Evaluate & Simplify Algebraic Expressions
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Transcript Evaluate & Simplify Algebraic Expressions
Evaluate & Simplify
Algebraic Expressions
Chapter 1.2
A First Definition of Exponents
We will say that, for any real number x and any natural number n
π₯π = π₯ β
π₯ β
β¦ β
π₯
where there are n factors of x. We call x the base and we call n the
exponent or the power.
For example, 97 = 9 β
9 β
9 β
9 β
9 β
9 β
9, where there are 7 nines
multiplied, or seven factors 9.
Even & Odd Exponents
β’ An important first property to note about exponents is how they affect
negative numbers
β’ By our definition, we know that
β3 4 = β3 β
β3 β
β3 β
β3 = +81
β’ Likewise, from the definition
β3 3 = β3 β
β3 β
β3 = β27
β’ In fact, we can show that raising a negative number to an even power
always produces a positive result; raising a negative number to an odd
power always produces a negative result
Even & Odd Exponents
But now we have a problem: how should we interpret
β24 ?
We could take it to mean β2 4 =
β2 β
β2 β
β2 β
β2 = 16
Or we could take it to mean β 24 =
β 2 β
2 β
2 β
2 = β16
We must choose an interpretation so that we all arrive at the same
answer
Even & Odd Exponents
β’ In making our choice we must consider that we will sometimes want
to show a perfect square as the square of a number
β’ That is, we might sometimes want to use 24 rather than 16
β’ But by the Inverse Property for Addition, we also have β16
β’ If 24 = 16 and β24 = 16, then we must use parentheses all the time if
we want to represent β16
β’ That is, we would always have to write β 24 = β16
Even & Odd Exponents
β’ To minimize the use of parentheses, we choose
β24 = β 24 = β16
β’ The way to interpret this is to say that the exponent does not affect the
sign
β’ So, in general, if n is an even whole number and x is any real number,
then
βπ₯ π = β π₯ π
β’ Why doesnβt this apply when n is an odd number?
Order of Operations
β’ The order of operations that you learned a few years ago is not a
consequence of the number properties
β’ The familiar order of operations results from our deciding to perform
multiplication first, then addition
β’ Also, we use parentheses to group operations that must be performed
first
β’ Finally, we choose to perform exponent operations
Order of Operations
So the order of operations (as you learned it before) is
1. Perform operations in grouping symbols first (parentheses ( ),
brackets [ ], braces { }, and operations that group within them, like
the square root symbol and the absolute value symbols)
2. Perform operations on exponents
3. Perform operations on multiplication (division)
4. Perform operations on addition (subtraction)
Evaluating
β’ We use the term variable to mean a letter that represents a number or
numbers when either the number(s) are not known (and we must find
them) or when the numbers are not specified
β’ An expression is a combination of numbers and/or variables under
one or more operations; it is not set equal to something else
β’ In an expression, the variables represent unspecified numbers
β’ If we specify numbers for our variables in an expression, we are then
able to evaluate the expression
Example 2: Evaluate an Algebraic Expression
Evaluate β4π₯ 2 β 6π₯ + 11 when π₯ = β3 (page 11)
β’ Note that this is an expression since it is not equal to something else
β’ This means that the variable doesnβt represent any particular number
β’ We are asked to find the value that results if π₯ = β3
β’ So we will substitute the value for the variable, then use the order of operations to evaluate
β’ β4 β3
2
β 6 β3 + 11 =
β’ β4 9 + 18 + 11 =
β’ β36 + 18 + 11 =
β’ β36 + 29 =
β’ β7
Guided Practice (Page 11)
Evaluate the expressions below
1. 63
2. β26
3. β2 6
4. 5π₯ π₯ β 2 when π₯ = 6
5. 3π¦ 2 β 4π¦ when π¦ = β2
6. π§ + 3 3 when π§ = 1
Guided Practice (Page 11)
Evaluate the expressions below
1. 63 = 216
2. β26 = β64
3. β2 6 = 64
4. 5π₯ π₯ β 2 when π₯ = 6; 5 6 6 β 2 = 120
5. 3π¦ 2 β 4π¦ when π¦ = β2; 3 β2 2 β 4 β2 = 20
6. π§ + 3 3 when π§ = 1; 1 + 3 3 = 43 = 64
Terms and Coefficients
β’ As we have done a few times before, letβs pretend that we donβt know
how to simplify the following
6π β 3π + 4π + 2π
β’ We will use the number properties to determine how to simplify this
(if it can be simplified)
β’ We will then look for a pattern so that we can simplify without having
to use the number properties
Terms and Coefficients
1.
