From Arithmetic to Algebra: Variables, Word Problems, Fractions and the Rules

CONTENTS OF CURRICULUM UNIT 17.05.09

  1. Unit Guide
  1. Overview
  2. Rationale
  3. Context
  4. Mathematics Background
  5. Teaching Strategies
  6. Classroom Activity
  7. Appendix 1 Implementing District Standards
  8. Notes
  9. Bibliography

Making Sense of Solving Equations Through Word Problems

Sally Yoo

Published September 2017

Tools for this Unit:

Mathematics Background

Prerequisite Skills

Conceptual understanding of the prerequisite skills such as simplifying expressions and the roles of the variable can affect student learning in solving algebraic equations. The paper, “Key Misconceptions in Algebraic Problem Solving” (7) discusses the importance of students having deep conceptual understanding of features in an equation such as the equal sign, variable, like terms, negative signs, and more. The article states that “misconceptions or gaps in conceptual knowledge of relevant features inhibit students’ performance and learning” of procedural algebra. Thus, students with the necessary prerequisite concepts will likely learn the concepts of procedural algebraic solving, while others may have “shallow” learning, which can lead to the mistakes I have seen in my classrooms year after year. Fortunately one of the key findings of the study is that if these same prerequisite skills are taught during the lesson, those students can also learn the skills for solving equations.

In this unit I will address the prerequisite skills by continually connecting manipulation of the algebraic equation to the computation used, and also referring back to the context of the word problem the equation came from.

Example: Sum of two consecutive numbers are 31. Find the numbers.

Arithmetic approach:

Since the numbers are consecutive, the larger number is one more than the smaller one. If we subtract 1 from it, we will get the smaller one. This will also subtract 1 from the sum, so the double of the smaller number is

31 – 1 =30.

Dividing by 2 then gives the smaller number.

30/2 = 15

Hence the two consecutive numbers are

15 and 16.

Algebraic approach:

Let x represent the smaller of the two numbers.

Since the numbers are consecutive, the other number is (x + 1).

Therefore, the word problem is translated to x + (x + 1) = 31.

(x +x) + 1 = 31 Associative Property.

2x + 1 = 31 Combine like terms.

2x + 1 + -1 = 31 +-1 Addition Property of Equality.

2x = 30 Combine like terms.

(1/2) 2x = (1/2) 30Multiplication Property of Equality

((1/2)2)x=15 Associative Property of Multiplication

1x = 15 Inverse Property of Multiplication

x =15 Identity Property of Multiplication

The smaller of the two numbers is 15.

Verify in the original equation.

15 + (15 + 1) = 31

15 + 16 = 31

31 = 31 verified

Through discussions, I plan to activate prior knowledge on combining like terms by showing that the x in the above equation is referring to the same number 15, the smaller number of the two numbers. Thus, we can combine the x and x, which results in 2x or two times the value of x, since both xs represent the same value: 15.  But the number 1 cannot be combined with x since they do not have the same value. In addition, the idea of relational equality, the left side of the equation having the same value as the right side, will be addressed in similar manner. In the above example, by substituting 15 for both variables x, the left side of the equation is equal to 31 just as the right side is equal to 31. Thus, relational equality is emphasized.

Reverse Engineering

“Reverse engineering” in this unit refers to connecting features of the given equation with the word problem it models. Students will make sense of the structure of the algebraic equation by examining each feature in context of the word problem before transforming the equation. When faced with word problems, students intuitively slow down and try to make sense of them. However students do not have the same expectations toward algebraic equations. They do not recognize the importance of investing time into understanding all the features of the equation. Instead they race through them until they come upon an equation with the same structure but which is written slightly differently. For those students, the slight difference in the way the problem is written makes it a totally new problem. Even though the problem is essentially the same as the problems they have successfully solved previously, these students do not know how to approach it.

I will use reverse engineering to explicitly teach students to examine the structure of the equation before transforming it to a simpler form so that they can generalize the concepts, instead of memorizing infinitely many procedures for all the different ways equations can be written. Since translating words to symbols was covered in the two previous curricular units, in this unit students will take the time to make sense of the equation by connecting all the relevant features of the equation to the text in the word problem, instead of deriving the equation from the word problem. For example: The sum of two consecutive numbers is 31. Find the numbers. The equation x + (x + 1) = 31 will be given to the students as a model for the above word problem. Students will focus on each feature of the equation through following questions: What does x represent? Why are there three terms (x, x, 1)? Why are all the terms being added? Why do they equal to 31? By understanding the reasons behind the structure of the equation, the students can think more deeply before transforming the equation.

