Download Substitution and Evaluation

Section 1.4
Substitution and Evaluation
Pre-Activity
Preparation
The admission prices to the family concert are as follows:
Adult tickets sell for $5 each
Children’s tickets sell for $3 each
Senior citizen tickets sell for $4 each
The revenue for ticket sales can be modeled by the following algebraic expression:
$5A + $3C + $4S
To determine the total revenue from ticket sales, replace the appropriate variable with the number of tickets
sold in each category.
In week 1, ticket sales were A = 50 (adults), C = 300 (children), and S = 50 (senior citizens). The revenue was:
($5 × 50) + ($3 × 300) + ($4 × 50)
= $250 + $900 + $200
= $1350
In week 2, A = 35 (adults), C = 250 (children), and S = 40 (seniors) tickets were sold. The revenue was:
($5 × 35) + ($3 × 250) + ($4 × 40)
= $175 + $750 + $160
= $1085
In both cases above, the process used was to multiply the given cost of the tickets by the number of tickets sold,
which varied week by week, and then add the amounts for each category.
Learning Objectives
• Build on the basic vocabulary of algebraic expressions
• Understand how Order of Operations works with algebraic expressions
• Evaluate algebraic expressions when given a value for each variable
Terminology
New Terms
Previously Used
to
Learn
coefficients
factor
algebraic expression
constant term
numerator
Order of Operations
constants
terms
substitution
denominator
variables
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Chapter 1 — Evaluating Expressions
58
Building Mathematical Language
Language
Definition
Example
algebraic
expression
A combination of variables, constants,
grouping symbols, and signs of operation
term
A term is a constant, variable, or product
of a number and variable(s), separated
from other terms in an expression by
addition or subtraction operations. (Recall
that subtraction is the same as addition of
the opposite of a number.)
substitution
The process of replacing variables with
numeric values.
–2x + y 2 – 4
x = 2 and y =3
–2( 2 ) + ( 3 ) 2 – 4
coefficient or
numerical coefficient
Usually the constant factor in a term
5 x 2y 3z
constant term
The term in an expression that has no
variable factor
–2x + y 2 – 4
–2x + (y 2 – 4)
–2x + y 2 – 4
(–2x) + (y 2) + (–4)
Term 1 is –2x
Term 2 is +y 2
Term 3 is –4
Order of Operations
As expressions become more complicated than the examples above, rules for evaluating these expressions
become necessary. When values are used in place of variables, there is a universally agreed upon
mathematical sequence, a specific order, in which each operation must be performed to evaluate the
resulting expression. Correctly using the Order of Operations ensures that you will obtain the same correct
value as other evaluators of a given mathematical expression.
Order of Operations
To evaluate an expression, the simplification process must follow the Order of Operations:
First, simplify the operations within Parentheses.
Then, simplify all numbers with Exponents (factors or terms with exponents).
Then, compute Multiplication and Division, left to right as they occur in each term.
Finally, compute Addition and Subtraction, left to right as they occur in the expression.
Without this convention, the expression 2 + 3 × 4 would have two possible values: 20 and 14. Using the
Order of Operations, you obtain the correct value of 14 since multiplication occurs before addition.
Try it!
Make-up a numerical expression that could have two or
more possible values without the order of operations.
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Section 1.4 — Substitution and Evaluation
Methodologies
Applying the Order of Operations to Evaluate an Expression
►
►
Example 1: Evaluate 3x2 – 5(y – 3) for x = –2 and y = –7
Example 2: Evaluate –2x + y( y – 4) for x = –3 and y = 2
Try It!
Steps in the Methodology
Step 1
Identify the
Variables (Va)
Step 2
Substitution
(R)
Step 3
Identify the
terms (T)
Make a list of each Variable and
its replacement value.
Example 1
Example 2
3x2 – 5(y – 3)
x replaced by –2
y replaced by –7
Replace the given value for the
indicated variable.
To prevent possible sign errors,
write the expression using
parentheses where the variable
should be and then insert the
proper value into the parentheses.
Identify the Terms of the
expression. Use braces { } or
brackets [ ] to separate the terms
of the expression.
???
Why can we do this?
3( x )2 – {5(( y ) – 3)}
3( x )2 – {5(( ) – 3)}
Substitute –2 for x and
–7 for y.
3(–2)2 – {5((–7) – 3)}
{3(–2)2 } – {5((–7) – 3)}
A
B
Both braces contain one
term each.
Special Case: When the expression
to be evaluated is a fraction, enclose
the numerator and denominator in
braces as if they were the “terms” of
the fraction. See Model 2.
