Chapter 1: Sets, Operations and Algebraic Language
1.4B: Evaluating Algebraic Expressions
Variable
A variable is something that varies, or changes. In math, a variable is a letter used to
represent a number in a mathematical phrase. The value of the number represented by
the variable in the mathematical phrase can change.)
Example: x. We use most letters except for O and I, which are easily confused with
numbers.
Term
A term is a number, a variable, or the product of numbers and variables.
Key Ideas
 Terms in an expression are always separated by a plus sign or minus sign.
 Terms in an expression are always either positive or negative.
 Numbers and variables connected by the operations of division and multiplication
are parts of the same term.
 You can move terms anywhere you want in an expression. If you move terms to
another expression (across the equal sign between the expressions), you have to
change the sign of the term.
 Terms are the building blocks of expressions.
Expression
A mathematical phrase
Variable Expression
A mathematical phrase that contains at least one variable, often called algebraic
expressionions
Equation
An equation is two expressions connected by an equal sign.
An algebraic expression (or variable expression) can be evaluated when each of its
variables is given a numerical value.
Evaluate each expression when m = -8 and n = +4.
(a)
mn
mn
Chapter 1, Section 1.4
Page 1
(b)  2n  m
Chapter 1: Sets, Operations and Algebraic Language
1.4B: Evaluating Algebraic Expressions
Solution:
Replace each variable with its assigned value, and then perform the indicated operations.
(a)
mn
mn
8 4
8 4
 12

4
 3

(b)
 2n  m 
 2 4  (8) 
 2 4  (8) 
 2 12
 24
Evaluating Variable Expressions. There are two general approaches to evaluating
variable expressions: 1) manually; and 2) using a graphing calculator.
 Manual Approach: Use substitution and the order of operations.
 Graphing Calculator Approach: Make an equation in which y equals the value
of the expression and input it into the Y= feature of the graphing calculator.
Then, use the 2nd TABLE feature to find the value of Y for a given value of X.
Note: Using the calculator method will often take longer than the manual solution.
Sample Regents Problem
If t  3 , then 3t 2  5t  6 equals
(1) -36
(3) 6
(2) -6
(4) 18
Two Solutions
Manual Solution
Chapter 1, Section 1.4
Page 2
Graphing Calculator Solution
Step 1.
Chapter 1: Sets, Operations and Algebraic Language
1.4B: Evaluating Algebraic Expressions
3t 2  5t  6
3