Justification

Furthermore, students will be expected to justify each statement by writing the properties for each subsequent equivalent equation until they become experts with the Properties of Equality.

  1. Addition Property of Equality: if a = b, then a + c = b + c.
  2. Multiplication Property of Equality: if a = b, then a(c) = b(c).
  3. Transitive Property of Equality: if a =b, and b =c, then a = c.

Here, a, b, and c stand for arbitrary numbers.

Because the process is so cumbersome, I did not stress the need to justify each step in the past. Students used the shortcuts, transforming equations without justification, before they became experts with the properties of equality, which lead to confusions about when to use the rules of operations versus properties of equality. In YNI 2017 Math Seminar, Dr. Howe, seminar leader, Professor Emeritus at Yale, and Professor of Mathematics Education at Texas A&M promoted the mantra that “Shortcuts are privileges of the experts”. When shortcuts are shown to students before they have a thorough understanding, they become bewildered and develop misconceptions. Promoting the shortcut before understanding is in place may block future learning. In order to avoid misconceptions, students need to see the logical flow and the process must be transparent to them. They need to understand that just as the Order of Operations is used to simplify numerical expressions and Rules of Arithmetic are used to simplify algebraic expressions, the Properties of Equality are used to manipulate equations.

Unlike the traditional approach, I will use only the three Properties of Equality mentioned above. Most textbooks treat the Addition Property of Equality and the Subtraction Property of Equality as separate entities. However, I will treat them as one property and take time to show students that subtraction is addition of the negative; thus my students will only need the Addition Property of Equality. Similarly, rather than the traditional way of using the Multiplication Property of Equality and Division Property of Equality by themselves, I will treat them as one property. Since division is multiplication by reciprocal, students only need to use the Multiplication Property of Equality. There are a couple of benefits to this approach. First, students will focus on the inherent inverse relationship between addition and subtraction, and multiplication and division. Second, students will have fewer Properties of Equality to learn.

Verification

My students will also verify the solution. According to G. Polya, author of ‘How to Solve It’ (8), verifying a solution by substituting it into the original problem and referring back to the context in the text of the problem is an important and necessary step. A solution needs to be interpreted in the context of the given situation to determine whether the obtained result makes sense in the context of the problem. Since a variable is a placeholder for a number in a given set, the obtained value must be a member of the predetermined set. If the solution is not a member of the set, then the solution would not make sense, and the plan must be reevaluated. My students will practice the cycle of devising a plan, reflecting on the process, and revising the plan until a solution that makes sense in the context of the situation is identified.

For example, in the problem, “The sum of two consecutive even numbers is 46. Find the numbers.”, if the variable represents the smallest even integer, then the solution must be an even number and also the smaller of the two numbers. If the solution found is not an even number, then it would be a signal for the student to rethink the problem and revise their plan. If the solution is not the smaller of the two even numbers, then that would also be a signal for the student to revise their thinking. This essential step is often skipped by the student but I will have my students practice it continually until it becomes a habit. The process of stopping to reflect becomes more and more valuable as problems become more complex. Famous mathematicians realize that solutions are often found through much trial and error, but most of my middle school math students come to my class believing “smart” students know the path to the solution right away. Thus when the path is not apparent immediately, they believe they are not smart and cannot do math. One of goals of this unit is to provide many opportunities for students to experience the need to revise their plans and try different approaches. Hopefully, my students will come to realize that the process of solving complex problems is often fraught with revision and that perseverance is a normal part of the process and a practice to be valued.

