlying Students’ Formulation of Simple Linear Equations”, Journal for Research in Mathematics Education , Vol. 24, No.3 (1993), 217-232
- G. Polya, How to Solve It: A New Aspect of Mathematical Method (Princeton and Oxford: Princeton University Press, 2004), 173-174
- National Governors Association Center for Best Practices, Council of Chief State School Officers, Common Core State Standards for Mathematics , (Washington D.C.: National Governors Association Center for Best Practices, Council of Chief State School Officers, 2010), 88-89
- Polya, How to Solve It, 7-8
- Swee Fong Ng and Kerry Lee, “The Model Method: Singapore Children’s Tool for Representing and Solving Algebraic Word Problems,” Journal for Research in Mathematics Education , Vol 40, No. 3 (2009), 291
- Jo Boaler, Lang Chen, et al. “Seeing as Understanding: The Importance of Visual Mathematics for our Brain and Learning,” (July 16, 2017) https://bhi61nm2cr3mkdgk1dtaov18-wpengine.netdna-ssl.com/wp-content/uploads/2016/04/Visual-Math-Paper-vF.pdf
- Ng and Lee, “The Model Method,” 282-313
- Lynn S. Fuchs, et al., “The Effects of Schema-Broadening Instruction on Second Graders’ Word-Problem Performance and Their Ability to Represent Word Problems with Algebraic Equations: A Randomized Control Study,” Elem Sch J. (2010), 4.
- Char Forsten, Step-by-Step Model Drawing: Solving Word Problems the Singapore Way (Peterborough, NH: Crystal Springs Books, 2010), 30-32
- Warren Wollman, “Determining the Sources of Error in a Translation from Sentence to Equation,” Journal for Research in Mathematics Education, Vol 14. No. 3 (1983), 170.
- Polya, How to Solve It, 135-136.
- Susanne Prediger, “How to develop mathematics-for-teaching and for understanding: the case of the equal sign”, J Math Teacher Educ (2010), 73-93.
- Margaret Tatem Kinzel, “Understanding Algebraic Notation from the Students’ Perspective”, The Mathematics Teacher, Vol. 92, No. 5 (1999), 437.
- Roger Howe, “From Arithmetic to Algebra,” Mathematics Bulletin—A Journal for Educators , 49 (2010), 13–21.
- Ng and Lee, “The Model Method,” 293.
- International Baccalaureate Organization, Mathematics Guide, For use from September 2014/January 2015 (Cardiff, Wales: International Baccalaureate Organization, 2014)
- “Cooperative Group Role Cards,” Read, Write, Think, (July 14, 2017) http://www.readwritethink.org/files/resources/lesson_images/lesson277/cooperative.pdf
- Wanda Collins, “Gallery Walk,” Math Concentration , (July 15, 2017) http://www.mathconcentration.com/profiles/blogs/using-the-gallery-walk-as-a-teacher-strategy

## Solver Title

## Generating PDF...

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## Rectangle Problems involving Algebraic Expressions Worksheets

Explore this assortment of worksheets to solve rectangle problems involving length, width, area, perimeter and diagonal with measures provided as algebraic expressions. These exclusive printable worksheets are aligned with the common core curriculum and are proposed for the use of middle school students. Recall and apply the properties of rectangle to solve the exercises provided in this section. Free worksheets are also included.

Find the Value of x using given Sides

One of the dimensions of each rectangle is expressed in a linear equation and the side opposite to it is given as a whole number. Applying the fact that opposite sides are equal, set up the equation and then solve for x.

Type: One-step and Two-step Equations

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Finding Length or Width

The opposite sides of the rectangle are represented as algebraic expressions. Equate both the expressions and find the value of x. Substitute the value of x in order to find the specified length.

Type: Multi-step Equations

Find the Value of x using given Diagonals

Engage students of grade 6 and grade 7 with this set of ready-to-print worksheets on finding the value of x by applying the property - The diagonals of a rectangle are equal and bisect each other.

Finding Length of the Diagonal

This compilation is based on the property; the two diagonals of a rectangle are congruent. Utilizing this property, set up the equation, solve for x and then find the length of the diagonal.

Finding Perimeter using Algebra

Enrich the knowledge of 7th grade and 8th grade students on finding the perimeter of a rectangle with these PDFs. The parameters of each rectangle are depicted as algebraic expressions. Solve for x and then find the perimeter.

Type: Two-step and Multi-step Equations

Finding Area using Algebra

Middle school learners can sharpen their equation solving skills with this set of phenomenal worksheets. Use the congruent property to solve for x, find the length and width and then find the area of the rectangle.

Length or Width from the Perimeter

The length and width are expressed as a linear equation. Set up the equations with the known perimeter to solve for x and then find the length and width of the rectangle.

Finding Length or Width from the given Area

Apply the area formula to set up the quadratic equation with the given area (whole numbers) and dimensions (expressions). Solve for x and then find the length and width. Beware of extraneous solutions.

