Section 1.5

Why?

Simplifying Algebraic Expressions

1.5

Part of mastering the language of mathematics involves writing algebra using the fewest symbols pos-sible. In the case of algebraic expressions, that means eliminating parentheses, reducing the number of operations to a minimum, and reducing duplicate instances of variables, so that, if you substitute particu-lar values for the variables, you will have the fewest possible number of steps to go through to evaluate the expression.

What Do You Already Know?	1.5

How does the Commutative Property differ from the Associative Property?

When do you use the Distributive Property?

Give an example where the Order of Operations must be used.

Goals	1.5

When you complete this activity, you should be able to:

1. Change an expression to an equivalent expression

2. Simplify an expression

3. Recognize when an expression is fully simplified

Building Mathematical Language	1.5
	
New Terms:	binary operator	like terms	unary operator

Key Concepts	1.5

Like terms are terms that have the same variables with the same exponents.

A unary operator operates on one number. Taking the negative of a number (–) is a unary operator. Taking the negative of 3 is –3.

A binary operator requires two numbers. Subtraction (–) is a binary operator. For example, 5 subtract 3 is (5 − 3).

Two expressions are equivalent if substituting any values for the variables in one expression yields the same result as substituting the same values for the same variables in the other expression.

A smplified algebraic expression: the sequence of terms begins with the highest power term on the left and has no parentheses, no double signs, and all fractions have been reduced.

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Chapter 1 — Expressions

Technique	1.5
	
Removing Parentheses

Removing the parentheses is a technique for rewriting an expression or equation.

Limitation: If there are more than two factors involved with parentheses, errors can occur.

Use the Distributive Property to substitute for parts of an expression that have parentheses and validate the signs when removing the parentheses.

Example 1: Remove the parentheses: 3(a – 2b) + 4(2a – b) – 5 3a – 6b + 8a – 4b – 5

Here 3(a – 2b) is equivalent to 3a – 6b and 4(2a – b) is equivalent to 8a – 4b by the Distributive Property. We replace 3(a – 2b) and 4(2a – b) with their equivalent expressions.

Example 2: Remove the parentheses  Please excuse the missing gifs (−).
2−3x(2−5x−4y)+7 
2−(3x(2−5x−4y))+7

2−(6x−15x2 −12xy)+7 2−6x+15x2 +12xy+7

The expression 3x(2 – 5x – 4xy) is replaced by the equivalent expression (6x – 15x2 – 12xy) and we validate the signs when removing the parentheses. In this process, –3x was distrib-uted over (2 – 5x – 4y). Note that this expression has not yet been fully simplified.

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1.5 Simplifying Algebraic Expressions

Methodology	1.5
	
Simplifying Algebraic Expressions

The methodology for solving the problem in Example 1 is shown in the following table. Students should complete Example 2 in the same table, following each step in the methodology.

Note: Algebraic expressions come in many forms, some with a few terms, and some with many terms. In working with algebraic terms, experts like to simplify the expression into its easiest interpretable form. The specification for such forms is to have only one of each type of term.

Example 1

Simplify:	5a − 3a(a2 + 2)

Example 2

2(x − 4) − 3x2 − (x − 4)

Your Turn

−3x + 2(5 − x) − x(5 + x) − 7

Steps
	Discussion
	
1 Remove parentheses
	Distribute through the expression so as to remove all parentheses in the expression.

Ex 1

5a − 3a3 − 6a
	Ex 2

2x − 8 − 3x2 − x + 4
	Your Turn

2 Check signs
	Review all coefficients for accuracy in signs, carefully checking for negatives and double negatives.

Ex 1

Checked.
	Ex 2

Checked.
	Your Turn

3 Identify the number of terms for simplified answer
	Look through every term to inventory how many like terms you will have and produce that number of groupings.

Ex 1

−3a3
5a, − 6a
	Ex 2
	
2x,−x −3x2 −8,4
	Your Turn

4 Set up buckets
	Set up the buckets with the variables from Step 3 using parentheses and addition signs.

Ex 1

(	)a3 + (	)a
	Ex 2

(	)x2 + (	)x + (	) constant
	Your Turn

55

Your Turn

Your Turn

Chapter 1 — Expressions

Steps
	Discussion
	
5 Rearrange the original expression
	Rearrange the original expression by grouping coefficients within the appropriate parentheses.

Ex 1
	
(−3a3) + (5a − 6a)
	Ex 2
	
(−3x2) + (2x − x) + (−8 + 4)

6 Validate all terms used
	Make sure that all terms in the previous expression are accounted for by lightly crossing out, term by term, each term that has been incorporated.

Ex 1

5a − 3a3 − 6a (−3a3 ) + (5a −6a )
	Ex 2

2x − 8 − 3x2 − x + 4

(−3x2 ) + (2x − x) + (−8 + 4)

7 Combine the terms
	Convention says that each variable has a single coefficient and that the computations needed to remove parentheses should be carried out.