Additionally, verifying by substitution into the original equation is a great opportunity to emphasize the concept of equivalence present in an algebraic equation, where two algebraic expressions are set to equal each other. An algebraic equation essentially is a question: it asks, of all the values that the variable might take, what particular value can be substituted into the two expressions so that the value on the right side of the equal sign “=” is the same as the value on the left side. This is a good opportunity to discuss the multiple meanings associated with “=” sign. From previous math classes, most students understand equal sign (=) only as signal to perform operations to find an answer. However, in algebra and higher math, it is essential to take the = sign as meaning that the two expressions on the either side of it should have the same value. For example 2x = y expresses the relationship between x and y, which can be interpreted as “the value of y is double the value of x” or “twice the value of x is equal to the value of y.” Thus, there is a defined relationship between the values of x and the values y that follow the rule described by 2x = y. The values of x and y are dependent on each other and each is responsible for how the other value changes: if the value of x changes, then the value of y changes in response to the change in x. This difficult concept of relational equivalency is introduced in 7th grade but needs to be mastered in 8th grade in order for a student to be successful in Algebra 1 and beyond.

Transitive Property of Equality

Less discussed in the classroom is the equivalence among the transformed equations. Using word problems, my students will understand that the original equation and all the transformed equations are essentially the same. The Transitive Property of Equality (if a = b and b = c, then a = c) is used almost unconsciously in conjunction with Properties of Operations to transform complex equations to simpler equations. My students can have a more robust understanding around equivalent equations if this process is made more transparent.

For example: Sum of two consecutive numbers are 31. Find the numbers.

Another way to explain the arithmetic approach:

31 is the sum of the two consecutive numbers. If 1 is taken away, then the sum is total of two of numbers that have the same value, the lower of the two consecutive numbers.

31 – 1 =30

Since 30 is two times the lower number, 15 is the lower number.

30/2 = 15

Therefore, the consecutive numbers are 15, 16

Let x represent the smaller of the two numbers. Then next consecutive number is (x + 1).

Therefore, the two consecutive numbers whose sum is 31 is represented by x + (x + 1) = 31.

(x +x) + 1 = 31 Associative Property.

2x + 1 = 31 Combine like terms

2x + 1 + -1 = 31 +-1 Addition Property of equality.

2x = 30 Combine like terms.

(1/2) 2x = (1/2) 30 Multiplication Property of equality

((1/2)2)x=15 Associative Property of Multiplication.

1x = 15 Inverse Property of Multiplication.

x =15 Identify Property of Multiplication.

The smaller of the two numbers is 15. Therefore, the consecutive numbers are 15, 16.

Verify in the original equation.

15 + (15 + 1) = 31

15 + 16 = 31

31 = 31 verified

The equation x + x + 1 = 31 is equivalent to the 2x + 1 = 31 by the Transitive Property of Equality because x + (x +1) = (x + x) +1= (1x + 1x) +1 = (1 + 1)x + 1 = 2x +1.Thus, x + (x +1) is simplified to 2x + 1 through a sequence of operations, of which the most important is combining like terms. Since x + x + 1 = 31 and x +x + 1 = 2x + 1, thus 2x + 1 = 31, we can conclude that the two equations x + x + 1 = 31 and 2x + 1 = 31 are essentially the same equations asking the same question.

Multiplication Property of Equality

The following one-operation word problems are intentionally sequenced to teach the Multiplication Property of Equality and highlight any misunderstanding of multiplicative relationships, which is taught in previous grades. Students need a deep conceptual understanding of the multiplicative relationship to fully grasp how the Multiplicative Property of Equality is used to transform the equations to uncover the unknown value. When the meaning of multiplication, division, and fractions are explicitly dealt with in a concrete setting, the misconceptions are more easily resolved. Because the computational demands are lower, students can focus on the structure of the equation and how each part functions during the transformation process.

Students will be asked to solve the following simple word problems using computation. They are designed and sequenced in such a way that most students will likely find the solution using computation. The purpose of the following word problems is to make the steps used in computation more transparent, so they can form a generalized rule. The steps in computational approach will be explicitly connected to the steps in algebraic approach. Rather than the conventional view of seeing arithmetic and algebraic as two completely different worlds, hopefully students come to realize that the approaches are closely connected.

Example: Equal Group: Number of group members unknown

Your friend tells you she bought 12 bananas. On this particular day, every bunch had the same number of bananas so she bought 2 bunches. She says to you “How many bananas do you think were in each bunch?

Arithmetic approach:

12 bananas are equally shared between the two bunches. Therefore, each bunch has 6 bananas.

12/2=6

Algebraic approach:

Let x represent the number of bananas in one bunch.

2x=12

 

(1/2)2x=(1/2)12 Multiplication Property of Equality.