Type: Quadratic Equations

Finding Unknown Angle using Algebra - Level 1

This bunch of printable level 1 worksheets present vertex angles as whole numbers and algebraic expressions. Determine the value of x.

Type: Two-step Equations

Finding Unknown Angle using Algebra - Level 2

Put your students' understanding of properties of rectangles to test with these astounding pdf worksheets. Find the value of x, when the vertex angle and angle formed by the diagonals are given as algebraic expressions.

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## Modeling, Functions, and Graphs

Katherine Yoshiwara

## Section A.3 Algebraic Expressions and Problem Solving

Example a.17 ..

- Choose a variable for the number of hours Loren works per week.
- Write an algebraic expression for the amount of Loren’s weekly earnings.
- Let \(h\) stand for the number of hours Loren works per week.
- The amount Loren earns is given by \begin{equation*} \blert{6\times (\text{number of hours Loren worked})} \end{equation*} or \(6\cdot h\text{.}\) Loren’s weekly earnings can be expressed as \(6h\text{.}\)

## Example A.18 .

Example a.19 ..

- Choose variables to represent the unknown quantities and write an algebraic expression for April’s weekly income in terms of her sales.
- Find April’s income for a week in which she sells $ \(350\) worth of cleaning products.
- Let \(I\) represent April’s total income for the week, and let \(S\) represent the total amount of her sales. We translate the information from the problem into mathematical language as follows: \begin{gather*} \blert{\text{Her income consists of }\$200 . . .\text{ plus }. . . 9\% \text{ of her sales}} \\ I \hphantom{consists of}= \hphantom{of}200 \hphantom{plus+}+ \hphantom{....}0.09 \hphantom{of her}S \end{gather*} Thus, \(I = 200 + 0.09S\text{.}\)
- We want to evaluate our expression from part (a) with \(S = 350\text{.}\) We substitute \(\alert{350}\) for \(S\) to find \begin{equation*} I = 200 + 0.09(\alert{350}) \end{equation*} Following the order of operations, we perform the multiplication before the addition. Thus, we begin by computing \(0.09(350)\text{.}\) \begin{align*} I \amp = 200 + 0.09(350)\amp\amp\blert{\text{Multiply }0.09 (350) \text{ first.}}\\ \amp = 200 + 31.5\\ \amp = 231.50 \end{align*} April’s income for the week is $ \(231.50\text{.}\)

## Remark A.20 . Calculator Tip.

Example a.21 ..

- Choose variables to represent the unknown quantities and write an expression for the cost of shipping Andrew’s painting.
- Find the shipping cost if Andrew uses \(2.9\) pounds of packing material.
- Let \(C\) stand for the shipping cost and let \(w\) stand for the weight of the packing material. Andrew must find the total weight of his package first, then multiply by the shipping charge. The total weight of the package is \(8.3 + w\) pounds. We use parentheses around this expression to show that it should be computed first, and the sum should be multiplied by the shipping charge of $ \(2.80\) per pound. Thus, \begin{equation*} C = 2.80(8.3 + w) \end{equation*}
- Evaluate the formula from part (a) with \(w = \alert{2.9}\text{.}\) \begin{align*} C \amp = 2.80(8.3 + \alert{2.9})\amp\amp\blert{\text{Add inside parentheses.}}\\ \amp = 2.80(11.2)\amp\amp\blert{\text{Multiply.}}\\ \amp = 31.36 \end{align*} The cost of shipping the painting is $ \(31.36\text{.}\)

## Remark A.22 . Calculator Tip.

Caution a.23 ., subsection problem solving, guidelines for problem solving..

- Identify the unknown quantity and assign a variable to represent it.
- Find some quantity that can be expressed in two different ways and write an equation.
- Solve the equation.
- Interpret your solution to answer the question in the problem.

## Subsection Supply and Demand

Example a.24 ..

- We are looking for the equilibrium price, \(p\text{.}\)
- The Coffee Connection would like the demand for its coffee to equal its supply. We equate the expressions for supply and for demand to obtain the equation \begin{equation*} 800 - 60p = 175 + 40p \end{equation*}
- Solve the equation. To get all terms containing the variable, \(p\text{,}\) on one side of the equation, we add \(60p\) to both sides and subtract \(175\) from both sides to obtain \begin{align*} 800 - 60p + \alert{60p - 175} \amp= 175 + 40p + \alert{60p - 175}\\ 625 \amp = 100p \amp\amp\blert{\text{Divide both sides by }100.}\\ 6.25 \amp = p \end{align*}
- The Coffee Connection should charge $ \(6.25\) per pound for its coffee.

## Subsection Percent Problems

Percent formula., example a.25 ..

