Lecture Notes Evaluating Algbraic Expressions page 1
Sample Problems
Evaluate each of the algebraic expressions when p = 7 and q = 3.
1. 15 p =
2. pq =
3. 4p q =
4. p 2q =
5. p
2
q
2
=
6. (p q)
2
=
7. 2q
2
=
8. (2q)
2
=
9. 15
p + q
5
=
10. (p + q)
2
(5q 2p)
4
=
Practice Problems
1. Evaluate each of the algebraic expressions when x = 6 and y = 8.
a) 19 y + x =
b) 19 (y + x) =
c) 2x
2
5y + 3 =
d) x
2
+ y
2
=
e) (x + y)
2
=
f) 3y x =
g) 3 (y x) =
h)
5x y
2
=
i) 5x
y
2
=
j)
x
2
5x + 4
y 3
=
2. Consider the expression
6a 3b ab + 2a
2
2a b
3. Evaluate this expression if
a) if a = 5 and b = 1
b) if a = 5 and b = 2
c) if a = 5 and b = 3
d) if a = 4 and b = 1
c
copyright Hidegkuti, Powell, 2008
Lecture Notes Evaluating Algbraic Expressions page 2
Sample Problems - Answers
1.) 8 2.) 21 3.) 25 4.) 1 5.) 40 6.) 16 7.) 18 8.) 36 9.) 13 10.) 99
Practice Problems - Answers
1. a) 17 b) 5 c) 35 d) 100 e) 196 f) 18 g) 6 h) 11 i) 26 j) 2
2.) a) 5 b) 5 c) 5 d) 4
Sample Problems - Solutions
Evaluate each of the algebraic expressions when p = 7 and q = 3.
1. 15 p = 8
Solution: Step 1. We re-write the expression with one modi…cation: we replace each variable by an empty
pair of parenthses.
Step 2. We insert the values into the parentheses. Now the problem becomes an order of operations
problem.
Step 3. We drop the unnecessary parentheses and work out the order of operations problem. (It may
appear awkward to create these parentheses but they will later become extremely helpful.)
Step 1. 15 p = 15 ( )
Step 2. = 15 (7)
Step 3. = 15 7
= 8
2. pq = 21
Solution:
Step 1. pq = ( ) ( )
Step 2. = (7) (3)
Step 3. = 21
3. 4p q = 25
Solution:
Step 1. 4p q = 4 ( ) ( )
Step 2. = 4 (7) (3)
Step 3. = 4 7 3 multiplication
= 28 3 subtraction
= 25
c
copyright Hidegkuti, Powell, 2008
Lecture Notes Evaluating Algbraic Expressions page 3
4. p 2q = 1
Solution:
Step 1. p 2q = ( ) 2 ( )
Step 2. = (7) 2 (3)
Step 3. = 7 2 3 multiplication
= 7 6 subtraction
= 1
5. p
2
q
2
= 40
Solution:
p
2
q
2
= ( )
2
( )
2
= (7)
2
(3)
2
= 7
2
3
2
exponents,
= 49 3
2
left to right
= 49 9 subtraction
= 40
6. (p q)
2
= 16
Solution:
(p q)
2
= [( ) ( )]
2
= [(7) (3)]
2
= (7 3)
2
subtraction in parentheses
= 4
2
exponentiation
= 16
7. 2q
2
= 18
Solution:
2q
2
= 2 ( )
2
= 2 (3)
2
= 2 3
2
exponentiation
= 2 9 multiplication
= 18
8. (2q)
2
= 36
Solution:
(2q)
2
= [2 ( )]
2
= [2 (3)]
2
= (2 3)
2
multiplication in parentheses
= 6
2
exponents
= 36
c
copyright Hidegkuti, Powell, 2008
Lecture Notes Evaluating Algbraic Expressions page 4
9. 15
p + q
5
= 13
Solution: From here on, we show computations in the form they should appear. Once you wrote
down the expression with little parentheses instead of the letters, you can insert the values into it. (In this
problem, we skipped the line 15
( ) + ( )
5
)
15
p + q
5
= 15
(7) + (3)
5
= 15
7 + 3
5
invisible parentheses!
= 15
10
5
division
= 15 2 subtraction
= 13
10. (p + q)
2
(5q 2p)
4
= 99
Solution:
(p + q)
2
(5q 2p)
4
= [(7) + (3)]
2
[5 (3) 2 (7)]
4
= (7 + 3)
2
(5 3 2 7)
4
addition in parentheses
= 10
2
(5 3 2 7)
4
multiplications in parentheses,
= 10
2
(15 2 7)
4
left to right
= 10
2
(15 14)
4
subtraction in parentheses
= 10
2
1
4
exponents, left to right
= 100 1
4
careful! 1
4
6= 4
= 100 1
= 99
For more documents like this, visit our page at https://teaching.martahidegkuti.com and c lick on Lecture Notes.
c
copyright Hidegkuti, Powell, 2008