Since  (1/2)2x=((1/2)2)x= (1)x=x, and (1/2)12=6, we can conclude by  the Transitivity Property of Equality that x=6

There are 6 bananas in each bunch

Verify by substituting into the original equation:

2(6) = 12

12 = 12 (verified)

The most important part is defining the variable: let x represent the number of bananas in one bunch. Once the variable is clearly defined, students can focus on the meaning of 2x, which means two times the number of bananas in each bunch since the friend bought two bunches. Thus the multiplicative relationship represented in the algebraic equation and in the text of the word problem can be explicitly connected. Furthermore, the connection between dividing by two in the arithmetic approach and the multiplying by half in the algebraic approach will be brought to the attention of the students to make the process transparent. Through questioning and class discussions, students would see that the computational step used in the arithmetic approach is also present in the algebraic approach.

For the multiplication property of equality, I will once again refer to the situation described in the word problem. Since 2x is two times the number of bananas in one bunch, half of 2x is the number of bananas in one bunch. Since 12 represents the twelve bananas from two bunches, half of 12 is the number of bananas from one bunch. Therefore, students can clearly understand and concretely see that the half multiplied to the left side of the equation is also multiplied to the right side of the equation to identify the number of bananas in one bunch. Thus, the rationale for the Properties of Equality used to maintain the balance between the two sides of the equation when transforming equations is shown explicitly.

Example: Comparison: Larger number unknown

Your teacher and your mom are talking at the registration when the teacher suddenly turns to you and asks how old your grandmother is. You forgot how old your grandmother is but you remember that your youngest aunt Kristie is 30 years old. Furthermore, your aunt Kristie is half the age of your grandmother. How old is your grandmother?

One possible approach:

Kristie is 30 years old.

Since 30 years is half of grandmother’s age, grandmother’s age is twice 30, or 60.

2(30)=60

Let p represent grandmother’s age.

30= 1/2 p .

(2)30=(2) 1/2 p Multiplication Property of Equality. (2) 30=((2) 1/2)p Associative Property of Equality.

60 = 1p Inverse Property of Multiplication.

60=p  Identify Property of Multiplication.

Since(2) 1/2 p=((2) 1/2)p=(1)p=p, and 2(30) = 60, we can conclude that p = 60 by the Transitive Property of Equality.

Grandmother is 60 years old.

Verify by substituting into the original equation.

30 = ½(60)

30 = 30 (verified).

The purpose of this example is to show students that the two above examples have multiplicative relationships represented in slightly different ways. Both equations actually have the same structure, Thus, the same property, Multiplication Property of Equality, is used to simplify an equation in the form of ax=b to the form x = c (where c = b/a).

Addition Property of Equality

Two operation word problems are arranged to teach the Addition Property of Equality and to address any misconceptions and gaps regarding prerequisite concepts of operations with negative signs, unlike terms, and the inverse property of addition. I will still write the equation that models the word problem since translating words to equations was covered by Rachelle Soroten, 2017 NYI fellow and author of the first curricular unit of the four complementary units. The problems are still simple enough that the arithmetic approach would be more efficient than using algebraic procedure. In fact, using arithmetic would avoid the usual mistakes I see around the operation with negative numbers and other prerequisite concepts. However, as students continue to compare the two approaches, they may begin to see the benefits of using an algebraic method. As problems become increasingly complex, more students may switch their approach from arithmetic to algebraic in the middle of the process when they experience difficulties. Thus, students are constructing their own knowledge around the limitations and benefits of both approaches through their personal experience rather than being told which method should be applied to which types of problem.

Example: Change (+): Array Group Size Unknown

There are 26 desks in your math class. Because the classroom is not a perfect rectangle, some desks are arranged in equal rows and other desks are arranged in a group of 6 desks.  You are sitting in the group of 6 desks. As you look around, you realize there are 5 rows. You started to wonder how many desks are in a row.

Arithmetic approach:

If the 6 desks in the group are taken away from the total, then 20 desks are arranged in rows.

26-6=20

Since 20 desks are arranged in 5 rows, 4 desks are in each row.

20/5=4

Therefore, 4 desks are in each row.

Algebraic approach:

Let x represent the number desks in each row.

 

5x+6=26.