- Let \(c\) represent the cost of the house last year.
- Express the current price of the house in two different ways. During the past year, the price of the house increased by \(4\%\text{,}\) or \(0.04c\text{.}\) Its current price is thus \begin{equation*} \stackrel{\text{Original cost}}{(1)c} + \stackrel{\text{Price increase}}{0.04c} = c(1 + 0.04) = 1.04c \end{equation*} This expression is equal to the value given for current price of the house: \begin{equation*} 1.04c = 100,000 \end{equation*}
- To solve this equation, we divide both sides by \(1.04\) to find \begin{equation*} c = \frac{100,000}{1.04}= 96,153.846 \end{equation*}
- To the nearest cent, the cost of the house last year was $ \(96,153.85\text{.}\)

## Caution A.26 .

Subsection weighted averages, example a.27 ..

- Let \(x\) represent the final exam score Kwan needs.
- Kwan’s grade is the weighted average of his test, homework, and final exam scores. \begin{equation*} \frac{\alert{0.50}(84) + \alert{0.20}(92) + \alert{0.30}x}{1.00}=90 \end{equation*} (The sum of the weights is 1.00, or 100% of Kwan’s grade.) Multiply both sides of the equation by \(1.00\) to get \begin{equation*} 0.50(84) + 0.20(92) + 0.30x = 1.00(90) \end{equation*}
- Solve the equation. Simplify the left side first. \begin{align*} 60.4 + 0.30x \amp = 90\amp\amp\blert{\text{Subtract 60.4 from both sides.}}\\ 0.30x \amp = 29.6\amp\amp\blert{\text{Divide both sides by 0.30.}}\\ x \amp = 98.7 \end{align*}
- Kwan needs a score of \(98.7\) on the final exam to earn a grade of \(90\text{.}\)

## Weighted Average.

Example a.28 ..

- Let \(p\) represent the number of pounds of LeanMeal needed.
- Simplify each side of the equation, using the distributive law on the right side, then solve. \begin{align*} 7.5 + 0.05p \amp = 4 + 0.08p \amp\amp\blert{\text{Subtract}~4+0.05p~\text{from both sides.}}\\ 3.5 \amp = 0.03p \amp\amp \blert{\text{Divid both sides by}~ 0.03.}\\ p \amp = 116.\overline{6} \end{align*}
- Delbert should mix \(116\frac{2}{3}\) pounds of LeanMeal with \(50\) pounds of JuicyBits to make a mixture that is \(8\%\) fat.

## Subsection Section Summary

Subsubsection vocabulary.

- Weighted average
- Equilibrium price
- Algebraic expression
- Evaluate an expression

## Subsubsection SKILLS

- Write an algebraic expression: #1–12
- Evaluate an algebraic expression: #1–12
- Write and solve an equation to solve a problem: #13–28

## Exercises Exercises A.3

Exercise group..

- Write an expression for Jim’s age in terms of Ana’s age.
- Use your expression to find Jim’s age when ana is 22 years old.
- Write an expression for the total cost of new wheels in terms of the number of wheels Rani must replace.
- Use your expression to find the total cost if Rani must replace 8 wheels.
- Write an expression for the total number of hours Helen must drive in terms of her average driving speed.
- Use your expression to find how long Helen must drive if she averages 45 miles per hour.
- Write an expression for the number of years before Ben gets his inheritance in terms of his present age.
- Use your expression to find how many more years Ben must wait after he turns 13 years old.
- Write an expression for the area of a circle in terms of its radius.
- Find the area of a circle whose radius is 5 centimeters.
- Write an expression for the volume of a sphere in terms of its radius.
- Find the volume of a sphere whose radius is 5 centimeters.
- Write an expression for the total bill for an item (price plus tax) in terms of the price of the item.
- Find the total bill for an item whose price is $490.
- Write an expression for the balance (initial deposit plus interest) in the account after one year in terms of the amount deposited.
- Find the total amount in the account after one year if $350 was deposited.
- Write an expression for the cost of a long-distance phone call in terms of the number of minutes of the call.
- Find the cost of a 27-minute phone call.
- Write an expression for the cost of the flight in terms of Mr. Owlsley’s weight.
- Find the cost if Mr. Owsley weights 162 pounds.
- Write an expression for the amount of rice Juan has consumed in terms of the number of weeks since he bought the bag.
- Write an expression for the amount of rice Juan has left in terms of the number of weeks since he bought the bag.
- Find the amount of rice Juan has left after 6 weeks.
- Write an expression for the elevation that Trinh has lost in terms of the distance she has cycled.
- Write an expression for Trinh’s elevation in terms of the number of miles she has cycled.
- Find Trinh’s elevation after she has cycled 9 miles.
- What are we asked to find in this problem? Assign a variable to represent it.
- Write an expression in terms of your variable for the distance Roger’s wife drives.
- Write an expression in terms of your variable for the distance Roger has cycled.
- Write an equation and solve it.
- Write an expression in terms of your variable for the distance Kate and Julie sailed.
- Write an expression in terms of your variable for the distance their father traveled.
- Write an expression in terms of your variable for the total cost incurred by each machine.
- Write an expression in terms of your variable for the total cost incurred by each refrigerator.
- Assuming the same rate of growth, what do you predict for the population of Midland next year?
- What was the population of Midland last year?
- Assuming the same rate of inflation, what do you predict for the price of a steak dinner next year?
- What did a steak dinner cost last year?
- Write algebraic expressions in terms of your variable for the amounts of each fertilizer the horticulturist uses. Use the table.
- Write expressions for the amount of potash in each batch of fertilizer.
- Write two different expressions for the amount of potash in the mixture. Now write an equation and solve it.
- Write algebraic expressions in terms of your variable for the amounts of each alloy the sculptor uses. Use the table.
- Write expressions for the amount of copper in each batch of alloy.
- Write two different expressions for the amount of copper in the mixture. Now write an equation and solve it.
- Write algebraic expressions for the total amounts Lacy’s pays its managers, its department heads, and its clerks.
- Write two different expressions for the total amount Lacy’s pays in salaries each year.
- Write algebraic expressions for the number of trucks and the number of cars that will meet emission standards.
- Write two different expressions for the total number of vehicles that will meet the standards.

