Math 1700 – Unit 1
Course Notes
Algebra 1
Student Name:
Student CNA-Q ID Number:
Unit 1 – Algebra I
Topic 1: Number Systems
1.1
Real numbers and properties
1.1.1 Define and give examples of the following:
1.1.1.1
Natural numbers
1.1.1.2
Integers
1.1.1.3
Rational numbers
1.1.1.4
Real numbers
1.1.1.5
Complex numbers
1.1.2 State and use the properties of real numbers
Number Systems
Real Numbers (R):
Real numbers are the set of numbers containing:

Natural Numbers

Whole Numbers

Integer Numbers

Rational Numbers

Irrational Numbers
Rational
Natural
1, 2, 3, 4, 5, 6…
Whole
0, 1, 2, 3, 4 5, …
Integer
…-3, -2, -1, 0, 1, 2, 3,…
…-3, -2, -1, 0, 1, 2, 3,…
and numbers like
-2.85,
,
,
, 3.858585…,
67.8549
Irrational: Numbers such as
2
Natural Numbers (N):
Natural numbers are the counting numbers. 1, 2, 3, 4, 5…
Whole numbers (W):
Whole numbers are the natural numbers, and zero. 0, 1, 2, 3, 4…
Integers (I):
Integers consist of the whole numbers and their negatives.
... −4, −3, −2, −1, 0, 1, 2, 3, 4…
Rationals (Q):
1.
Numbers which can be written as fractions
Examples:
2.
1
22
,  ,
3
7
6
6 ,
1
5 
5
5

1
1
Decimal numbers which terminate (They stop; have an exact value.)
Examples: 3.7,
3.
1
,
2
0.845, −0.165,
6.25 , 46.345
Decimal numbers which have an infinite repeating pattern
Examples: 1.3333….= 1.3 , 0.6666666…= 0.6 , 2.141414….= 2.14 ,
5.123412341234…= 5.1234
Irrationals ( Q ):
These are numbers such as
5 ,  7 ,  (constant), e (constant) , 45.68982…
1.
They cannot be written as fractions.
2.
They do not have terminating decimals.
3.
They do not have a repeating decimal pattern.
Check these values on your calculator.
(More about roots later)
π = 3.1415926535897932384626433832795... You
cannot write a simple fraction that equals Pi exactly.
3
(Check your answers below.)
1. List the number sets to which  4 belongs.
2. List the number sets to which 50 belongs.
3. Circle the numbers which are rational.
4
0
8
55
9
12
3
1.676767…
3.4
7.3
987.385681397…
17
4
12.3
1. Integer, Rational, Real. (Natural numbers start at 1.
Whole numbers start at 0.)
2. Irrational, Real.
(Non-repeating, non-terminating decimal)
3. The rational numbers are:  4 , 0,
17
4
, 55, 9,
12
3
,
1.676767… , 3.4, and 7.3 . The other numbers are irrational.
4
Study the following table before you complete the next exercise. The
sets to which each number belongs are already checked for you.
Number
Natural
Whole Integer Rational Irrational
Real
7.68581…
15
8

9
2
.345
0
2
3
1. Check the sets to which each number belongs.
Number
a)
b)
Natural
Whole
Integer Rational Irrational
Real
 69
9
c)
48.6767…
d)
7
10
e)
17
f)
6
g)
8.75
5
2. To which sets does each number belong? Use the symbols: N, W, I, Q, Q , R
a) 
4
5
b)
56
________________________________________________________
c)
15
________________________________________________________
d)
3.679679679... ___________________________________________________
e) 0
3.
________________________________________________________
________________________________________________________
Place each number in the correct set(s) below.
4
55
22
7
8
12
3
3.4
987.3856813…
a) Natural
__________________________________________________
b) Whole
__________________________________________________
c) Integer
__________________________________________________
d) Rational
__________________________________________________
e) Irrational __________________________________________________
6
4.
Check the sets to which each number belongs.
Number Natural
a)
b)
c)
Whole
Integer Rational Irrational
Real
8
 8
8
4
e)
6.125
g)
36
h)
5
8
i)
5e
j)
22
7
Imaginary Numbers:
 Since the square of a positive number or a negative number is positive, it is
not possible to square any real number and have a negative result.
 6    6    6   36
o Example:
2
 6    6    6   36
2
 We need a number system to include square roots of negative numbers
because these numbers have solutions which are NOT real
 For example, there is no real solution to 16 or 36
 Mathematicians created a new system of numbers using the imaginary unit, i,
defined as i  1 . With this new system of numbers, radicals of negative
numbers can now be simplified!
o Examples: 16  (16)(1)  16  1  4i
36  (36)(1)  36  1  6i
7
Complex Numbers:
 can be written in the form a  bi where a and b are real numbers (including 0)
and i is an imaginary number
 Complex numbers contain two 'parts':
o one part is real; another part is imaginary

