9.2.3B Algebraic Fractions

Grade: 
9-12
Subject:
Math
Strand:
Algebra
Standard 9.2.3

Generate equivalent algebraic expressions involving polynomials and radicals; use algebraic properties to evaluate expressions.

Benchmark: 9.2.3.4 Arithmetic: Algebraic Fractions

Add, subtract, multiply, divide and simplify algebraic fractions.

For example: [math]\frac{1}{1-x}+\frac{x}{1+x}[/math] is equivalent to [math]\frac{1+2x-x^{2}}{1-x^{2}}[/math].

Overview

Big Ideas and Essential Understandings 

Standard 9.2.3 Essential Understanding

Generating equivalent expressions and giving reasons as to why two expressions are equivalent provides a cornerstone for making sense of the structure of algebraic expressions. In elementary school students learn multiplication facts by building on facts that they already know. Students who calculate [math]6\times 8[/math] by thinking of the problem as [math]5\times 8+8[/math] are using the distributive property to reason that both expressions are equivalent. Students who calculate the value of the expression [math]4\times 7\times 25[/math] by first multiplying the [math]4[/math] and the [math]25[/math] to get [math]100[/math] then multiply this result by [math]7[/math] are using the associative and commutative properties of multiplication to generate equivalent expressions. Instruction at the elementary and middle school levels should pay explicit attention to these properties and how they are used to generate equivalent expressions.

Instruction at the high school should highlight these properties as students begin to work with polynomials. Students need to be able to explain that the expression [math](2x+3)(5x+4)[/math] is equivalent to [math]10x^{2}+23x+12[/math] because of the distributive property in much the same way that they used the distributive property to determine that the expression [math]23\times 54[/math] is equivalent to [math]1242[/math] when they were in elementary school. Students should be able to draw area models to explain how the distributive property works using both numerals and polynomials.

In elementary and middle school students learn to generate equivalent fractions by multiplying by special forms of one. For example, the fraction [math]\frac{2}{3}[/math] is equivalent to the expression [math]\frac{2}{3}\times 1[/math] which is equivalent to [math]\frac{2}{3}\times \frac{2}{2}[/math] which is equivalent to [math]\frac{4}{6}[/math]. Students at the middle school level should be able to explain the reasons why the expressions are equivalent. At the secondary level students should be able to provide the same reasoning to explain why [math]\frac{3x(x-2)}{5(x-2)}[/math] is equivalent to [math]\frac{3x}{5}[/math] but this reasoning is more advanced since these two expressions need to be equivalent for all values that the variable can take on. The expression [math]\frac{3x(x-2)}{5(x-2)}[/math] is not really equivalent to [math]\frac{3x}{5}[/math] since the expression [math]\frac{3x(x-2)}{5(x-2)}[/math] is not defined for [math]x=2[/math] while the expression [math]\frac{3x}{5}[/math] is defined when [math]x=2[/math]. The two expressions are only equivalent if the domain is restricted to not include [math]2[/math].

Students are introduced to radical notation in middle school and are asked to estimate the value of [math]\sqrt{20}[/math] to the nearest whole number. Students at the high school are asked to think of these numbers as objects in themselves and to generate equivalent expressions with these new numbers. Some of the properties are the same for these new numbers as they were for rational numbers (e.g. [math]2\sqrt{3}+5\sqrt{3}[/math] is equivalent to [math]7\sqrt{3}[/math] because of the distributive property just as [math]\frac{2}{17}+\frac{5}{17}[/math] is equivalent to [math]\frac{7}{17}[/math] for the same reason) while some new properties are introduced that are unique to this new set of numbers (e.g. [math]\sqrt{20}[/math] is equivalent to [math]2\sqrt{5}[/math] because of the property [math]\sqrt{a}\cdot\sqrt{b}=\sqrt{a\cdot b}[/math]).

Overall the idea of equivalent expressions has been a focus of the mathematics in grades K to 12. It is very important that students continue to focus on the reasons that make two expressions equivalent rather than memorize isolated sets of rules.

 

Benchmark Cluster 

All Standard Benchmarks

9.2.3.1

Evaluate polynomial and rational expressions and expressions containing radicals and absolute values at specified points in their domains.

9.2.3.2

Add, subtract and multiply polynomials; divide a polynomial by a polynomial of equal or lower degree.