## Factoring Algebraic Expressions Word Problems

Factorization of algebraic means finding the factors of the given expression which refers to finding two or more expressions whose product is the given expression. This process of finding two or more expressions whose product is the given expression is known as the factorization of algebraic expressions. A factor is a number that divides the given number without any remainder. It simply means expressing a number as a multiplication of two other numbers. Similarly, in Algebra we write the algebraic expressions as a product of their factors. The only difference here is that an algebraic expression involves numbers and variables combined with an arithmetic operation like addition or subtraction.

1. Read the word problem and determine the greatest common factor of the terms. 2. Keep the greatest common factor outside the brackets, divide the polynomial terms by this factor and write the remaining expression inside the brackets. 3. Verify your answer by multiplying the factors to get the original expression.

A rectangular garden is being separated into 6 equal sections for different types of flowers. The area of the garden is (12x + 30) square meters. What is the area of each section of the garden?

1. Find the greatest common factor of the two terms. The greatest common factor of 12 and 30 is 6.

2. Keep the greatest common factor outside the brackets, divide the polynomial terms by this factor and write the remaining expression inside the brackets. (12x + 30) = (6 ● 2x) + (6 ● 5) 6(2x + 5)

3. Verify your answer by multiplying the factors to get the original expression. 6(2x + 5) 12x + 30

Therefore, the area of each section of the garden is (2x + 5) square meters.

Practice Factoring Algebraic Expressions Word Problems

## Practice Problem 1

## Practice Problem 2

## Practice Problem 3

Factor – is a number that divides the given number without any remainder.

Linear Expression – an algebraic expression in which the variable is raised to the first power, and variables are not multiplied or divided.

Term – either a single number or variable, or numbers and variables multiplied together. Terms are separated by + or − signs.

Like terms – Terms that have the same power for the same variables. In like terms, one can only change the numerical coefficient.

Unlike terms – Terms that have different variables or the same variables raised to different powers.

Distributive Property – to multiply a sum or difference by a number, multiply each term inside the parenthesis by the number outside the parenthesis.

Constant – a term without a variable.

Variable – In algebra, a symbol (usually a letter) standing in for an unknown numerical value in an equation or expression.

Coefficient – is an integer that is multiplied with the variable of a single term or the terms of a polynomial.

Pre-requisite Skills Identify Factors Greatest Common Factor-GCF Arithmetic Sequences Geometric Sequences The Distributive Property Simplifying Complex Algebraic Expressions Add Linear Expressions Subtract Linear Expressions

Related Skills Solve One-Step Equations by Addition and Subtraction Solve One-Step Equations by Multiplying and Dividing Solve Equations with Rational Coefficients Solve Two-Step Equations Solve Complex Equations Solve Equations with Rational Coefficients Solve Two-Step Equations Write Two-Step Equations Solve Two-Sided Equations with Rational Coefficients Solve Multi-Step Equations

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## 9.6: Solve Applications of Quadratic Equations

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## Learning Objectives

By the end of this section, you will be able to:

- Solve applications modeled by quadratic equations

Before you get started, take this readiness quiz.

- The sum of two consecutive odd numbers is \(−100\). Find the numbers. If you missed this problem, review Example 2.18.
- Solve: \(\frac{2}{x+1}+\frac{1}{x-1}=\frac{1}{x^{2}-1}\). If you missed this problem, review Example 7.35.
- Find the length of the hypotenuse of a right triangle with legs \(5\) inches and \(12\) inches. If you missed this problem, review Example 2.34.

## Solve Applications Modeled by Quadratic Equations

We solved some applications that are modeled by quadratic equations earlier, when the only method we had to solve them was factoring. Now that we have more methods to solve quadratic equations, we will take another look at applications.