5x+6+(-6)=26+(-6) Addition Property of Equality.

5x=20 Combine like terms.

(1/5)5x=(1/5)20 Multiplication Property of Equality.

((1/5)5)x=4 Associative Property of Multiplication.

1x = 4 Identity Property of Multiplication.

x = 4

4 desks are in each row.

Verify by substituting into the original equation:

5(4) + 6 = 26

20 + 6 = 26

26 = 26 verified

As noted above the equation 5x + 6 = 31, which models the word problem, will be given to the students. The purpose of the above example is for the students to notice the structure of the equation and connect each feature to the text.  This is the “reverse engineering” discussed above. Thus, students understand that + 6 represents the six desks that are not part of the array, 5x represents five rows times some the unknown number of desks in each row, and 31 represents the total number of desks in the classroom. The aim is for students to see that each part of the equation has a purpose and a concrete meaning. Once students understand the role of each feature, they can more easily understand why 5x cannot be combined with 6. My hope is that by spending time connecting each feature of the equation to the context in the word problem, misconceptions such as transforming 5x + 6 = 31 to 11x = 31 can be avoided.

To teach the Addition Property of Equality, I would bring students’ attention to the fact that the same operations are performed in the arithmetic approach and the algebraic approach. In the arithmetic method, 31 is decreased by 6 to take away the extra desks that were not part of the array. The same procedure is present in the algebraic approach. Since 6, the six desks, is not part of the array, -6 is added to 5x + 6 so that the expression can be simplified 5x: five rows times the number of desks in a row. Since 31 represents the total desks in the classroom, 31 would also have to decrease by 6: the six desks eliminated in 5x + 6. Thus students can observe how the Addition Property of Equality is used to simplify the equation. Students can understand the rationale for needing to simplify the equation. They would realize that if they could focus only on the arrays, then finding the number of desks in a row is much simpler.

Once the students simplify the equation to 5x = 25, they can use the Multiplication Property of Equality, the concept already addressed in discussing the one-step equation. By comparing the arithmetic approach and the algebraic approach, students will notice that the same operations are performed in both approaches. In computation, students took away 6 from 31, then divided the result, 25, by 5 to obtain 5, five desks in a row. In algebraic approach, 6 is taken away from the both sides of equal sign (=), then divide by 5 both sides of the equal sign. The same operations are used for both approaches.

Example: Change (-): Compare – Bigger Unknown

Your grandmother is telling you a story of why she only has 6 hats even though she used to have a lot them. On her 40th birthday, there was a fire because not all the birthday candles were put out.  She was able to salvage only one-third of her hats. Then she lost 2 more hats most recently. How many hats did she have before the fire?

Arithmetic approach:

Right after the fire, grandmother had 8 hats since she currently has 6 hats and lost 2 hats after the fire.

(6+2)=8

Since 8 hats is one-third of the total hats, she had three times as many hats before the fire.

8×3=24

Therefore, grandmother had 24 hats before the fire.

Algebraic approach:

Let t represent the number of hats grandmother had before the fire.

6=  1/3 t-2

6+2=( 1/3 t-2)+2 Addition Property of Equality.

8 = 1/3 t+(-2+2) Associative Property of Addition.

8= 1/3 t + 0 Inverse Rule of Addition.

8= 1/3 t Identify Property of Addition.

(3)8=(3) 1/3 t Multiplication Property of Equality.

24 =((3) 1/3 )t Associative Property of Multiplication.

24 = 1t Inverse Property of Multiplication.

24 = t Identity Property of Multiplication.

Grandmother had 24 hats before the fire.

Verify by substituting into the original equation:

6= 1/3 (24)-2

6 = 8 – 2

6 = 6  verified

The aim of the above example is to show that the Addition Property of Equality is used for equations that are written slightly differently but in effect, have the same structure. This is an opportunity to reinforce the concept that subtraction is the same as adding the opposite. In other words, (1/3)t-2 is effectively the same as(1/3)t + -2. Therefore structurally, the two above examples are the same. Therefore, both of the equations can use the Addition Property of Equality to simplify the equation ax +b = c to ax = d (with d = c – b). Then, the Multiplication Property of Equality can be used to simplify the equation ax =d to the equation x = e.