Examples:
2  5i
2i
4  3i
7  0i (a  7 and b  0)
(a  0 and b  2)
Graphical Illustration of Number Systems:
8
Check the sets to which each number belongs.
Number
a)
9  10i
b)
3i
c)
49
e)
 100
g)
4i
h)
2
Real
Imaginary
Complex
Natural Whole Integer Rational Irrational Real
Imaginary Complex
10
5
7 31
 4
10i
2e
9
Properties of Real Numbers
 Before we begin our study of algebra, we need to review the fundamental laws
of algebra, most of which we use naturally without thinking about them.
Name the property of real numbers that is shown in each statement.
Example
 5   
1
 1
5
Property
Inverse- Multiplication
1 4  3  1  4  3
Associative – Addition
23  32
Commutative – Multiplication
70  7 07
Identity – Addition
3  2  5   3  2  3  5 
Distributive
10
1.
Name the property of real numbers shown in each statement.
Example
Property
 2   3   3   2
7  1  1 7
6 a  b   6a  6b
4x   4x   0   4x   4x
 6  5   2  6   5  2
2.
The area of each rectangle below can be represented in two ways:
as the area of a single rectangle, or as the sum of the areas of the two
rectangles.
(a)
Find the area of each rectangle in two different ways.
(b)
What property of real numbers did you use?
11
Unit 1 – Algebra I
Topic 2: Order of Operations
1.2 Signed numbers: Variable expressions
1.2.1 Perform the four basic operations with signed numbers
1.2.2 Evaluate expressions involving exponents and the four basic
operations
Operations with Integers
Addition of Integers
Rule: If two integers have the same sign, add their absolute values and give the sign of the
+6 +8 +14
original numbers for the answer.
Examples:
6  8  14
(6)  (8)  14
No sign in front
indicates positive.
If the integers have different signs, find the difference between the two numbers and give
the sign of the number which has the largest absolute value.
Examples:
(7)  (5)  2
(8)  (5)  3
Two like signs become a plus.
Answer: 10
Two unlike signs become a minus.
Answer: 2
Subtraction of Integers
Rule: To subtract, add the opposite, (or negative), of the number being subtracted.
Every integer has an opposite.
Examples:
–3 is the opposite of 3 .
6  (8)  6  (8)  14
5 is the opposite of –5.
 4  2   4  (2)  6
a)
8  (4)
b)
8  (4)
c)
8  4
d)
7  (4)
e)
1  (8)
f)
5  (6)
g)
1  ( 4)
h)
3  ( 4)
i)
5  ( 4)
12
Rules for Multiplication and Division
Positive x Positive
( )( )
OR
Positive  Positive
()  ()
Positive x Negative
( )()
OR
Positive  Negative
(  )  ( )
Negative (  )
Negative x Positive
( )( )
OR

Negative (  )
Negative x Negative
( )( )
OR
Negative  Negative
( )  ( )
Positive (+)
Positive (+)
Negative
Positive
( )  (  )
Brackets can be used to
show multiplication.
A dot “ ” can be used to
show multiplication.
5  3
a)
8  4
b)
(8)(4)
c)
d)
(7)  (4)
e)
1 8
f)
g)
(81)  (3)
h)
(9)(4)(1)
i)
28
7
k)
(18)  (3)
l)
18
3
j)
(8)  (1)
(5)(6)(2)
13
Order of Operations
Operations in math refer to addition, subtraction, multiplication, and division.
Some numbers have exponents. For example:
73
3 is called the exponent.
7 is called the base.
7 3 is in exponential form.
7 3 means 7  7  7
54  5  5  5  5
Note: These dots mean “times”, or
“multiply by”. Don’t confuse them with
the decimal point, which is placed lower.
7.2 means “7 point 2”
To perform calculations on numbers we must follow the BEDMAS order.
B
Complete the operations inside the BRACKETS first.
E
Next, do the EXPONENTS.
D
M
Then, DIVIDE and MULTIPLY, in the order they appear, from left to right.
A
S
Finally, ADD and SUBTRACT, in the order they appear, from left to right.
1) Evaluate 5  3  2 using the Order of Operations.

Step 1:
Multiply
5 + 3
Step 2:
Add
5 + 6
Answer:
2
11
2) Evaluate 2  8  2  2  3 using the Order of Operations.
Step 1:
Divide first; then multiply
28  22  3
Step 2:
Add
2  4  6
Answer:
12
14
3) Evaluate 9 – 5 ÷ (8 – 3) x (–3)2 + 6 using the Order of Operations.
Step 1:
Brackets
9 – 5 ÷ (8 – 3) x (–3)2 + 6
Step 2:
Exponent
9 – 5 ÷ 5 x (–3)2 + 6
Step 3:
Divide
9 – 5 ÷ 5 x 9 + 6
Step 4:
Multiply
9–1 x 9 + 6
Step 5:
Subtract
9 – 9 + 6
Step 6:
Add
0+6
Answer:
Know your calculator!!
Use brackets when squaring a
negative number. To square
, use
. (Answer: 9)
6
1. Evaluate using the Order of Operations. Show your work.
a) 52  2  3  32
b) 40  3  8  4  7
c) (7  4  5)  8  2  11
d) (3  23 )  (3  12   2)
2
e) (6  2)3  8   4 
f) 40  3  8  4  7
15
g) 30  62  3  (7  5)3
h) 2 1  2  9  9
i) (2  4)2  (2)3  7    11  16  2
j) (6)2  (2)3  4  16  4 
k) (2)2  (2)3  4  16  2
l) (4  2)3  4   6  4 
m) ( 8)2  (3)3  7   11  18  2
n) 3   2  10  2  3  4  6  3   2)  8 
16
Unit 1 – Algebra I
Topic 3: Units of Measurement
1.7
Units of measurement
1.7.1 Use the metric system (SI)
1.7.2 Change units (dimensional analysis)
Metric System
 The Metric System, developed in the late 1700s to standardize units of
measurement in Europe, is the primary system of measurement used in much
of the world today.
 It is called the International System of Units or SI. Three common units in
the system are metre, gram and liter,
Quantity
length
mass
volume
Base Unit
meter
gram
liter
Symbol
m
g
L
 Prefixes are used to denote sizes of units.
Six of the most common prefixes are:
kilo (k)
=
1000
A kilometre means 1000 meters.
hecto (h)
=
100
A hectolitre means 100 litres.
deca (da)
=
10
A decagram means 10 grams.
deci (d)
=
A decimetre means
of a metre.
centi (c)
=
A centilitre means
of a litre.
milli (m)
=
A milligram means
of a gram.
base unit:
metre, liter, or gram
17
To convert between the various sizes of each unit, move back and forth
along the prefixes to move the decimal point OR multiply/divide by
multiples of 10.
To convert from a smaller unit to a larger unit, move the decimal to the
left, OR divide by 10, 100, or 1000 - depending on the number of spaces
between the units.
To convert from a larger unit to a smaller unit, move the decimal to the
right OR by 10, 100, or 1000 - depending on the number of spaces
between the units.
m
millimetres
metre
dam
centimetres
hm
decametres
kilometres
km
decimetres
Multiply
hectometres
Divide
dm
cm
mm
Convert 56.3 decimetres (dm) to millimetres (mm).
Solution:
km
hm
dam
m
dm
cm
Place your pencil on dm. Count the number of spaces to mm
Move the decimal in 56.3 two spaces to the right.
56.3
56.3 0
mm
2 spaces right.
OR Multiply 56.3 by 100.
5630.
The decimal has moved to the end of the number. Thus, we don’t need to write it.
Answer:
5630 mm
18
Convert 7236.4 metres (m) to kilometres (km).
Solution:
km
hm
dam
m
dm
cm
Place your pencil on “m”. Count the number of spaces to km
Move the decimal in 7236.4 three spaces to the left.
7236.4
Answer:
mm
3 spaces to the left.
OR Divide 7236.4 by 1000.
7.2364
7.2364 km
Convert each measurement to the required unit.
1.
46 millimetres (mm) to metres (m)
1. _____________________
2.
76.2 centimetres (cm) to decametres (dam)
2. _____________________
3.
4.6 m to km
3. _____________________
4.
458 milligrams (mg) to decigrams (dg)
4._____________________
5.
0.02 grams (g) to centigrams (cg)
5._____________________
6.
0.457 g to mg
6._____________________
7.
3.87 kilolitres to milliliters
7._____________________
8.
0.957 liters (L) to milliliters (mL)
8._____________________
9.
662 cL to L
9._____________________
19
Time
 We will concentrate on conversions between seconds, minutes, hours, and days
 The diagram below shows the relationship between these units
Unit Cancellation
How Many Kilograms Are in 1 532 Grams?
Step A: Show the relationship between
kilograms and grams.
A.
1 kg  1000 g
Step B: both sides of the equation are divided
by 1000 g.
B.
1 kg
1 000 g