Ex 1

−3a3 + 5a − 6a −3a3 + (−a) −3a3 − a
	Ex 2

−3x2 + (2x − x) + (−8 + 4) −3x2 + (x) + (−4)
−3x2 + x − 4
	Your Turn

8 Validate by substitution
	Pick a number other than 0,1, and –1 to substitute for the variable in both the original and simplified expressions. If both values are equivalent, you have validated your work.

Ex 1
	Let x = 2	And
5a − 3a(a2 + 2)	−3a3 − a 5(2) − 3  2(22 + 2)	−3(23) − 2 10 − 6(4 + 2)                    −3(8) − 2 10 − 6(6)                           −24 − 2 10 − 36                             −26 ü −26
	Ex 2
	
Let x = 2	And
2(x − 4) − 3x2 − (x − 4)	−3x2 + x − 4 2(2 − 4) − 3  22 − (2 − 4)	−3(22) + 2 − 4 2(−2) − 3  4 − (−2)	−3(4) + 2 − 4 −4 −12 + 2	−12 + 2 − 4 −14	−14 ü

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1.5 Simplifying Algebraic Expressions

Steps
	Discussion
	
8 Validate by substitution (continued)
	Pick a number other than 0,1, and -1 to substitute for the variable in both the original and simplified expressions. If both values are equivalent, you have validated your work.

Your Turn

Addressing Common Errors	1.5

The following table shows some of the most common errors that learners tend to make with the kinds of problems covered in this section.

ERROR:	Incorrectly distributing a negative sign

Incorrect Process

−(x−3) = −x−3

Correct Process

−(x−3) = −x+3

Resolution

Correctly distribute the negative sign realizing the negative sign represents –1.

Validation

Substitute 5 for x in both expressions.

−(5 −3) = −2 and − 5 + 3 = −2

ERROR:	Double operation symbols

Incorrect Process

−(x−3) = −x−−3

Correct Process

−(x−3) = −x−(−3) = −x+3

Resolution

Use symbols correctly.

Validation

Substitute 7 for x in both expressions.

−(7−3) = −2 and −7+3= −2

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Chapter 1 — Expressions

Preparation Inventory	1.5

I can...	OR Here's my question...

recognize like terms.

add like terms.

subtract like terms.

use the Order of Operations.

Critical Thinking Questions	1.5

1. What are equivalent expressions?

2. What changes can you make to an expression to yield an equivalent expression?

3. Why do you remove parentheses?

4. How do you remove parentheses?

5. How do you ensure that the signs are correct?

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1.5 Simplifying Algebraic Expressions

6. How do you ensure that all the terms are used?

7. Why shouldn’t you use zero in validating that terms are equivalent?

8. How might you use graphing or the table feature of a calculator to validate that your simplified expression in one variable is equivalent to the original expression?

Demonstrate Your Understanding	1.5

Simplify each of the following expressions.

1. 2x(5 − x2) + x3

2. a2 − 4a(a + 3) − 6

3. 12 − 3(a − 5) − 7a

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Chapter 1 — Expressions

4. –3x(5 – x) + 3x – 3(x + 2)

5. 4x2 – 3x – 8 + 7x(x – 2) + 5

6. –(x – 2) + 3(2 – x) + 7x3 – x + 2x(x + 3)

Create and Solve the Hardest Problem	1.5

Your Hardest Problem

Problem Solution

What makes a problem where you must simplify an algebraic expression a hard problem to solve?

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1.5 Simplifying Algebraic Expressions

Identify, Describe, & Correct the Errors	1.5

Identify the error you find in each of the following worked solutions and describe the error made. Solve the problem correctly in the space provided.

Problem 1

Simplify:
5−2(x−1)
	
Worked Solution (Find the error!)
	
5−2(x−1) 5−2x−1 4−2x

Problem 2

Simplify and validate:

2x−3x+5x2 +4x−5x+7
	
Worked Solution (Find the error!)

2x−3x+5x2 +4x−5x+7 (5x2)+(2x−3x−5x)+(7) 5x2 −6x+7
Validate: Let x =0.
2x−3x+5x2 +4x−5x+7
2  0−3  0+5  02 +4  0+7 0−0+0+0+7
7 And
5x2 −6x+7
5  02 −6  0+7 0−0+7
7

Correct Process	Describe the Error

Correct Process

Describe

the Error

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Chapter 1 — Expressions

Problem 3	Correct Process

Simplify:
−3x2 −2x(3−x)+7
	
Worked Solution (Find the error!)

−3x2 −2x(3− x)+7 −3x2 −6x−2x2 +7 −5x2 −6x+7

Describe

the Error

What Did You Learn About Learning Math?	1.5

1. What did you learn about working with algebraic expressions?

2. How did you use the methodologies and common errors in working through the Demonstrate Your Understanding Problems?

3. What methods will you use to remind yourself of important concepts dealing with simplifying algebraic expressions?

4. What learning skills did you use in this section?