9.2.3.3

Factor common monomial factors from polynomials, factor quadratic polynomials, and factor the difference of two squares.

For example: [math]9x^{6}-x^{4}=(3x^{3}-x^{2})(3x^{3}+x^{2})[/math].

9.2.3.4

Add, subtract, multiply, divide and simplify algebraic fractions.

For example: [math]\frac{1}{1-x}+\frac{x}{1+x}[/math] is equivalent to [math]\frac{1+2x-x^{2}}{1-x^{2}}[/math].

9.2.3.5

Check whether a given complex number is a solution of a quadratic equation by substituting it for the variable and evaluating the expression, using arithmetic with complex numbers.

For example: The complex number [math]\frac{1+i}{2}[/math] is a solution of [math]2x^{2}-2x+1=0[/math], since [math]2\left ( \frac{1+i}{2}\right ) ^{2}-2\left ( \frac{1+i}{2}\right ) +1=i-(1+i)+1=0[/math].

9.2.3.6

Apply the properties of positive and negative rational exponents to generate equivalent algebraic expressions, including those involving nth roots.

For example: [math]\sqrt{2}\times\sqrt{7}=2^{\frac{1}{2}}\times7^{\frac{1}{2}}=14^{\frac{1}{2}}=\sqrt{14}[/math]. Rules for computing directly with radicals may also be used: [math]\sqrt[3]{2}\times\sqrt[3]{x}=\sqrt[3]{2x}[/math].

9.2.3.7

Justify steps in generating equivalent expressions by identifying the properties used. Use substitution to check the equality of expressions for some particular values of the variables; recognize that checking with substitution does not guarantee equality of expressions for all values of the variables.

************************************************************************

Benchmark Group B

9.2.3.4

Add, subtract, multiply, divide and simplify algebraic fractions.

For example: [math]\frac{1}{1-x}+\frac{x}{1+x}[/math] is equivalent to [math]\frac{1+2x-x^{2}}{1-x^{2}}[/math].

What students should know and be able to do [at a mastery level] related to this benchmark

  • Students should be able to add, subtract, multiply, and divide fractions with numerators and denominators that are polynomials.
  • Students should be able to find equivalent forms for algebraic fractions by multiplying by special forms of one (i.e. [math]\frac{x+1}{x-3}=\frac{x+1}{x-3}\cdot 1=\frac{x+1}{x-3}\cdot\frac{x-1}{x-1}=\frac{x^{2}-1}{x^{2}-4x+3}[/math]) or factoring special forms of one (i.e. [math]\frac{x^{2}-9}{x^{2}-8x+15}=\frac{(x+3)(x-3)}{(x-5)(x-3)}=\frac{(x+3)}{(x-5)}\cdot\frac{(x-3)}{(x-3)}=\frac{(x+3)}{(x-5)}\cdot 1=\frac{x+3}{x-5}[/math]).
  • Students should understand that two algebraic fractions like [math]\frac{x+1}{x-3}[/math] and [math]\frac{(x+1)(x-1)}{(x-3)(x-1)}[/math] are equivalent only if the domain of [math]\frac{x+1}{x-3}[/math] is restricted to not include [math]x=1[/math].

Work from previous grades that supports this new learning includes:

  • In middle school students learn to add, subtract, multiply, and divide fractions using algorithms.
  • Students in middle school know how to create equivalent fractions by multiplying by special forms of one (i.e. [math]\frac{2}{3}=\frac{2}{3}\cdot 1=\frac{2}{3}\cdot\frac{5}{5}=\frac{10}{15}[/math]) or factoring special forms of one (i.e. [math]\frac{18}{27}=\frac{2\cdot 9}{3\cdot 9}=\frac{2}{3}\cdot\frac{9}{9}=\frac{2}{3}\cdot 1=\frac{2}{3}[/math]).
  • Students in middle school should understand that division by zero is undefined.
Correlations 

Correlation

NCTM Standards:

(p.296, PSSM) Instructional programs from Pre-K through grade 12 should enable all students to

2.) Represent and analyze mathematical situations and structures using algebraic symbols. In grades 9-12 all students should

  • Understand the meaning of equivalent forms of expressions, equations, inequalities, and relations;
  • Write equivalent forms of equations, inequalities, and systems of equations and solve them with fluency - mentally or with paper and pencil in simple cases and using technology in all cases;
  • Use symbolic algebra to represent and explain mathematical relationships;
  • Use a variety of symbolic representations, including recursive and parametric equations, for functions and relations;
  • Judge the meaning ,utility, and reasonableness of the results of symbol manipulations, including those carried out by technology.