1000 g
1000 g
C.
1 kg
 1
1000 g
D.
1536 g  ? kg
E.
1536 g 
F.
1536 g 
G.
1.536 g  1536 kg
Step C: Show how the value of 1 kg/1000 g is
the equal to the number 1. This step is
important in the unit cancellation method.
When you multiply a number or variable by 1,
the value is unchanged.
Step D: Restates the example problem.
Step E: Multiply both sides of the equation by
1 and substitute the left side's 1 with the value
in step C.
Step F is the unit cancellation step. The gram
unit from the top (or numerator) of the fraction
is canceled from the bottom (or denominator)
leaving only the kilogram unit.
Dividing 1536 by 1000 yields the final answer
in step G. There are 1.536 kg in 1536 grams.
1 kg
 ? kg
1000 g
1 kg
 ? kg
1000 g
20
1.
Convert 40 minutes to seconds.
Solution:
2.
60sec
60sec
 40 min 
 2400sec
1min
1min
Convert 1200 hours to days.
Solution:
3.
40min
1200hr 
1 day
1 day
 1200 hr 
 50 days
24hr
24 hr
Convert 6000 seconds to hours
Solution:
14400 sec 
1min
1hr

 4 hours
60 sec
60 min
Conversions with Area and Volume
 The area or volume units for a conversion can be found by squaring or cubing the
base units, respectively.
100  cm 
3
3
1 min
3
3
100cm 
 What is 
 ?
 1min 


3
10000cm3

1min3
1. Convert 50 mL to liters.
50 mL 
1L
1L
50 L
 50 mL 

 0.05 L
1000 mL
1000
1000 mL
2. Convert 38.2 m/s to kilometers per hour.
38.2 m
1 km
60 s 60 min 38.2 m
1 km
60 s
60min







1s
1000 m 1 min
1h
1h
1 s
1000 m 1 min
38.2  1 km  60  60

1  1000  1  1h
 138 km/h
21
3
3. Convert 575 g/cm to kilograms per cubic meter
3
575 g
575 g
1 kg
1003 cm
1 kg
1003 cm3





1 cm3 1000 g
13 m3
13 m3
1 cm3 1000 g

575  1kg  1003
1  1000  13m3
 575 000 kg/m3 or 5.75  105 kg/m3
Calculate each unit conversion:
1.
Convert 50 mL/min to L/hr.
2.
Convert 38.2 m/s to kilometers per hour.
3.
Change 200 mm/s to meters per second.
4.
Change 1.32 km/h to meters per minute.
22
5.
Change 9.80 m/s2 to centimeters per minute squared.
6.
Convert 575 g/cm3 to kilograms per cubic meter.
7.
Convert 7.45 g/cm2 to kilograms per millimeter squared.
8.
Convert 1.45 m/s2 to km/h2 .
9.
Convert 0.08595 kg/m 3 to mg/cm3 .
10.
Convert 36 g/min2 to kg/hr 2
23
Unit 1 – Algebra I
Topic 4: Laws of Integral Exponents
1.3
Exponential laws
1.3.1
Simplify expressions by using the laws of exponents
1.3.2
Evaluate exponential expressions