Let’s first summarize the methods we now have to solve quadratic equations.

## Methods to Solve Quadratic Equations

- Square Root Property
- Completing the Square
- Quadratic Formula

As you solve each equation, choose the method that is most convenient for you to work the problem. As a reminder, we will copy our usual Problem-Solving Strategy here so we can follow the steps.

## Use a Problem-Solving Strategy

- Read the problem. Make sure all the words and ideas are understood.
- Identify what we are looking for.
- Name what we are looking for. Choose a variable to represent that quantity.
- Translate into an equation. It may be helpful to restate the problem in one sentence with all the important information. Then, translate the English sentence into an algebraic equation.
- Solve the equation using algebra techniques.
- Check the answer in the problem and make sure it makes sense.
- Answer the question with a complete sentence.

We have solved number applications that involved consecutive even and odd integers, by modeling the situation with linear equations. Remember, we noticed each even integer is \(2\) more than the number preceding it. If we call the first one \(n\), then the next one is \(n+2\). The next one would be \(n+2+2\) or \(n+4\). This is also true when we use odd integers. One set of even integers and one set of odd integers are shown below.

\(\begin{array}{cl}{}&{\text{Consecutive even integers}}\\{}& {64,66,68}\\ {n} & {1^{\text { st }} \text { even integer }} \\ {n+2} & {2^{\text { nd }} \text { consecutive even integer }} \\ {n+4} & {3^{\text { rd }} \text { consecutive even integer }}\end{array}\)

\(\begin{array}{cl}{}&{\text{Consecutive odd integers}}\\{}& {77,79,81}\\ {n} & {1^{\text { st }} \text { odd integer }} \\ {n+2} & {2^{\text { nd }} \text { consecutive odd integer }} \\ {n+4} & {3^{\text { rd }} \text { consecutive odd integer }}\end{array}\)

Some applications of odd or even consecutive integers are modeled by quadratic equations. The notation above will be helpful as you name the variables.

## Example \(\PageIndex{1}\)

The product of two consecutive odd integers is \(195\). Find the integers.

Step 1 : Read the problem

Step 2 : Identify what we are looking for.

We are looking for two consecutive odd integers.

Step 3 : Name what we are looking for.

Let \(n=\) the first odd integer.

\(n+2=\) the next odd integer.

Step 4 : Translate into an equation. State the problem in one sentence.

“The product of two consecutive odd integers is \(195\).” The product of the first odd integer and the second odd integer is \(195\).

Translate into an equation.

\(n(n+2)=195\)

Step 5 : Solve the equation. Distribute.

\(n^{2}+2 n=195\)

Write the equation in standard form.

\(n^{2}+2 n-195=0\)

\((n+15)(n-13)=0\)

Use the Zero Product Property.

\(n+15=0 \quad n-13=0\)

Solve each equation.

\(n=-15, \quad n=13\)

There are two values of \(n\) that are solutions. This will give us two pairs of consecutive odd integers for our solution.

\(\begin{array}{cc}{\text { First odd integer } n=13} & {\text { First odd integer } n=-15} \\ {\text { next odd integer } n+2} & {\text { next odd integer } n+2} \\ {13+2} & {-15+2} \\ {15} & {-13}\end{array}\)

Step 6 : Check the answer.

Do these pairs work? Are they consecutive odd integers?

\(\begin{aligned} 13,15 & \text { yes } \\-13,-15 & \text { yes } \end{aligned}\)

Is their product \(195\)?

\(\begin{aligned} 13 \cdot 15 &=195 &\text{yes} \\-13(-15) &=195 & \text { yes } \end{aligned}\)

Step 7 : Answer the question.

Two consecutive odd integers whose product is \(195\) are \(13,15\) and \(-13,-15\).

## Exercise \(\PageIndex{1}\)

The product of two consecutive odd integers is \(99\). Find the integers.

The two consecutive odd integers whose product is \(99\) are \(9, 11\), and \(−9, −11\).

## Exercise \(\PageIndex{2}\)

The product of two consecutive even integers is \(168\). Find the integers.

The two consecutive even integers whose product is \(128\) are \(12, 14\) and \(−12, −14\).

We will use the formula for the area of a triangle to solve the next example.

## Definition \(\PageIndex{1}\)

Area of a Triangle

For a triangle with base, \(b\), and height, \(h\), the area, \(A\), is given by the formula \(A=\frac{1}{2} b h\).

Recall that when we solve geometric applications, it is helpful to draw the figure.

## Example \(\PageIndex{2}\)

An architect is designing the entryway of a restaurant. She wants to put a triangular window above the doorway. Due to energy restrictions, the window can only have an area of \(120\) square feet and the architect wants the base to be \(4\) feet more than twice the height. Find the base and height of the window.

## Exercise \(\PageIndex{3}\)

Find the base and height of a triangle whose base is four inches more than six times its height and has an area of \(456\) square inches.