Step 4
Parentheses (P)
Simplify within Parentheses to
a single number for each set of
braces. Once the numbers within
parentheses are simplified, the
braces can be eliminated.
Note: Make a note of the operation
completed from this point forward.
A
B
Removing parentheses in term B:
{3(–2)2} – {5(–7 –3)}
= {3(–2)2} – {5(–10)}
Remove braces:
= 3(–2)2 – 5(–10)
P
Validate: (–10) + (3) = –7 
continued on next page
Chapter 1 — Evaluating Expressions
60
Steps in the Methodology
Step 5
Evaluate Exponents.
Exponents (E)
Step 6
Multiplication
and Division
(MD)
Perform Multiplication and
Division as they appear
from left to right in each
term.
Example 1
(–2)2 = (–2)(–2) = 4
= 3(4) – 5(–10)
E
= 12 – (–50)
M
Validate:
12
3
Step 7
Addition and
Subtraction
(AS)
Perform Addition and
Subtraction as they appear
from left to right.
Reduce fractions to lowest
terms.
Example 2
=4 
-50
-10
=5 
12 – (–10)
= 12 + 10
Answer: 22
S
Validate:
22 + (– 10) = 12 
Step 8
Validation (V)
Arithmetic Validation is
done step-by-step.
Process validation is best
done by reworking the
problem after a short break
in time.
3x2 – 5(y – 3)
x&y
Va
= 3(–2)2 – {5((–7) – 3)} R
= {3(–2)2}– {5((–7) – 3)} T
= 3(–2)2 – 5(–7 – 3)
= 3(–2)2 – 5(–10)
P
= 3 • 4 – 5(–10)
E
= 12 –(–10)
M
= 12 + 10
= 22
S
All operations worked in
correct order.
???
Parentheses can be inserted into any expression to clarify the meaning as
long as they DO NOT change the order of operations. Since addition and
subtraction are at the bottom of the order of operations hierarchy, identifying the terms and enclosing
them in braces ensures that those operations will be performed last. In fractions, the fraction bar can be
assumed to be a symbol of enclosure. The same is true for square root symbols; assume that everything
under the radical is enclosed within parentheses.
Why can we do this?
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Section 1.4 — Substitution and Evaluation
Models
Model 1
What is the value of 2a – 3ab when a = –1 and b = 5?
Step 1
Variables
In the expression 2a – 3ab, replace a with –1 and b with 5.
Step 2
Substitution (R)
2( ) – 3( )( ) =
= 2(–1) – 3(–1)(5)
Step 3
Identify the Terms
Step 4
Skip this step, no Parentheses to evaluate
Step 5
Skip this step, no Exponents to evaluate
Step 6
Multiplication
= –2 – (–15)
Step 7
Subtract
= –2 +15
= {2(–1)} – {3(–1)(5)}
= 13
Step 8
Validate by redoing the steps
Validate:
Validate: 13 – 15 = –2 
and
2(–1) – 3(–1)(5) =
= –2 – (–15)
= –2 + 15
= 13 
RTMS, correct order
Try it!
Find the value of the above expression if a = 2 and b = –4
-2
=2 
-1
-15
= -3;
5
-3
=3 
-1
Chapter 1 — Evaluating Expressions
62
Model 2
Evaluate 2x2 + (2y – 3) for x = 4 and y = –1
Step 1
Variables
Replace x with 4 and y with –1.
Step 2
Substitution (R)
2( )2 + [2( ) – 3] =
= 2(–1)2 + [2(–1) – 3]
Step 3
Identify the Terms
= {2(–1)2} + {2(–1) – (3)}
Step 4
Parentheses
In order to clear the parentheses for the second term, we need to apply
the Order of Operations, in this case, multiplication and subtraction
(Steps 6 and 7) to the expression within the parentheses. Once we
have done so, and have a single number for each term, we continue
on to Steps 5 through 8.
= {2(–1)2} + {2(–1) – (3)}
= {2(–1)2} + {(–2) – (3)}
= 2(–1)2 + (–5)
Step 5
Exponents
(–1)2 = (–1)(–1) = 1
= 2(1) + (–5)
Step 6
Multiplication
= 2 + (–5) Step 7
Addition
= 2 + (–5)
= –3
Answer: –3
Step 8
Validate by redoing the steps
2(–1)2 + [2(–1) – 3] =
= {2(–1)2} + {2(–1) – (3)}
= {2(–1)2} + {(–2) – (3)}
= 2(–1)2 + (–5)
= 2(1) + (–5)
= 2 + (–5)
= –3 
Multiplication
Subtraction
Validate: 2 ÷ 1 = 2 
RTPEMA, correct order
Validate: –2 ÷ 2 = –1 
Validate: –5 + 3 = –2 
63
Section 1.4 — Substitution and Evaluation
Model 3: Special Case
Evaluate
3x - 2
when x = –4.