In mathematics, we often have a number multiplied by itself several times
To show this type of product, we use the notation a n where a is the number and n is the
number of times it appears
In the expression a n , the number a is called the base , and n is called the exponent
is called the base and
is called the exponent
Exponents of Positive Numbers
52 means 5  5  25
53 means 5  5  5  125
54 means 5  5  5  5  625
The exponent tells us how many times to
multiply a number by itself.
52 means  (5  5) or  1   5  5    25
52 does not mean (5)  ( 5) It is not the square of 5
because the negative sign is not raised to the power of 2.
53 means   5  5  5  or  1   5  5  5    125
54 means   5  5  5  5  or  1   5  5  5  5    625
24
Exponents of Negative Numbers
(2)2 means
(2)3 means
(2)4 means
 2    2
 4
 2   2   2
 8
 2   2    2   2
 16
Since “ 2 ” is inside the bracket, the negative (as well as “2”)
is raised to the indicated power.
(2)2 means   2   2 or  1  4   4
(2)3 means   2   2   2 or  1  8  8
(2)4 means   2   2   2   2 or  1  16   16
Know your calculator!!
Use brackets when squaring a negative
number. For example, to square
use
. (Answer: 81)
Evaluate.
1) 34
2) 32
3) (4) 2
4) (4)3
5) 42
Solutions:
1) 34  3  3  3  3  81
2) 32    3  3   9
3) (4)2   4   4  16
4) (4)3   4   4   4   64
5) 42    4  4  16
25
1. Evaluate.
a) 63
b) 6 2
c) (6) 2
d) (3)4
e) (3)3
f) (3)4
g) (1)2  (7)2
h) (2)3  (2)2
i) (7)3  (1)2
Rule #1: Multiplication Rule
The Multiplication Rule of Exponents
For any integers m and n, and any real number a
To multiply expressions with the same base,
keep the base and add the exponents.
x3  x2
means
x  x  x  x  x  x5
Using the Multiplication Rule, ADD exponents.
x3  x2

x 3 2

x5
26
Simplify.
1) x  x
5
3) 2 x  3x  x
2) 3  3
2
3
9
4
6
4) (3x )( x )(2 x )
8
3
11
 1 18 3 
x y   (15 x 6 y 1 )
3

5) 
Brackets with no sign in between indicate
multiplication. A “dot” is optional between
the brackets.
Solutions:
1) x  x  x
5
3
53
3) 2 x  3x  x  6 x
8
11
23
2) 3  3  3
 x8
2
8  11  1
Multiply numerical coefficients.
 6 x 20
3
9
 35  243
4) 3x  x  2 x  6 x
4
6
9  4  6
 6 x1  6 x
Add all exponents, including the negatives.
1
 1 18 3 
x y  15x 6 y 1  
 15  x18  x 6  y 3  y 1  5 x12 y 2
3
3

5) 
Remember, we add exponents
of terms with like bases only.
Simplify.
a) x  x
9
______________________________________
b) x  x  x
3
2
4
c) x y  x x
17
5
13
______________________________________
4
______________________________________
27
d) (8 x y )(5 x y )
______________________________________
e) (5ax)  (4ax )  (3ax)
______________________________________
2
6
3
9
3
f)
px19  p x3  3x5
7 4
3 5
______________________________________
3 3
g) (5 x z )(2 x z )( x z )
h)
______________________________________
1 14 4
x  x  16 x5
2
7
______________________________________
3
11 1
i) (15 p r )(2 p r )(3 p r )
______________________________________
2 17
ax  6ax 10  3a 1x 5
3
______________________________________
j)
2
5 1
k) 10 p r  p r  3 p r
3
______________________________________
 1 15 4 
x y    45 x 5 y10 
3

l) 
______________________________________
1 8 6
x z   6 x 10 z 3   x3 z 2  ______________________________________
6

m) 
6 1
n) 6 p r
o)
a x
2
6
 p 1r 8  (3) p9r 5
y 4   4 a 1 x 5 y10

______________________________________
______________________________________
28
Rule #2: Division Rule
Rule 2:
The Division Rule of Exponents
For any integers m and n, and any nonzero number a
To divide expressions with the same base,
keep the base and subtract the exponents.
x5
x2
x x x  xx
xx
means

x3
Using the Division Rule, SUBTRACT exponents.
x5  3  x2
Simplify.
37
2) 2
3
x8
1) 3
x
4)
24 x15 y11
6 x 2 y 3
5)
6a 5 x15 y11
24a 4 x 2 y11
3)
x7 y9
xy 6
6)
16 p 6 x 9
40 p 2 x 6
Solutions:
x8
83
5
1) 3  x  x
x
37
72
 35 or 243
2) 2  3
3
3)
x7 y9
 x7  1 y9  6  x6 y3
6
xy
4)
24 x15 y11
  4 x15  2 y11  3   4 x13 y 8
2 3
6 x y
5)
a5  4 x15  2 y11
6a5 x15 y11
ax13


24a 4 x 2 y11
4
4 y11
6)
16 p 6 x 9
2 p 6  2 x9  6 2 p 4 x3


40 p 2 x 6
5
5
6 divides down into 24. The answer,
4, is placed in the denominator.
Coefficients are reduced
to lowest terms.
29
Simplify.
36
a) 2
3
c)
a19b3
a11b 2
x8
b) 5
x
d)
x13 y17 z 7
x11 y 5 z 2
p10 q10
e) 10 10
p q
24 p10 q10 r14
f)
8 p10 q10 r14
15a13c 7
g)
5a11c 7
12 p10
h) 
18 p 6
20 p10 q11
i)
45 p 9 q10
13 p 7 q10 r 14
j)
23 p 6 q 9 r13
k)
12 p10 q10 r 14
4 p 6 q 9 r10
l)
50m10 n10 p5
5m6 n9
30
Rule #3: Exponent to an Exponent Rule
The “Exponent to an Exponent” Rule
Rule 3:
For any integers m and n, and any real number a
When an exponent is raised to another exponent, the exponents are
multiplied.
( x3)4
means
x3  x3  x3  x3  x12  Multiplication rule 
Using the “Exponent — Exponent ” rule, MULTIPLY exponents.
( x3 )4  x3  4  x12
Rule #4: Product to an Exponent Rule
Rule 4:
The “Product to an Exponent” Rule
For any integer m, and any real numbers a and b,
When a product is raised to a power, each term
is raised to that power.
(2 x3)4
means
2 x3  2 x3  2 x3  2 x3  16 x12
Using the “Product to an Exponent” rule, MULTIPLY exponents.
(21 x 3 )4  21 4 x 3  4  24 x12  16 x12
31
Simplify.
1) ( x5 )2
2) ( x7 y 2 )3
3) (3a9b4 )2
4) (2 x5 y 4 z)3
5) ( x4 y 2 r 6 )5
Solutions:
1) ( x5 )2  x10
3) (3a9b4 )2  32 a18b8  9a18b8
2) ( x7 y 2 )3  x21 y 6
4) (2 x5 y 4 z)3  (2)3 x15 y12 z3   8x15 y12 z 3
5) ( x4 y 2r 6 )5  (1)5 x20 y10r 30   x20 y10r 30
The coefficient is –1.
Simplify.
a) ( x8 )2
b) (32 )3
c) (2m4 n3 )3
d) (2a 4 )3
32
e) ( x4 )2
f) ( x4 )3
g) (3x 4 y 2 z 3 )4
h) (2a5b4c)3
i) (2 p3q 2 r 5 s)2
j) (2 x3 y 6 z 5 )4
k) ( x 4 y3 )7
l) (2 xy)4
Simplify.
a) (5xy 4 )  (3x3 y)
b) 3ax9  a 7 x3
33
c) (4 x6 y3 z)(2 x3 y 2 z 2 )
d)
1 17 3
x yz  12 x 10 y 5 z 1
3
2w11
14w6
e)
32a13b7 c9
8a11c 7
f)
g)
8 p 7 q10 r 12
36 p 7 q 9 r 7
h) (5x2 y3 )3
i) (2 x2 y3 z 4 )4
j) (2a5b4c)3
34
Rule #5: Quotient to an Exponent Rule
Rule 5:
The Quotient to a Power Rule for Exponents
For any integer m, and any real numbers a and b,
When a quotient is raised to an exponent, the numerator and
denominator are each raised to that exponent.
 3 x4 