The height of the triangle is \(12\) inches and the base is \(76\) inches.

## Exercise \(\PageIndex{4}\)

If a triangle that has an area of \(110\) square feet has a base that is two feet less than twice the height, what is the length of its base and height?

The height of the triangle is \(11\) feet and the base is \(20\) feet.

In the two preceding examples, the number in the radical in the Quadratic Formula was a perfect square and so the solutions were rational numbers. If we get an irrational number as a solution to an application problem, we will use a calculator to get an approximate value.

We will use the formula for the area of a rectangle to solve the next example.

## Definition \(\PageIndex{2}\)

Area of a Rectangle

For a rectangle with length, \(L\), and width, \(W\), the area, \(A\), is given by the formula \(A=LW\).

## Example \(\PageIndex{3}\)

Mike wants to put \(150\) square feet of artificial turf in his front yard. This is the maximum area of artificial turf allowed by his homeowners association. He wants to have a rectangular area of turf with length one foot less than \(3\) times the width. Find the length and width. Round to the nearest tenth of a foot.

## Exercise \(\PageIndex{5}\)

The length of a \(200\) square foot rectangular vegetable garden is four feet less than twice the width. Find the length and width of the garden, to the nearest tenth of a foot.

The length of the garden is approximately \(18\) feet and the width \(11\) feet.

## Exercise \(\PageIndex{6}\)

A rectangular tablecloth has an area of \(80\) square feet. The width is \(5\) feet shorter than the length.What are the length and width of the tablecloth to the nearest tenth of a foot?

The length of the tablecloth is approximately \(11.8\) feet and the width \(6.8\) feet.

The Pythagorean Theorem gives the relation between the legs and hypotenuse of a right triangle. We will use the Pythagorean Theorem to solve the next example.

## Definition \(\PageIndex{3}\)

Pythagorean Theorem

- In any right triangle, where \(a\) and \(b\) are the lengths of the legs, and \(c\) is the length of the hypotenuse, \(a^{2}+b^{2}=c^{2}\).

## Example \(\PageIndex{4}\)

Rene is setting up a holiday light display. He wants to make a ‘tree’ in the shape of two right triangles, as shown below, and has two \(10\)-foot strings of lights to use for the sides. He will attach the lights to the top of a pole and to two stakes on the ground. He wants the height of the pole to be the same as the distance from the base of the pole to each stake. How tall should the pole be?

## Exercise \(\PageIndex{7}\)

The sun casts a shadow from a flag pole. The height of the flag pole is three times the length of its shadow. The distance between the end of the shadow and the top of the flag pole is \(20\) feet. Find the length of the shadow and the length of the flag pole. Round to the nearest tenth.

The length of the flag pole’s shadow is approximately \(6.3\) feet and the height of the flag pole is \(18.9\) feet.

## Exercise \(\PageIndex{8}\)

The distance between opposite corners of a rectangular field is four more than the width of the field. The length of the field is twice its width. Find the distance between the opposite corners. Round to the nearest tenth.

The distance between the opposite corners is approximately \(7.2\) feet.

The height of a projectile shot upward from the ground is modeled by a quadratic equation. The initial velocity, \(v_{0}\), propels the object up until gravity causes the object to fall back down.

## Definition \(\PageIndex{4}\)

The height in feet, \(h\), of an object shot upwards into the air with initial velocity, \(v_{0}\), after \(t\) seconds is given by the formula

\(h=-16 t^{2}+v_{0} t\)

We can use this formula to find how many seconds it will take for a firework to reach a specific height.

## Example \(\PageIndex{5}\)

A firework is shot upwards with initial velocity \(130\) feet per second. How many seconds will it take to reach a height of \(260\) feet? Round to the nearest tenth of a second.

## Exercise \(\PageIndex{9}\)

An arrow is shot from the ground into the air at an initial speed of \(108\) ft/s. Use the formula \(h=-16 t^{2}+v_{0} t\) to determine when the arrow will be \(180\) feet from the ground. Round the nearest tenth.

The arrow will reach \(180\) feet on its way up after \(3\) seconds and again on its way down after approximately \(3.8\) seconds.

## Exercise \(\PageIndex{10}\)

A man throws a ball into the air with a velocity of \(96\) ft/s. Use the formula \(h=-16 t^{2}+v_{0} t\) to determine when the height of the ball will be \(48\) feet. Round to the nearest tenth.

The ball will reach \(48\) feet on its way up after approximately \(.6\) second and again on its way down after approximately \(5.4\) seconds.

We have solved uniform motion problems using the formula \(D=rt\) in previous chapters. We used a table like the one below to organize the information and lead us to the equation.

The formula \(D=rt\) assumes we know \(r\) and \(t\) and use them to find \(D\). If we know \(D\) and \(r\) and need to find \(t\), we would solve the equation for \(t\) and get the formula \(t=\frac{D}{r}\).

Some uniform motion problems are also modeled by quadratic equations.