5-x
Step 1
Variables
3 x - 2 Replace x with –4
5-x
Step 2
Substitution (R)
=
Step 3
Identify the Terms
with braces
{3( -4) - 2}
=
{5 - ( -4)}
Steps 4
through
7
Follow the Order
of Operations for
the numerator
and denominator
separately
???
Why do we do this?
Step 8
Validate by redoing the steps
=
3( ) - 2
3( -4) - 2
=
5 -( )
5 - ( -4)
Take a closer look
Numerator
no parentheses, no exponents
{3(–4) – 2}
= {–12 – 2}
M
= –14
S
{3( -4) - 2}
{5 - ( -4)}
Validate:
{-14}
{5 + 4}
=
-14
9
=−
3
= -4 
-14 + 2 = -12 
{-12 - 2}
=
{5 - ( -4)}
=
-12
Take a closer look
Denominator
no parentheses, no exponents
{5 – (–4)} = 5 + 4 = 9
S
Validate: 9 + (–4) = 5 
14
9
3( −4) − 2
=
5 − ( −4)
=
−12 − 2
5+ 4
= –
14
9
RTMAS, correct order
???
Why not?
1. Leaving answers as improper fractions is acceptable in algebra, especially if the fraction is a negative
number. Ask your instructor which form is to be used for your class.
2. It is customary to move the negative sign to indicate that the number is negative.
Both
-14
14
14
and
are equivalent to - .
9
-9
9
Chapter 1 — Evaluating Expressions
64
Addressing Common Errors
Issue
Substituting
incorrectly
Incorrect
Process
Evaluate x2 – y
when x = – 5
and y = –2
–5 – 2 =
= –25 – 2
= –27
2
Evaluate when
a = –1, b = –3,
c = 1 and d = –2:
ab + cd
–1 – 3 + 1 – 2
= –5
Incorrectly
identifying
the terms
of an
expression
Not
noticing the
difference
between a
subtraction
sign and
the sign of
a negative
number
Simplify:
52– 2 (–3 + 9)
= 52 – 2 (6)
= 25 – 2 (6)
= 23 (6)
= 138
Find the value of
the expression if
x = –2.
x2 – (– 5 + x)
= (–2)2 – 5 + (–2)
=4–5–2
= –3
Resolution
Show the
substitution step by
using parentheses
around the
variables.
In this case,
substituting
incorrectly changed
products (ab) and
(cd) into terms.
Use parentheses
to show the
substitution.
Terms of an
expression are
separated by
addition and
subtraction signs.
Insert brackets to
separate the terms.
Simplify each term
separately.
Make sure to
identify and mark
each term by
enclosing it in
braces. If there are
double parentheses
in the intermediate
steps, keep them
there until you
have accounted for
each term in the
expression.
Correct
Process
Validation
Evaluate x2 – y when
x = – 5 and y = –2
RT, correct substitution
terms identified
( )2 – ( )
= (–5)2 – (–2)
= 25 – (–2)
= 25 + 2
= 27 R
E
25 ÷ –5 = –5 
27 –2 = 25 
RTES, correct order
S
RT, correct substitution
terms identified
Evaluate when
a = –1, b = –3,
c = 1 and d = –2:
ab + cd
3 ÷ (–3) = –1 
–2 ÷ 1= –2 
1 – (–2) = 3 
= ( )( ) + ( )( )
=(–1)(–3) + (1)(–2)
R
= {(–1)(–3)} + {(1)(–2)} T RTMA, correct order
= {3} + {–2}
M
=1
A

Simplify:
6 – 9 = –3
[52] – [2 (–3 + 9)]
25 ÷ 5 = 5 
12 ÷ 6 = 2 
= [5 2] – [2 (6)] P
13 + 12 = 25 
E
= 25 – [2 (6)]
PEMS, correct order
M
= 25 – 12
S
= 13
Find the value of the
expression if x = –2.