6
2
3  3   5   3   6
3  9    15   6
2
 5
27  15  6
12  6
18
Step 2
When the input variable is -3, the output is 18.
Regents Problem
If x = –4 and y = 3, what is the value of x  3y 2 ?
(1) –13
(3) –31
(2) –23
(4) –85
One Solution
This one cannot be solved with the calculator approach because there is more than one
variable. It must be solved manually using substitution.
x  4, y  3
x  3y2
   3 
2
 4   3  3 
4  3  9 
2
4  27
31
Homework Questions
1
Brett was given the problem: “Evaluate 2x 2 + 5 when x = 3.” Brett wrote that the
answer was 41. Was Brett correct? Explain your answer.
2
If t  3 , then 3t 2  5t  6 equals
Chapter 1, Section 1.4
Page 3
Chapter 1: Sets, Operations and Algebraic Language
1.4B: Evaluating Algebraic Expressions
(1) -36
(2) -6
3
(3) 6
(4) 18
What is the value of the expression 2 x 3 y when x  2 and y  3?
(1) -192
(3) -48
(2) -108
(4) 48
4
If x = –4 and y = 3, what is the value of x  3y 2 ?
(1) –13
(3) –31
(2) –23
(4) –85
5
If a = 3 and b = -1, what is the value of ab  b 2 ?
(1) -2
(3) -4
(2) 2
(4) 4
Homework Solutions
REGENTS QUESTIONS
1
SOLUTIONS
Brett was given the problem: “Evaluate Brett was not correct.
2x 2 + 5 when x = 3.” Brett wrote that the
2x2  5
answer was 41. Was Brett correct? Explain
your answer.
2(3) 2  5
2(9)  5
18  5  23
Brett made a mistake on the third line,
multiplying 2 and 3 to get 6 first. He
then squared 6 for 36 and added 5 for
41. He failed to apply the exponent
before multiplying.
2
If t  3 , then 3t 2  5t  6 equals
(1) -36
(3) 6
(2) -6
(4) 18
Chapter 1, Section 1.4
Page 4
(4)
Chapter 1: Sets, Operations and Algebraic Language
1.4B: Evaluating Algebraic Expressions
3t 2  5t  6
3( 3) 2  5( 3)  6
3(9)  15  6
27  15  6
12  6  18
3
4
What is the value of the expression 2 x 3 y
when x  2 and y  3?
(1) -192
(3) -48
(2) -108
(4) 48
If x = –4 and y = 3, what is the value of
x  3y 2 ?
(1) –13
(3) –31
(2) –23
(4) –85
(3)
2x3 y
2( 2) 3 (3)
48
(3)
x  3y 2
4  3(3) 2
4  3(9)
4  27  31
5
If a = 3 and b = -1, what is the value of
ab  b 2 ?
(1) -2
(3) -4
(2) 2
(4) 4
(3)
ab  b2
(3)( 1)  ( 1)2
3  1
4
All Regents Questions
REGENTS QUESTIONS
1
SOLUTIONS
Brett was given the problem: “Evaluate Brett was not correct.
2x 2 + 5 when x = 3.” Brett wrote that the
2x2  5
answer was 41. Was Brett correct? Explain
your answer.
2(3) 2  5
2(9)  5
18  5  23
Brett made a mistake on the third line,
multiplying 2 and 3 to get 6 first. He
then squared 6 for 36 and added 5 for
Chapter 1, Section 1.4
Page 5
Chapter 1: Sets, Operations and Algebraic Language
1.4B: Evaluating Algebraic Expressions
41. He failed to apply the exponent
before multiplying.
2
If t  3 , then 3t 2  5t  6 equals
(1) -36
(3) 6
(2) -6
(4) 18
(4)
3t  5t  6
2
3( 3) 2  5( 3)  6
3(9)  15  6
27  15  6
12  6  18
3
4
(3)
What is the value of the expression 2 x 3 y
when x  2 and y  3?
(1) -192
(3) -48
(2) -108
(4) 48
2x3 y
2( 2) 3 (3)
48
If x = –4 and y = 3, what is the value of
x  3y 2 ?
(1) –13
(3) –31
(2) –23
(4) –85
(3)
x  3y 2
4  3(3) 2
4  3(9)
4  27  31
5
If a = 3 and b = -1, what is the value of
ab  b 2 ?
(1) -2
(3) -4
(2) 2
(4) 4
(3)
ab  b2
(3)( 1)  ( 1)2
3  1
4
6
7
If x  2 and y  3, what is the value of
2 x 2  3xy  2 y 2 ?
(1) -20
(3) 8
(2) -2
(4) 16
1 2
xy is
2
(3) -4
(4) -8
If x = 4 and y = -2, the value of
(1) 32
(2) 8
Chapter 1, Section 1.4
Page 6
(3)
2(2)  3(2)( 3)  2( 3) 2
2
8  18  18
8
(2)
Chapter 1: Sets, Operations and Algebraic Language
1.4B: Evaluating Algebraic Expressions
1 2
xy
2
1
(4)( 2) 2
2
2(4)  8
8
What is the value of
y = –3?
(1) –2
(2) 2
1) Evaluate:
x2  4 y
, if x = 4 and
2
(4)
x  4y
2
2
4  4(3)
2
16  12 28

 14
2
2
2
(3) 10
(4) 14
52 ÷ (-22 + 32) + 24 · (1/4) =
2) Evaluate: 122 - 42 ÷ (-1/2) + 2 · (-3)2 =
3) a = -3
b=7
5a - 12b + 9 · 3
2b - 3a + 1
4) Evaluate 3y2 + 8x = , when x = 3 and y= -2
a) 12
b) 36
c) 60
d) 0
5) (112 + 20 ∙ ¾ ) ÷ 4 – 5 ∙ 7 =
6) Evaluate the expression when a = -5, b = 4, and c = -2
3b  16  2c  3a
2a 2  9c  4b


Chapter 1, Section 1.4
Page 7