## Example \(\PageIndex{6}\)

Professor Smith just returned from a conference that was \(2,000\) miles east of his home. His total time in the airplane for the round trip was \(9\) hours. If the plane was flying at a rate of \(450\) miles per hour, what was the speed of the jet stream?

This is a uniform motion situation. A diagram will help us visualize the situation.

We fill in the chart to organize the information.

We are looking for the speed of the jet stream. Let \(r=\) the speed of the jet stream.

When the plane flies with the wind, the wind increases its speed and so the rate is \(450 + r\).

When the plane flies against the wind, the wind decreases its speed and the rate is \(450 − r\).

The speed of the jet stream was \(50\) mph.

## Exercise \(\PageIndex{11}\)

MaryAnne just returned from a visit with her grandchildren back east. The trip was \(2400\) miles from her home and her total time in the airplane for the round trip was \(10\) hours. If the plane was flying at a rate of \(500\) miles per hour, what was the speed of the jet stream?

The speed of the jet stream was \(100\) mph.

## Exercise \(\PageIndex{12}\)

Gerry just returned from a cross country trip. The trip was \(3000\) miles from his home and his total time in the airplane for the round trip was \(11\) hours. If the plane was flying at a rate of \(550\) miles per hour, what was the speed of the jet stream?

Work applications can also be modeled by quadratic equations. We will set them up using the same methods we used when we solved them with rational equations.We’ll use a similar scenario now.

## Example \(\PageIndex{7}\)

The weekly gossip magazine has a big story about the presidential election and the editor wants the magazine to be printed as soon as possible. She has asked the printer to run an extra printing press to get the printing done more quickly. Press #1 takes \(12\) hours more than Press #2 to do the job and when both presses are running they can print the job in \(8\) hours. How long does it take for each press to print the job alone?

This is a work problem. A chart will help us organize the information.

We are looking for how many hours it would take each press separately to complete the job.

## Exercise \(\PageIndex{13}\)

The weekly news magazine has a big story naming the Person of the Year and the editor wants the magazine to be printed as soon as possible. She has asked the printer to run an extra printing press to get the printing done more quickly. Press #1 takes \(6\) hours more than Press #2 to do the job and when both presses are running they can print the job in \(4\) hours. How long does it take for each press to print the job alone?

Press #1 would take \(12\) hours, and Press #2 would take \(6\) hours to do the job alone.

## Exercise \(\PageIndex{14}\)

Erlinda is having a party and wants to fill her hot tub. If she only uses the red hose it takes \(3\) hours more than if she only uses the green hose. If she uses both hoses together, the hot tub fills in \(2\) hours. How long does it take for each hose to fill the hot tub?

The red hose take \(6\) hours and the green hose take \(3\) hours alone.

Access these online resources for additional instruction and practice with solving applications modeled by quadratic equations.

- Word Problems Involving Quadratic Equations
- Quadratic Equation Word Problems
- Applying the Quadratic Formula

## Key Concepts

- Translate into an equation. It may be helpful to restate the problem in one sentence with all the important information. Then, translate the English sentence into an algebra equation.
- Solve the equation using good algebra techniques.
- For a triangle with base, \(b\), and height, \(h\), the area, \(A\), is given by the formula \(A=\frac{1}{2}bh\).

- For a rectangle with length,\(L\), and width, \(W\), the area, \(A\), is given by the formula \(A=LW\).

- The height in feet, \(h\), of an object shot upwards into the air with initial velocity, \(v_{0}\), after \(t\) seconds is given by the formula \(h=-16 t^{2}+v_{0} t\).

## Grade 7 Mathematics Module: Solving Problems Involving Algebraic Expressions

This Self-Learning Module (SLM) is prepared so that you, our dear learners, can continue your studies and learn while at home. Activities, questions, directions, exercises, and discussions are carefully stated for you to understand each lesson.

Each SLM is composed of different parts. Each part shall guide you step-by-step as you discover and understand the lesson prepared for you.

Pre-tests are provided to measure your prior knowledge on lessons in each SLM. This will tell you if you need to proceed on completing this module or if you need to ask your facilitator or your teacher’s assistance for better understanding of the lesson. At the end of each module, you need to answer the post-test to self-check your learning. Answer keys are provided for each activity and test. We trust that you will be honest in using these.

Please use this module with care. Do not put unnecessary marks on any part of this SLM. Use a separate sheet of paper in answering the exercises and tests. And read the instructions carefully before performing each task.

If you have any questions in using this SLM or any difficulty in answering the tasks in this module, do not hesitate to consult your teacher or facilitator.

In this lesson, we will use algebraic expression to find the values of the things or unknown in real life situations.

This module was designed and written with you in mind. It is here to help you master Solving Problems Involving Algebraic Expressions. The scope of this module permits it to be used in many different learning situations. The language used recognizes the diverse vocabulary level of students. The lessons are arranged to follow the standard sequence of the course. But the order in which you read them can be changed to correspond with the textbook you are now using.