= (–2)2 – ( –7)
x – (– 5 + x)
= 4 – (– 7)
–7– (–2) = –5 
2
= (–2)2 – (–5 + (–2))
= {(–2)2}–{(–5+(–2))}
= (–2)2 – ( –7)
= 4 – (– 7)
= 11
R
T
P
E
S
(–2)(–2) = + 4 
= 11
11 + (–7) = 4 
RTPES, correct order
65
Section 1.4 — Substitution and Evaluation
Incorrect
Process
Issue
Not
reducing
fractions
to lowest
terms
Evaluate when
x = –1:
2 x -1
5-x
=
2( ) - 1
5 -( )
=
2(-1 ) -1
5 - ( - 1)
=
{ 2(-1 ) -1 }
{ 5 - ( - 1) }
=
{-3}
{66}
Correct
Process
Resolution
The braces
enclosing the
“terms” of the
fraction can be
removed when
the contents are
simplified to a single
number or simplified
expression.
Final answers
should be in
reduced fraction
form.
The steps are correct
until:
Validation
=
2(− 1 ) − 1
5 − ( − 1)
R
{-3}
{6}
=
T
=
-3
6
{ 2(− 1 ) − 1 }
{5 − ( − 1) }
=
M
=
-1
, fully reduced
2
{(− 2 ) − 1 }
{5 − ( − 1) }
−2 ÷ −1 = 2 
=-
1
2
=
{−3}
{6}
S
Numerator: − 3 + 1 = −2 
Denominator: 6 + ( −1) = 5 
Reduce:
=
−3
6
1
2
=
−1
1
=–
2
2
RTMS, correct order
Preparation Inventory
Before proceeding, you should be able to meet each of the following conditions:
You have memorized the Order of Operations and can apply it correctly
You can use each new term correctly to identify or describe algebraic components or processes
You can evaluate an algebraic expression given any value(s) for the variable(s)
Section 1.4
Activity
Substitution and Evaluation
Performance Criteria
• Evaluatng algebraic expressions for given values of the variables
– substition of correct values
– correct application of the Order of Operations
– documention of the process
– accurate calculations
– validation of the answer
Critical Thinking Questions
1. When is “1” the numerical coefficient of a term?
2. In any single term, what relationship does the numerical coefficient have to the variable parts of the term?
(What does the coefficient mean?)
3. What are the steps in the Order of Operations?
66
Section 1.4 — Substitution and Evaluation
4. Why can different values be substituted in the same algebraic expression?
5. Why is there a need for a specific order of operations?
6. How do you identify the terms of an expression?
7. What is the difference between a negative sign and a subtraction sign?
67
Chapter 1 — Evaluating Expressions
68
Tips
for
Success
Taking time between working a problem and validating the same problem allows you to view your work
more objectively. This will increase your chances of identifying any mistakes you may have made.
Demonstrate Your Understanding
1. Evaluate 3m–2mn when given the following values:
Values
a)
m = 2, n = –2
b)
m = –2, n = 2
c)
m = –1, n = –3
Worked Solution
Validation
69
Section 1.4 — Substitution and Evaluation
Values
d)
Worked Solution
Validation
m = 5, n = 0
2. Evaluate 2 x 2 - 4( x + 3) for the following values of x:
Values
a)
–1
b)
3
c)
0
d)
–2
Worked Solution
Validation
Chapter 1 — Evaluating Expressions
70
3. Evaluate 3 xy 2 - 2 x 2 y for the following values:
Values
a)
x = 1, y = –3
b)
x = –1, y = –2
c)
x = 3, y = –2
d)
x = –1, y = –1
Worked Solution
Validation
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Section 1.4 — Substitution and Evaluation
4. Evaluate
x +3
for the following values:
y 2 -1
Values
a)
x = 1, y = –3
b)
x = –1, y = –3
c)
x = –5, y = –4
d)
x = –2, y = –2
Worked Solution
Validation
Chapter 1 — Evaluating Expressions
72
Identify
and
Correct
the
Errors
In the second column, identify the error(s) in the worked solution or validate its answer.
If the worked solution is incorrect, solve the problem correctly in the third column and validate your answer.
Worked Solution
1) When m = –1
and n = –2 evaluate:
2m – 3mn
2 – 1 – 3 – 1 – 2= –5
2) Evaluate when a = –3
a -5
a +9
( ) -5
( ) +9
( -3) - 5
=
( -3) + 9
-8
=
6
=
3) Evaluate for x = –3
- x2 - 3x + 4
= -( -3) 2 - 3( -3) + 4
= 9 -9 + 4
= 4
4) Evaluate when a = 3
and b = –2:
2a – ab
2( ) – ( )( )
2(3) – (3)(–2)
=6–6=0
3) Evaluate for a = –4
and b = 4:
a2 – b2
( )2 – ( )2
= (–4)2 – (4)2
= {(–4)2} – {(4)2}
= 16 – 16 = 0
Identify Errors
or Validate
Correct Process
Validation