 4b2 

3


 3 x4 
3
 3  x4 
3

   4  
4b2
3
3
b2
3
3
444 27 x12
3

3

3

x


4 4 4 b222 64b6
Using the “Quotient to an Exponent” rule, MULTIPLY exponents.
(21 x 3 )4  21 4 x 3  4  24 x12  16 x12
Simplify.
 x
1)  
5
 3x3 
2)  4 
 5y 
2
2
 m5 n3 
3)  
2 
 2m n 
Solutions:
 3x3 
2)  4 
 5y 
2
x2
 x
1)   
25
5
3
 m5 n 3 
3)   2  
 2m n 
 m n 
 2m n 
5 3 3
2
3
 
m15 n9
23 m6 n3
 
m9 n 6
8
 3x 
5 y 
3 2
2

4 2
3
9 x6

25 y 8
(Simplest form)
The negative can be placed with
EITHER the numerator OR the
denominator (not both).
35
Simplify.
2
a)  
 y
4
 8 x3 
b)   
 3 
 3x5 
c)  10 
 2y 
2
3
 m5 n 5 
d)   2 
 5x 
3
 m3 n 2 
e)   2 
 2m n 
 x3 y 4 
f) 
2 3 
 2 x y 
4
5
36
Rule #6: Zero Exponent Rule
Rule 6:
The Zero Exponent Rule
,
For any real number a, where
When a number or expression (other than 0), is raised to the
power of zero, the value is 1.
Recall: when we divide a number (or expression) by itself, we get 1.
x3
1
3
x
5
 1,
5
e.g.
x3
x3
also equals
x3  3  x0
Thus,
x0  1
Rule #7: Negative Exponent Rule
Rule 7:
The Negative Exponent Rule
For any real number a, where
, and for any integer
When the base “ ” is raised to a negative integer power, we can
rewrite it as 1 over the base, raised to a positive power.
3
According to the rule, 5 
Thus,
1
1

3
1
5
53
 1
53
 53
1
1
53
Shortcut:
1
 53
3
5
37
Simplify. Give a value where possible.
a) 90
b) 9 0
c) (9 xyz )0
d) 9xyz 0
e) 20  xy 0  32
f)
3
4 0
Solutions:
0
0
a) 90  1
b) 9   1  9   1  1   1
c) (9 xyz)0  1
d) 9 xyz 0   9 xy(1)   9 xy
e) 20  xy 0  32   (1)  ( x)(1)  (9)   9 x f)
3
3
3
3



 3
0
0
4
1 4
11 1
Simplify. Write all answers with positive exponents.
a) 2 4
b) x 3
e) (3)3
f)
3
d)  
4
c) 2x 4
1
32
g)
3 x 7
5 x 5
h)
2
21 x6 y 7 z 1
x 5 y 2
Solutions:
a) 24 
c) 2 x 4 
1
1

4
2
16
3
d)  
4
2
x4
e) (3)3 
1
1
1

or 
3
(3)
27
27
7  5
g)
b) x 3 
3x 7 3x 

5 x 5
5
3x 2
3


5
5x2
f)
2
1
x3
2
16
4
   
9
3
1
 32  9
2
3
21 x6 y 7 z 1 x 6  ( 5) y 7  ( 2)
x11 y 5
x11
h)



x 5 y 2
2z
2z
2 y5 z
38
When a fraction has a negative exponent, flip the
fraction and the exponent becomes positive.
Here’s another way to simplify expressions with
negative exponents.
First, rewrite with positive exponents!!
Switch terms with negative exponents from the numerator to the
denominator (or denominator to numerator) to make the
exponents positive. For example,
Now that all exponents are
positive, we can simplify.
Multiply
Divide
and
and
.
.
Add exponents.
Subtract 2 from 7.
Place the answer from division wherever the term
with the larger exponent was located.
Since y 7 was located in the denominator, place
y5
in the denominator.
39
1. Simplify. Write all answers with positive exponents.
a) 5 2
b) 5 x 2
d) (5)3
e)
3
g)  
 5x 
i)
x 4 y 4
21
3
c) (5)3
1
32
f)
3
h)   
2

j)
4 x 8
5 x 5
3

When you flip
fractions, remember to
change the sign of the
exponent only!!
32 x5 y 3
x 1
40
2.
k)
4a 4 a 7b 1
a 1b 2
l) x7 y8  x 4 y 0 y 3
m)
8 p 4 q 7 r 1
12 p 1q 8 r 2
n) a 1b3c  a 4b3c 1
For each expression, remove the brackets using the “product to a power
rule” and then simplify. Write all answers with positive exponents.
2 4 2
Example: Simplify (3 xy )( x y )
(2 x 3 y 4 ) 1
3 xy  x 4 y 8