The module is comprised of only one lesson:

- Solving Problems Involving Algebraic Expressions

After going through this module, you are expected to:

- solve problems involving algebraic expressions.

## Grade 7 Mathematics Quarter 2 Self-Learning Module: Solving Problems Involving Algebraic Expressions

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To solve an algebraic word problem: Define a variable. Write an equation using the variable. Solve the equation. If the variable is not the answer to the word problem, use the variable to calculate the answer. It's important for us to keep in mind how we define our variables.

Algebraic Expressions and Word Problems Related Topics: More Lessons for Grade 7 Math Worksheets Share this page to Google Classroom Examples, solutions, videos, worksheets, games and activities to help Algebra 1 or grade 7 students learn how to write algebraic expressions from word problems. Beginning Algebra & Word Problem Steps Name what x is.

Part 1 Understanding the Basics Download Article 1 Understand the difference between an algebraic expression and an algebraic equation. An algebraic expression is a mathematical phrase that can contain numbers and/or variables. It does not contain an equals sign and cannot be solved.

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.

Table 1.4.1. Terms88 in an algebraic expression are separated by addition operators and factors89 are separated by multiplication operators. The numerical factor of a term is called the coefficient90. For example, the algebraic expression x2y2 + 6xy − 3 can be thought of as x2y2 + 6xy + ( − 3) and has three terms.

An algebraic expression is a series of terms - numbers, variables, or a product of numbers and/or variables - separated by signs of addition and subtraction. In some cases, an algebraic...

Evaluate the expression ( a − b) 2 If a = 3 and b = −5, at a = 3 and b = −5. Solution. Following "Tips for Evaluating Algebraic Expressions," first replace all occurrences of variables in the expression ( a − b) 2 with open parentheses. (a − b)2 = (() − ())2 ( a − b) 2 = ( () − ()) 2. Secondly, replace each variable with its ...

Here is a set of practice problems to accompany the Rational Expressions section of the Preliminaries chapter of the notes for Paul Dawkins Algebra course at Lamar University.

MA.7.EE.B.A: Use variables to represent quantities in a real-world or mathematical problem, and construct simple equations and inequalities to solve problems by reasoning about the quantities. MA.7.7.EE.B.4.A: Solve word problems leading to equations of the form px+q=r and p (x + q)=r, where p, q, and r are specific rational numbers.

r + 15 = 2s, where r = the number of points Ron scored and s = the number of points Shawn scored. Any two different variables are acceptable. Problem : Translate the following word statement into an algebraic expression or equation: "John's age divided by 3 is equal to John's age minus 14." = j - 14, where j = John's age.

To solve an algebraic expression, simplify the expression by combining like terms, isolate the variable on one side of the equation by using inverse operations. Then, solve the equation by finding the value of the variable that makes the equation true. ... Study Tools AI Math Solver Popular Problems Study Guides Practice Cheat Sheets ...

The rational expression will be undefined when the denominator is zero. Since x + 3 = 0 when x = − 3, then -3 is a critical point. Step 3. Use the critical points to divide the number line into intervals. Step 4. Above the number line show the sign of each factor of the rational expression in each interval.

Explore this assortment of worksheets to solve rectangle problems involving length, width, area, perimeter and diagonal with measures provided as algebraic expressions. These exclusive printable worksheets are aligned with the common core curriculum and are proposed for the use of middle school students. Recall and apply the properties of ...

🔗 A.3 Algebraic Expressions and Problem Solving 🔗 You are familiar with the use of letters, or variables, to stand for unknown numbers in equations or formulas. Variables are also used to represent numerical quantities that change over time or in different situations.

Learn how to solve problems involving algebraic expressions.Grade 7 MathematicsLearning Task 3 Solve the following. Do this in a separate sheet of paper.1. ...

1. Find the greatest common factor of the two terms. The greatest common factor of 12 and 30 is 6. 2. Keep the greatest common factor outside the brackets, divide the polynomial terms by this factor and write the remaining expression inside the brackets. (12x + 30) = (6 2x) + (6 5) 6 (2x + 5) 3. Verify your answer by multiplying the factors to ...

It may be helpful to restate the problem in one sentence with all the important information. Then, translate the English sentence into an algebra equation. Solve the equation using good algebra techniques. Check the answer in the problem and make sure it makes sense. Answer the question with a complete sentence. Area of a Triangle

116 10K views 2 years ago Lesson for Grade 7 Mathematics This lesson is intended for learners to learn the concept of "Solving Problems Involving Algebraic Expressions". ...more ...more...

The module is comprised of only one lesson: Solving Problems Involving Algebraic Expressions After going through this module, you are expected to: solve problems involving algebraic expressions. Grade 7 Mathematics Quarter 2 Self-Learning Module: Solving Problems Involving Algebraic Expressions Math7_Quarter2_Module7

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