 

Common Core State Standards (CCSM)

High School: Algebra

Seeing Structure in Expressions

Interpret the structure of expressions.

·         A-SSE.1. Interpret expressions that represent a quantity in terms of its context.

  • Interpret parts of an expression, such as terms, factors, and coefficients.
  • Interpret complicated expressions by viewing one or more of their parts as a single entity.For example, interpret P(1+r)n as the product of P and a factor not depending on P.

·         A-SSE.2. Use the structure of an expression to identify ways to rewrite it. For example, see [math]x^{4}-y^{4}[/math] as [math](x^{2})^{2}-(y^{2})^{2}[/math], thus recognizing it as a difference of squares that can be factored as [math](x^{2}-y^{2})(x^{2}+y^{2})[/math].

Write expressions in equivalent forms to solve problems.

·         A-SSE.3. Choose and produce an equivalent form of an expression to reveal and explain properties of the quantity represented by the expression.

a. Factor a quadratic expression to reveal the zeros of the function it defines.

b. Complete the square in a quadratic expression to reveal the maximum or minimum value of the function it defines.

c. Use the properties of exponents to transform expressions for exponential functions. For example the expression [math]1.15^{t}[/math] can be rewritten as [math](1.15^{\frac{1}{12}})^{12t}\approx 1.012^{12t}[/math] to reveal the approximate equivalent monthly interest rate if the annual rate is 15%.

 

Arithmetic with Polynomials & Rational Expressions

Perform arithmetic operations on polynomials.

·         A-APR.1. Understand that polynomials form a system analogous to the integers, namely, they are closed under the operations of addition, subtraction, and multiplication; add, subtract, and multiply polynomials.

Rewrite rational expressions.

·         A-APR.6. Rewrite simple rational expressions in different forms; write a(x)/b(x) in the form q(x) +r(x)/b(x), where a(x), b(x), q(x), and r(x) are polynomials with the degree of r(x) less than the degree of b(x), using inspection, long division, or, for the more complicated examples, a computer algebra system.

·         A-APR.7. (+) Understand that rational expressions form a system analogous to the rational numbers, closed under addition, subtraction, multiplication, and division by a nonzero rational expression; add, subtract, multiply, and divide rational expressions.

Misconceptions

Student Misconceptions 

Student Misconceptions and Common Errors

  • Students will incorrectly add algebraic fractions (e.g. [math]\frac{3}{x-1}+\frac{2}{1-x}=\frac{5}{x-1}[/math]) because they fail to realize that subtraction is not commutative (i.e. [math]x-1[/math] is not equivalent to [math]1-x[/math]).
  • Students will incorrectly "cancel out" terms instead of factors when simplifying algebraic fractions (i.e. [math]\frac{x-5}{x+2}=\frac{x-5}{x+2}=\frac{-5}{2}[/math].
  • A common strategy to correctly solve equations involving algebraic fractions (rational equations) is to multiply both sides of the equation by the least common denominator to correctly generate an equivalent equation that have denominators of one and is easier for students to solve symbolically.

[display]\frac{20}{x+2}+\frac{15}{x+3}=8[/display]

[display](x+2)(x+3)\frac{20}{x+2}+(x+2)(x+3)\frac{15}{x+3}=(x+2)(x+3)8[/display]

[display](x+3)20+(x+2)15=(x+2)(x+3)8[/display]

Some students incorrectly apply this same "clearing the fraction" strategy when working with expressions.

(1)

[display]\frac{20}{x+2}+\frac{15}{x+3}[/display]

 

(2)

[display](x+2)(x+3)\frac{20}{x+2}+(x+2)(x+3)\frac{15}{x+3}[/display]

Expression in (2) is not equivalent to expression in (1)

(3)

[display](x+3)20+(x+2)15[/display]

 

  • Students incorrectly subtract fractions by forgetting to subtract the entire expression in the numerator of the fraction on the right of the subtraction symbol.

(1)

[display]\frac{5x+10}{(x+2)(x+3)}-\frac{3x+6}{(x+2)(x+3)}[/display]

 

(2)

[display]\frac{5x+10-3x+6}{(x+2)(x+3)}[/display]

Expression in (2) is not equivalent to expression in (1). Student did not subtract the entire numerator from the fraction on the right of the equal sign.