21 x 3 y 4
3 x 5 y 7
 1 3 4
2 x y

2 3 x 5 y 4
x3 y7
6 x2
 3
y
Remove brackets. (Multiply exponents.)
Combine terms in the numerator.
(Add exponents.)
Move terms with negative exponents
from numerator to denominator (or vice
versa) to make all exponents positive.
Multiply 2 and 3 to give 6.
2
Subtract exponents 5 and 3 to give x .
y3 .
7
3
Place y in the denominator, since y
Subtract exponents 7 and 4 to give
(having the larger exponent) is in the
denominator.
41
( p 6 )1
a)
( p 4 )3
x 2
d)
(2 x 3 ) 2
g) (3h5k 2 )2 (h1k 4 )3
i)
(2 x 3 y 4 ) 2
x 3 y
b)
e)
2
 3xy 
c)
 4x y 
( x) 2
f)
( x 3 y ) 1
1 3
3
2
h)
2
j)
x 2 y 6
(2 x 3 y 4 ) 2
1
ab 1c 2 
1
p 1q 1 
3
3
42
3x 2 y 4 ( x 2 y 5 ) 1
k)
12 x 6 y 8
 3m9 n 4 
m)   2 2 
 2m n 
 2 3
o)
l)
(3x 3 y 4 )4
n)
32 x 3 y
3
( x ) ( 2 y )
(2 x 5 ) 2 (3 y ) 0
(4 xy )( x 3 y 4 ) 3
(2 x 3 y ) 2
4 2
p)
 3x7 y 5 
 3 2 
 2x y 
2
 9 x 8 y 9 
 4 7 
 2x y 
2
43
Name of Rule
Rule
Example
The Multiplication Rule
am  an  am  n
x2  x3  x5
The Division Rule
am
mn

a
an
x7
5

x
x2
The “Power to a
Power” Rule
(a m )n  a mn
( x 5 ) 2  x 10
The “Product to a
Power” Rule
(ab)m  a m bm
(2 x 2 y 5 )3  8 x 6 y15
The “Quotient to a
Power” Rule
The Zero Exponent
Rule
The Negative Exponent
Rule
a
 
b
m
a

m
bm
(a)0  1
(a)
n
 1n
a
3
 3x 
27 x15
 10  
30
2
y
8
y


5
(2 x 2 y 3 )0  1
( x)
3
1

( x )3
44
Unit 1 – Algebra I
Topic 5: Algebraic Expressions
1.4
1.2.3
Addition and subtraction of algebraic expressions
1.4.1 Add and subtract like terms
1.4.2 Simplify algebraic expressions by removing grouping
symbols and collecting like terms
1.4.3 Add and subtract algebraic expressions
Evaluate variable expressions
Simplifying Algebraic Expessions
In algebraic expressions, the parts that are separated by plus or minus signs are called terms. For
example, in the expression 3 x 2  4 x  6 , the terms are 3x 2 , 4 x, and 6.
Like terms contain the same variables, raised to the same power or exponent.
3x and 6x are like terms.
They both contain x.
6c2 and 19c2 are like terms.
They both contain c2.
2xy3 and  10xy3 are like terms.
They both contain xy3.
km2x5 and 7km2x5 are like terms.
They both contain km2x5.
4x and 5x2 are unlike terms.
x and x2 have different exponents.
5x and 17xy are unlike terms.
x and xy are not alike.
The number in front of a variable is called a numerical
coefficient, or simply a coefficient.
Coefficient
Variable
Coefficient
3x
3x
means 3 times x.
Exponent
10xy3
10xy3 means 10 times x times y .
When there is no number in front of a
variable, the coefficient is 1.
3
45
To simplify an algebraic expression, add
or subtract the coefficients of like terms.
Simplify each algebraic expression.
2) 7 x 2  5 x 2
1) 3x  5x
Answers:
12x 2
8x
5) 7 x  5 x 2 y   2 x  3 x 2 y 
4) 7 x3 y  5 x3 y
Answer:
7 x  5 x 2 y  2 x  3x 2 y
2x3 y
Answer: 7 x  2 x  5 x 2 y  3 x 2 y
3) 9x  x
10x
Take note of like terms.
Distributive Property
Combine like terms.
Add/subtract coefficients.
5x  2x2 y
6) 8 x 2  7 ax 2  9  3 y 2  1  ( 2ax 2 )
8 x 2  7ax 2  2ax 2  9  1  3 y 2
Answer:
8 x  9ax  8  3 y
2
2
2
Combine these two signs to become subtract (  ).
Combine like terms and write down the other terms.
Add and subtract coefficients of only the like terms.
Simplify each algebraic expression.
a) 3mn 2  9mn 2
b) 5 p 2  2 p  8  4 p 2  5 p  (6)
c) 8x2  xy  15x 2  7 xy
d) 4a 2  a  a 2
46
e) 3an 2  an 2
f) 11 p 2  2 px  8  4 p 2    5 p – 6 px  2 
g) 3c2  4c2  (6cd )  5  2cd  c
h) p 2  p 2  p 2  p  p
i) x2  y 2  2 x2  2 y  y 2  3x 2
j)
k)
 8 yz
2
 14yz    yz  2 yz 2 
 y  x  z   y  x  z
l) 5.2 p 2  8  4.5 p 2 – 6 p  3
m) 2 xy  (2 xy)  3 y 2  3 y 2  4 xy
n)
o) 13x 5 y 4  2 x 3 y 2   10 x 3 y 2  9 x 5 y 4  7 
p)
 2m n  8.2mn    9m n  7.9mn  2 
2
2
5m n  2mn  6m    9m n  mn 
2
2
2
47
Evaluating Algebraic Expressions
 An algebraic expression usually contains one or more variables.
 When we assign number values to these variables, we can evaluate the
algebraic expression.
 In order to evaluate an algebraic expression you need to use the Order of
Operations - BEDMAS (studied earlier).
1. Evaluate the algebraic expression 3n  5 for n  2 .
Solution: Substitute the value 2 for
3n  5
3(2)  5
65
Answer:
n.
( 3n means 3 multiplied by n)
Multiply first.
Then add.
11
2. Evaluate the algebraic expression 2 x 2  5 x  4 for x  3 .
Solution: Substitute the value 3 for x .
2 x2  5x  4
Answer:
(3)2  (3)(3)  9
On your calculator, type ( 3) and
2
press x . Be sure to use brackets.
Note:
2(3)2  5(3)  4
Do the exponent first, to give 9.
2(9)  5(3)  4
Multiply next.
18  15  4
Add and subtract last.
29
48
1. Evaluate each algebraic expression for x  3, y  2, and z  1 .
a) 3x  5 y  4 z
b) x 2  8 x  5 y
c) x 2  5 y 2
d) x 2  y 2  z 2
2. Evaluate each algebraic expression for a  3.4 and b  2.6 . Round answers to one decimal
place.
a) 3a  2b
b) 2a 2  3b2
c) a 2  b2
d) b2  a 2
e) 4a 2  2b
f) 6b  3a 3  2b 2
49
Unit 1 – Algebra I
Topic 6: Linear Equations
1.5
Linear equations and formulas
1.5.1 Solve a linear equation in one unknown
1.5.2 Solve literal equations and rearrange formulas
Solving Linear Equations
 When a number or variable is added or subtracted to one side of an equation, it
can be ‘moved’ to the other side by changing its sign
 Move all the terms that contain a variable to one side and those that contain a
number only to the other side
1)
2)
x  9   15
x   15  9
x  6
11x  16  10 x  7
11x  10 x   7  16
x   23
on the left
side of the
equal sign
becomes
on
the right side.
on the right
becomes
on the left.
on the left
becomes
on
the right.
 The Division Property states that for any numbers a, b, and c, if a = b, then
a b