2.
3.
4.
5.
6.
7.
8.
9.
6π β 3π + 4π + 2π =
6π + β3π + 4π + 2π =
6π + 4π + β3π + 2π =
(6π + 4π) + (β3π + 2π) =
π 6 + 4 + π β3 + 2 =
π 10 + π β1 =
10π + β1 π =
10π + βπ =
10π β π
1.
2.
3.
4.
5.
6.
7.
8.
9.
Given
Def. of Subt.
Comm. Prop. Add.
Assoc. Prop. Add.
Dist. Prop. (factoring)
Addition
Comm. Prop. Add.
Theorem
Def. of Subt. (reverse)
Terms and Coefficients
β’ How can we use the steps from the previous slide to come up with a
shortcut that allows us to simplify this simplification?
β’ 6π β 3π + 4π + 2π
β’ Note that we ended up adding 6 and 4, and also adding β3 and 2
β’ Why were we able to add them?
β’ In an expression that includes addition (subtraction), the parts of the
expression separated by + or β are called terms of the expression
β’ Terms that have the same variable and exponent combination are
called like terms
Terms and Coefficients
β’ We can simplify our expression by looking for like terms
β’ The number parts of the terms are called coefficients
β’ So we need merely add the coefficients and keep the variable
β’ 6π β 3π + 4π + 2π =
β’ Since the first and third terms are like terms, we add the coefficients
and keep the variable
β’ Likewise, the second and fourth terms are like terms, so we add the
coefficients and keep the variable
β’ So we get 6π β 3π + 4π + 2π = 10π β π
Example 4: Simplify by Combining Like
Terms
a) 8π₯ + 3π₯
b) 5π2 + π β 2π2
c) 3 π¦ + 2 β 4(π¦ β 7)
d) 2π₯ β 3π¦ β 9π₯ + π¦
Example 4: Simplify by Combining Like
Terms
a) 8π₯ + 3π₯ = π₯ 8 + 3 = π₯ 11 = 11π₯
b) 5π2 + π β 2π2 = π2 5 β 2 + π = π2 3 + π = 3π2 + π
c) 3 π¦ + 2 β 4 π¦ β 7 = 3π¦ + 3 2 + β4 π¦ + β4 β7
= 3π¦ + 6 + β4π¦ + 28 = β1 π¦ + 34 =
β 1 β
π¦ + 34 = βπ¦ + 34
Example 4: Simplify by Combining Like
Terms
d)
3π₯ β 3π¦ β 9π₯ + π¦ = β6π₯ β 2π¦
Guided Practice (Page 12)
8. Identify the terms, coefficients, and like terms in the expression
2 + 5π₯ β 6π₯ 2 + 7π₯ β 3
Simplify
9. 15π β 9π
10. 2π β 1 + 6π + 5
11. 3π3 + 5π2 β π3
12. 2π2 + π β 7π β 5π2
13. 8 π₯ β 3 β 2(π₯ + 6)
14. β4π¦ β π₯ + 10π₯ + π¦
Guided Practice (Page 12)
8. Identify the terms, coefficients, and like terms in the expression
2 + 5π₯ β 6π₯ 2 + 7π₯ β 3
Terms: 2, 5π₯, β6π₯ 2 , 7π₯, β3
Coefficients: 5, β6, 7
Like terms: 5π₯ and 7π₯; 2 and β 3
Guided Practice (Page 12)
Simplify
9. 15π β 9π = 6π
10. 2π β 1 + 6π + 5 = 8π + 4
11. 3π3 + 5π2 β π3 = 2π3 + 5π2
12. 2π2 + π β 7π β 5π2 = β3π2 β 6π
13. 8 π₯ β 3 β 2 π₯ + 6 = 6π₯ β 36
14. β4π¦ β π₯ + 10π₯ + π¦ = β3π¦ + 9π₯
Exercise 1.2
β’ Page 13, #5-31 odds, 37-49 odds, 52-55 (25 total)