(3)

[display]\frac{2x+16}{(x+2)(x+3)}[/display]

This expression is the student's final answer and is not equivalent to the expression in (1) but is equivalent to the expression in (2).

Vignette

In the Classroom

The following vignette is a classroom lesson introducing subtraction of algebraic fractions. The teacher wants to make sure that her students know reasons for why two expressions are equivalent and connect the steps they learned for subtracting fractions with steps involving fractions with variables.

 

Ms. L:  Please answer the question on the warm-up and be prepared to defend your answers.

Warm-Up

Which of the following expressions are equivalent to [math]\frac{3}{8}[/math]?

 

A

B

C

D

E

[display]\frac{9}{24}[/display]

[display]\frac{3+5}{8+5}[/display]

[display]\frac{3x}{8x}[/display]

[display]\frac{7}{8}-\frac{1}{2}[/display]

[display]\frac{1}{5}+\frac{2}{3}[/display]

 

 

 

[After 5 minutes.]

Ms. L:     Ron, please share what you think about the expression in part A.

Ron:       I think it is equivalent, because nine twenty-fourths is the same as three-eighths.

Ms. L:     So it just is because it is?

Ron:       No, because nine is three times three and twenty-four is three times eight. The three's cancel because they are a special form of one.

Ms. L:     OK, so two fractions are equivalent if you can show how one fraction can be multiplied by a special form of one to get the other.

Ron:       Yes.

Ms. L:     Loan, is the expression under B equivalent to three-eighths.

Loan:      No. The fraction in part B is equal to eight thirteenths, which is more than one-half. Three-eighths is less than one half.

Ms. L:     So you can't just cancel out the fives and get three over eight?

Loan:      No. You can just tell from this example plus when you multiply a number by one you get the same number or value as in expression A, but addition does not seem to work that way.

Ms. L:     OK, what about expression C?

James:     This is true; you just cancel out the x's. I mean you factor out a special form of one and it magically disappears.

Ms. L:     I am not sure about magic but I agree that you get an equivalent expression because you are multiplying by the identity, one.

Dawn:    I am not sure that C is equivalent. What if x equals zero. Then three x over 8 x is undefined because you can't divide by zero.

Ms. L:     James do you have any response to Dawn?

James:     Dawn, you are very picky. I am sure it is equivalent if x is not zero but I am not sure when it is zero. I think it just cancels out the same way.

Ms. L:     Three times zero over eight times zero is still zero divided by zero which is undefined and not one so it is not equivalent. If we want three x over eight x to be equivalent to three-eighths we need to also say that x cannot be zero.

James:     That is weird but it makes sense.

Ms. L:     Sammie, what do you think about expression D?

Sammie: I got a common denominator and subtracted the fractions. The answer is three-eighths so the expression is equivalent.

Ms. L:     Good. Today we are going to subtract algebraic fractions and use the same ideas that Sammie just described. Sammie multiplied both fractions by special forms of one to generate a new subtraction expression that is equivalent. She then subtracted the numerators and left the denominators to get her answer. When we work with algebraic fractions it is just as important to have reasons for what you do to generate equivalent expressions. Kim, what did you get for expression E?

Kim:       It is equivalent because one plus two is three and five plus three is eight.

James:     That can't be right since you do not have a common denominator.

Dawn:    It is not even close, since two-thirds is bigger than one-half and you are adding one-fifth to make it even bigger. Three-eighths is smaller than one-half.

Ms. K:    Good. Checking the numbers is a good strategy for equivalence. Please look at the following problem. [Writes: Subtract. [math]\frac{5}{x+2}-\frac{2}{x+3}[/math]]

               When Sammie subtracted the fractions before, she found a common denominator and then generated equivalent fractions. Think about what the common denominator should be for these two fractions. Talk to your partner about what you think it should be.

               [waits 3 minutes.]

               Dawn, what do you think the common denominator should be?

Dawn:    James and I think it should be [math](x+2)(x+3)[/math]. The only way to change the [math](x+2)[/math] is by multiplying. I think the only thing you can do is multiply it by [math](x+3)[/math]. When Sammie did it she only had to multiply one of the fractions by a special form of one to get a common denominator, this one looks like we have to do both.