c c
 In other words, for any equation, if both sides of the equation are divided by the
same real number, then both sides will still be equal.
Solve each equation.
1)
6 x  4  2 x  24
6 x  2 x  24  4
4 x  20
4 x 20

4
4
x  5
Move
to the
left and
to
the right and
change their
signs.
Divide each
side by 4 to
complete.
2)
5  7 x  12  29  2 x  8
7 x  2 x  29  8  5  12
5 x  28
Move variables
to the left and
numbers to the
right.
Divide by 5 to
5 x 28

complete.
5
5
28
3
x
 5 or 5.6
5
5
50
Solve each equation.
a) 4 x  4  3x  6
b) 2 x  9  x  20
c) 15 x  2  5  14 x
d) 12  3x   15  2 x
e) 5 x  4  3x  12
f) 6 x  3  5 x  11  8 x  25
g) 2 x  6  3x  11  24  12 x  2
h) 9 x  48  5 x   7 x  70
51
i) 7 x  2  3x  5  8 x  13
j) 8.3x  7.4  2 x  2  4.2 x  5.8
For some equations, it is easier to solve them by leaving variables
on the right side and placing numbers on the left.
Example:
Since 2x is already on the right side, leave it
there and move the numbers to the left.
Switch around at the end to get x = 37.
k) 56  7 x  14
l) 72  9 x  1
m) 9  12  6 x  15
n) 22  14  5 x  40
52
Solving Linear Equations with Brackets
 Some equations will contain brackets. To solve these equations, the Distributive
Property (Lesson 1). Remember how the Distributive Property works:
4( x  8)  4x  32
Multiply 4 by x and
. Be
careful of the signs!!
Multiply –3 by x and
.
3( x  9)   3x  27
Solve each equation.
1)
5( x  3)  35
5 x  15  35
2)
First, remove the
brackets.
5 x 50

5
5
x  10
5 x  6 x  8  5 x  7 x  10
5 x  6 x  5 x  7 x  10  8
5 x  35  15
5 x  50
5 x  2(3 x  4)  5 x  (7 x  10)
Then solve the
equation.
x2
is the same
as
Multiply each term inside
the bracket by
Solve each equation.
a) 3( x  5)  2 x  8
b) 3x  2( x  5)  4( x  2)
53
c) 7 x  6  8  (6 x  4)
d) 5(4 x  2)   3(8 x  4)  3x
e) 3(7 x  2)  5(4 x  1)  17
f) 6 x  ( x  2)  9 x  14
g) 2( x  3)  3x   4(3x  6)  2
h) 3( x  1)  4( x  2)  2
54
Solving Equations with Fractions
 To solve an equation with fractions, change it into an equation without fractions.
 Multiply each term of the equation by the Lowest Common Denominator (LCD)
Solve each equation.
1)
2)
x
5
x
5
5
x
5
5
x
x
5
 6
The denominator is 5.
Multiply each term by 5.
 65
 65
 30

2x

3
The denominators are 5 and 3. LCM (5, 3) = 15
2
1

3
5
 x
 2x 
 2
1
15    15    15    15  
5
 3 
 3
5
5
3
5
Multiply each term by 15.
3
x
 2x 
2
1
15    15    15    15  
5
 3 
3
5
3 x  10 x  10  3
7 x  13
7 x
13

7
7
x 
13
6
or  1
7
7
;
55
3)
1 x
2x  3
3


4
2
4
The denominators are 2 and 4. LCM (2,4) = 4
 1 x 
 2x  3 
3
4
  4
  4 
 4 
 2 
 4
1
2
 1 x 
4
 