Deriving and Solving Multi-step Equations

For the following multi-step word problems, reverse engineering will not be used because the previous curricular unit written by Rachelle Soroten stopped at deriving algebraic equations with simple two-step problems. Thus, with the introduction of the problem involving multi-step operations, my students will be expected to derive algebraic equations. However, they will also be expected to solve using both arithmetic and algebraic approaches. My hope is that, since we spent a significant amount of time connecting the algebraic equation to the texts in the word problems, my students will have developed the habit of reading word problems more closely. I hope they will come to the realization that word problems require patience and perseverance but are very much approachable and not impenetrable. In fact, word problems help with understanding the rules that govern procedural algebra.

Example: Consecutive Odd Numbers: All Numbers Unknown

The sum of the three consecutive odd numbers is 99. What are the numbers?.

Arithmetic approach:

The average of three numbers whose sum is 99 is 33.

99/33=3.

However, if the numbers are consecutive odd numbers, then their average is equal to the middle number.

Since 33 is the middle number, the smaller consecutive odd number is 2 smaller (33 – 2), and the bigger consecutive odd number is 2 greater (33 + 2).

Therefore, the consecutive odd numbers are 31, 33, and 35.

Algebraic approach:

Let s represent the smallest of the three odd numbers. Since the numbers are odd numbers, the other numbers are (s +2) and (s + 4). However, variable s can represent the middle or the highest number. If s represents the middle number, then the other numbers would be (s – 2) and (s + 2). If s represents the highest number, then other numbers would be (s – 2) and (s – 4).

s+(s+2)+(s+4)=99.

3s+6=99  Commutative Property of Addition and Combine like terms.

3s+6+ -6=99+ -6 Addition Property of Equality.

3s=93 Combine like terms.

(1/3)3s=(1/3)93 Multiplication Property of Equality. ((1/3)(3))s= 31 Associative Property of Multiplication.

1s = 31 Inverse Property of Multiplication.

s = 31 Identify Property of Multiplication.

31 is the smallest consecutive odd number. The consecutive odd numbers are 31, 33, and 35.

Verify by substituting into the original equation:

31+31+2+31+4=99

31 + 33 + 35 = 99

99 = 99 verified

After students have found the three consecutive numbers using arithmetic, they will work on deriving an algebraic equation that models the situation. One of the most important aspects of writing an algebraic equation is defining the variable including the unit. For the above example, a solution of 32, which is an even number, may not signal to the students that the solution does not make sense if they did not clearly state the variable must represent an odd number. Thus, in this case, the students most likely did not pay close attention to the fact the numbers are odd numbers and are not just consecutive numbers. Without clearly defining the variable, students would struggle in making sense of the solution in the context of the problem. If students are not taking time to make sense of the solution, they are not taking time to reflect and revise their approach during the problem solving process. When students take time to reflect on their work, they are learning that the process involves revision, attention to detail, and perseverance. These are many of the mathematical habits described in Common Core Mathematical Standards. (See Appendix 1)

Once the variable is clearly defined, students can refer to their computation to help set up the algebraic equation. If the variable is defined as the smallest odd integer, 31, then the subsequent odd number would be 33, which is 31 + 2. The largest odd number would be 35, which is 31 + 4.  Structurally the numerical equation (31 + 31 +2 + 31 + 4 = 99) is the same as the algebraic equation (s + s +2 +s + 4 = 99). If the students defined the variable as the largest odd number, 35, then the preceding consecutive odd numbers would be 33, which is 35 – 2, and 31, which is 35 – 4. Thus, the algebraic equation would change to s + s – 2 + s – 4 = 99. Showing different equations that model the same situation emphasizes the importance of defining a variable and the impact the definition can have on writing algebraic equations.

Simplifying the Multi-step Equations to ax + b =c

Another reason for the above example is to teach students to simplify complex equations to simple two – step equations. Multi-step equations with variables on both sides of the equation are addressed later in this unit. Students will use rules of operations, such as combining like terms and distributive property, to rewrite the equation in ax + b = c form. In the above example, students will rewrite s+s+2+s+4=99 to the simpler form 3s+6=99. After the equation is simplified to ax + b =c, students will transform the equation using the Addition Property of Equality and the Multiplication Property of Equality. In the past, some of my students did not have a clear understanding of the purpose of the Rules of Arithmetic and the Properties of Equality, nor did they understand the differences between them. By explicitly showing that the Rules of Arithmetic are used to rewrite complex equations to the simpler ax + b =c form, and the Properties of Equality are used to transform ax + b =c to the form x = d, essentially finding the value of the variable, the roles of the two sets of the rules should become much clearer to the student.