Ms. K:    OK. So if we multiply the first fraction by the special form of one, [math]\frac{x+3}{x+3}[/math] and the second fraction by a different special form of one, [math]\frac{x+2}{x+2}[/math] we get the following expression.

               [Writes: [math]\frac{5}{x+2}\cdot\frac{x+3}{x+3}-\frac{2}{x+3}\cdot\frac{x+2}{x+2}[/math]]

               Is this expression equivalent to the original expression?

James:     Yes, all you did was multiply each fraction by a special form of one.

Ms. K:    Alright, is this expression equivalent?

               [Writes: [math]\frac{5x+15}{(x+2)(x+3)}-\frac{2x+4}{(x+3)(x+2)}[/math]]

Sammie: Yes, all you did was multiply the fractions and you also used the distributive property on the numerators.

Loan:      Where did the parentheses come from in the denominator?

Ms. K:    There are implied parenthesis in both the numerator and denominator of all fractions. If we did not include them then we would not be multiplying the entire denominators from both fractions.

Loan:      OK. You also did that in the top as well.

Ms. K:    Yes. Talk to your partner and see if you think the following expression is equivalent?

               [Writes: [math]\frac{5x+15-2x+4}{(x+2)(x+3)}[/math]]

               [Waits 3 minutes.]

Loan:      I do not think so. If there are implied parenthesis in the numerator then you need to subtract the entire numerator not just the two x's. I think it should be minus four.

Dawn:    Yes, I agree. You did it when you multiplied, but not here when you added.

Ms. K:    So if I change the plus to a minus then would it be equivalent?

               [Changes the expression to [math]\frac{5x+15-2x-4}{(x+2)(x+3)}[/math]]

Loan:      Yes. So the final answer or last equivalent fraction should be [math]\frac{3x+19}{(x+2)(x+3)}[/math].

Ms. K:    Good. Just make sure that when you generate equivalent expressions while working the problems you have a good reason as to why the expressions are equivalent.

 

Resources

Instructional Notes 

Teacher Notes

  • Students should realize that operations with algebraic fractions are very similar to operations involving fractions with no variable parts. Furthermore the justifications for the manipulation of an expression remain the same.
  • One of the most important ideas related to algebraic fractions is the concept of equivalent algebraic fractions. Students need to realize that they can change the form of an algebraic fraction by multiplying by special forms of one. For example, the fraction [math]\frac{x+1}{x-3}[/math] can be changed into an equivalent form such as [math]\frac{(x+1)(x+5)}{(x-3)(x+5)}[/math]. It should be noted that the expression [math]\frac{(x+1)(x+5)}{(x-3)(x+5)}[/math] is not defined when [math]x=-5[/math] while the expression [math]\frac{x+1}{x-3}[/math] is defined when [math]x=-5[/math] and therefore the expressions [math]\frac{(x+1)(x+5)}{(x-3)(x+5)}[/math] and [math]\frac{x+1}{x-3}[/math] are only equivalent if we restrict the domain to not include [math]x=-5[/math] for both expressions. The fraction [math]\frac{(x+1)(x+5)}{(x-3)(x+5)}[/math] is equivalent to [math]\frac{(x+1)}{(x-3)}[/math], when [math]x\neq-5[/math], because [math]\frac{x+5}{x+5}=1[/math].
  • Students should realize that simplifying rational expressions involves factoring special forms of one. The expression [math]\frac{2x-10}{x^{2}-3x-10}[/math] can be re-written in factored form as [math]\frac{2(x-5)}{(x+2)(x-5)}[/math] which can then be re-written as a multiplication of two fractions as [math]\frac{2}{x+2}\cdot\frac{x-5}{x-5}[/math]. Since the expression [math]\frac{x-5}{x-5}=1[/math], the expression [math]\frac{2}{x+2}\cdot\frac{x-5}{x-5}[/math] can be thought of as [math]\frac{2}{x+2}\cdot 1[/math] and by the multiplicative identity property is equivalent to [math]\frac{2}{x+2}[/math] if [math]x\neq5[/math]. The common error of cancelling terms instead of factors can be addressed by focusing on finding factors of one.

 

Instructional Resources 

Instructional Resources

This activity explores the relationship that can be described by rational expressions involving light reflections.