The Tipping Point

Due to the gradual increase in complexity, each student will reach a tipping point, which will be different for different students, where they come to rely more heavily on the algebraic procedures to solve problems. As the computations get more complicated, students will see the need for symbols to represent the situation. Their reliance only on computational strategy will gradually shift to include procedural algebraic strategies. Therefore, the challenge for more fluent algebraic procedural problem solvers would be to solve the problems using the arithmetic approach and show the operational connections between the two approaches. Interestingly, the process of switching to the arithmetic approach from the algebraic approach may be extremely difficult for the students. However, this process will deepen their understanding of number relationships. Furthermore, students may have a new found appreciation for the beauty of Algebra.

Example: Different Prices: Both Prices unknown: multi–step equation with distributive property a(x + b) = c.

Movie tickets cost: $14 for adults and $8 for children. The movie theater sold 40 tickets for a total of $416. How many adult tickets were sold?

Arithmetic approach:

If only children tickets were sold then the total cost would be $320.

8(40) = $320

The difference in the two totals is $96.

$416 - $320  = $96

The difference between adult tickets and children tickets is $6.

$14 - $8 =$6

16 adult tickets were sold.

96/6 = 16.

Algebraic Approach:

Let x represent the number of adult tickets sold.

 

14x+8(40-x)=416

14x+320-8x=416 Distributive Property

6x+320=416 Combine like terms

6x+320-320=416-320 Addition Property of Equality

6x=96 Combine like terms

(1/6)6x=(1/6)96 Multiplication Property of Equality

x = 16

16 adult tickets were sold.

Verify by substituting into the original equation:

14(16)+8(40-16)=416

224 + 320 – 128  = 416

544 – 128 = 416

416 = 416 verified

The above problem is a good example of the tipping point. Many students find the algebraic approach easier to understand compared with the arithmetic approach. The concepts used to solve the problem using the algebraic approach are similar to the ones used in previous problems, thus the difficulty level does not significantly increase. Because the arithmetic approach is much more difficult than the previous problems, I expect most students to solve the above problem successfully using the algebraic approach but struggle with the arithmetic approach. By the way, the arithmetic solution given above uses what was traditionally known as the method of false position, which was a well-recognized technique in the era before symbolic algebra (roughly, before 1600), when algebra word problems had to solved by direct reasoning, without the use of symbols.

I will continue to emphasize that all complex equations can be simplified to ax + b = c by combining like terms and using the distributive property. The Properties of Equality can then be used to find the solution.

Example: Catch up problem: Time unknown: multi-step equations with variables in both sides of the (=) ax + b =cx + d

Larry has $250 in a bank. He takes out $20 every week.  Jessica has $40 in her savings account and she deposits $15 each week. When will they both have the same amount in the savings account?

Arithmetic approach:

Difference in the initial amount is $210.

$250 - $40 = $210 in favor of Larry

Weekly difference in deposits is $35.

$15 – (-$20) = $35 in favor of Jessica

It will take 6 weeks for Jessica to catch up to Larry since $210/$35 = 6

Algebraic approach:

Let k represent the number of weeks until the accounts are equal.

250-20k=40+15k

250-20k+20k=40+15k+20k Addition Property of Equality.

250=40+35k Combine like terms

250-40=40-40+35k Addition Property of Equality

210=35k Combine like terms

(1/35)210=(1/35)35k Multiplication Property of Equality

6 = k

It will take 6 weeks for Larry and Jessica to have the same amount.

Verify in the original equation

250-20(6)=40+15(6)

250-120=40+90

130=130 verified

The purpose of the above problem is to show students that if the two sides of the equation are simplified to ax +b = cx + d, then one possible efficient step would be to use the Properties of Equality to transform it to ex = f.  This will be the first problem where students are using the Addition Property of Equality on variable expressions. In all previous examples the Addition Property of Equality was used only on numerical expressions. Students would need to be reminded that variable expressions and numerical expressions both represent values thus the same property can be used in similar ways to transform the equation.

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