New Vocabulary 

New Vocabulary

  • Algebraic Fraction: Algebraic fractions are fractions with variables in the numerator or denominator, such as [math]\frac{36}{x}[/math]. Others include [math]\frac{x^{2}}{y}[/math] or [math]\frac{5x}{y^{3}}[/math]. Since division by 0 is impossible, variables in the denominator have certain restrictions. The denominator can never equal 0.
  • Special Form of One: Expressing the value of one whole in fraction form such as or [math]\frac{9}{9}[/math] or [math]\frac{3}{3}[/math].
Professional Learning Communities 

Professional Learning Communities

Reflection - Critical Questions regarding the teaching and learning of these benchmarks.

  • Do my students know what makes rational expressions equivalent?
  • Do I emphasize justification for equivalent expressions as students operate with algebraic fractions?
  • Are my students making connections between operations involving algebraic fractions and operations involving fractions with no variable parts.

Materials - suggest articles and books

  • Schoenfeld, A. H. & Arcavi, A. (1988). On the meaning of variable. Mathematics Teacher, September.
  • Chalouh, L. & Herscovics, N. (1988). Teaching algebraic expressions in a meaningful way. In 68th Yearbook of the National Council of Teachers of Mathematics, pp. 233-246. Reston, VA: 2008
References 

References

  • Common Core State Standards
  • Ellis, M. W., & Bryson, J. L. (2011). Absolute value equations and inequalities: A conceptual approach. Mathematics Teacher, 104 (8), 592-595.
  • Ruddell, M.R., & Shearer, B.A. (2002). "Extraordinary," "tremendous," "exhilarating, "magnificent": Middle school at-risk students become avid word learners with the vocabulary self-collection strategy (VSS). Journal of Adolescent & Adult Literacy, 45, 352-363.

Assessment

Assessment

Example Item 1

 

Which one of the following expressions is equivalent to [math]\frac{3x}{6x+12}[/math]?

Assume that [math]x\neq-2[/math].

A. [math]\frac{x}{2x+4}[/math]

B. [math]\frac{1}{14}[/math]

C. [math]\frac{1}{3x+12}[/math]

D. [math]\frac{3}{18}[/math]

 

Source: Made Up

Correct Answer: A

Cognitive Level: Application

Example Item 2

11.              Which of the following expressions is equal to [math]\frac{1}{x+2}-\frac{2}{x+1}[/math]  ?

A. [math]\frac{-1}{2x+3}[/math]

B. [math]\frac{-x-3}{x^{2}+2}[/math]

C. [math]\frac{-1}{x^{2}+3x+2}[/math]

D. [math]\frac{-x-3}{x^{2}+3x+2}[/math]

E. [math]\frac{-x+5}{x^{2}+3x+2}[/math]

 

Source: NAEP Grade 12, 2009, Question M2-11

Correct Answer: D

Cognitive Level: Hard (27.52% correct)

Example Item 3

 

Which one of the following expressions is equivalent to [math]\frac{3}{x-1}[/math]?

Assume that [math]x\neq1[/math].

A. [math]\frac{3}{x-1}[/math]

B. [math]\frac{-3}{x-1}[/math]

C. [math]\frac{-3}{1-x}[/math]

D. [math]\frac{4}{x}[/math]

 

Source: Made Up

Correct Answer: B

Cognitive Level: Application

Differentiation

Struggling Learners 

Struggling Learners

  • One of the most difficult ideas for students involves knowing the difference between terms and factors. The expression [math]x^{2}+3x-10[/math] has three terms. Students who struggle often time cancel terms instead of factors. Students need to see how the factored form [math](x+5)(x-2)[/math], which contains two factors, allows them certain freedoms to generate equivalent expressions.
English Language Learners 

English Language Learners

  • Use sorting activities to differentiate the vocabulary of factor and term of a polynomial expression.
Extending the Learning 

Extending the Learning

  • Students should be able to make sense of addition and subtraction involving algebraic fractions by thinking of each term as a function. This functions approach to addition and subtraction emphasizes that equivalent expressions are expressions that generate equal values for all values of the variable in the domain.

Parents/Admin

Classroom Observation 

Administrative/Peer Classroom Observation

Students are: (make list)      

Teachers are: (make list)

  • Making sense of each step they use to develop equivalent expressions.
  • Asking students to justify their steps.
  • Talking with each other to make sense of problems.
  • Asking guiding questions to get students to focus on sense-making versus following rules.
  • Using technology to match tables and graphs of expressions to check for equivalence.
  • Modeling how to use tables and graphs to make sense of equivalent expressions.