 4 
 2x  3 
4
 
 2 
Multiply each term by 4.
1
3
4 
4
1(1  x)  2(2 x  3)  1(3)
Watch the signs
carefully!!
1  x  4x  6  3
–2 is outside the
bracket.
5 x  5  3
5 x  3  5
5 x  8
5 x
8

5
5
8
3
x   or  1 or  1.375
5
8
;
Note: In the steps above,
expanded to give
is
.
=
(Distributive property)
Solve each equation.
a)
3x
6
7
b)
2x
 8
3
56
c)
2
x  4x  2
5
d)
e)
8x
6
2

 2x 
5
5
5
f)
g)
x5
x4

2
3
h)
5x
 4  14
7
3x
5x 1
4 

5
4
2
3y
5

5
2

1
y

2
5
57
i)
x
2x
8 
3
7
9
k) 7 
x 5
3
3x  5


2
2
2
j)
6x  1
2x

 3
5
3
l)
x  1 2x  1
x5


2
3
4
58
Literal Equations
To solve an equation or formula for a given variable, follow these guidelines:
1)
If there are fractions, multiply both sides of the equation by the LCD.
2)
Using addition or subtraction, to move all terms involving the variable that you are solving
for are on one side of the equation and all other terms are on the other side.
3.
Divide both sides of the equation by the coefficient of the variable that you are solving for,
leaving only the required variable on one side of the equation.
Solve each equation for the indicated variable.
9
C  32, for C
5
1) I = Prt, for r
2) F 
Solution:
Solution:
Switch the equation around to put r
on the left side.
Switch the equation around to
put C on the left side.
9
C  32  F
5
Divide each side by Pt to get “r” by itself.
Prt  I
5
Prt
I

Pt
Pt
9
C  5  32  5  F
5
9C  160  5 F
Pr t
I

Pt
Pt
r 
I
Pt
9C  5 F  160
9C
5 F  160

9
9
C 
5 F  160
9
or C 
5F
160

9
9
Either answer is acceptable.
Solve each equation for the indicated variable.
a) I = Prt, for t
b) P = 4s, for s
59
c) V = lwh, for w
e) A
 B  C = 180, for B
g) A = P + Prt, for r
i) y 
k) S =
3
xz , for z
4
1 2
gt , for g
2
d) P = 2l  2w, for l
f) y = mx  b , for m
h) 2 x  3 y  6, for y
j) V =
1 2
 r h, for h
3
l) y =
2
x  8 , for x
3
60
Unit 1 – Algebra I
Topic 7: Word Problems – 1 Variable
1.6 Word Problems
1.6.1
Translate English phrases into algebraic expressions
1.6.2
Solve word problems involving one unknown
1.6.3
Express several unknowns in terms of one variable
1.6.4
Solve word problems involving several unknowns
Algebraic Expressions


A variable is a letter which is used to represent an unknown number or quantity.
The letters x, y, and z are commonly used to represent variables, but any letters can be
used.
If we don’t know the distance from Qatar to Montreal, we could say that the distance is x.
If we don’t know Aisha’s age, we could say that her age is x.
If Fatima is 6 years older than Aya (whose age is x), we say that Fatima’s age is x + 6.
In each of these examples, “x” is a variable.
An algebraic expression is a number or variable, or a combination of numbers and variables,
connected by arithmetic operators.
Examples of algebraic expressions:
x7
5y 8
2x
3x 2  5 x  2
Words that mean Addition
Examples of written phrases
Add
A number added to 6
Sum
The sum of a number and 6
Plus
A number plus 6
More than
6 more than a number
Increased by
A number increased by 6
Older than
6 years older than a person’s age
4x  3 y3  6
Algebraic Expression
x6
61
Words that mean Subtraction
Examples of written phrases
Subtract
A number subtract 8
Minus
A number minus 8
Difference
Decreased by
The difference between a
number and 8
A number decreased by 8
Less than
8 less than a number
Younger than
8 years younger than a
person’s age
Algebraic Expression
x 8
Note: "Less than" “younger than” and “subtracted from” appear
backwards in the algebraic expression from how it’s
written in words!
Words that mean Multiplication
Examples
Multiplied by
A number multiplied by 5
5x
Times
3 times a number
3x
Twice (two times, doubled)
Twice a number
2x
Product (the answer when
numbers are multiplied)
Of (used for fractions)
The product of 7 and a
number
One-half of a number
Words that mean Division
Examples
Divided by
A number divided by 15
Divided into
A number divided into 15
Quotient (the answer when
numbers are divided)
The quotient of a number and 9
Algebraic Expression
7x
1
x
2
Algebraic Expression
x
15
15
a
y
9
Any letter or variable can be used to
represent an unknown number.
62
Write an algebraic expression for each.
Answers
1)
4 more than 6 times a number
1) 6x  4
2)
8 less than 3 times a number
2) 3x  8
3)
18 divided by the product of 4 and a number
3)
4)
6 years older than twice Jane’s age (x)
4) 2x  6
5)
The sum of c and d, divided by the product of 6 and y
5)
6)
9 times the sum of x and y
6) 9( x  y)
18
4x
cd
6y
Note: Brackets are needed to answer #6.
Do you need brackets for: “the sum of 9 times x
and y” ?
1. Write an algebraic expression for each. Use
a) The sum of
x
and 8
x
to represent "a number".
a) ___________________
b) The product of 4 and
y
b) ___________________
c) The quotient of 7 and
b
c) ___________________
d) The product of 5 and
y , increased by 8
d) ___________________
e) Three times the sum of
x
and 4
f) 7 more than three times a number
e) ___________________
f) ___________________
63
g) The product of
h) The sum of
r
r
s
and
and
, decreased by 2
s , divided by 5
g) ___________________
h) ___________________
i) Three times a number, increased by 16
i) ___________________
j) The product of 9 and a number, decreased by six
j) ____________________
k) Twice a number, decreased by twenty-nine
k) ___________________
l) The quotient of 7 times a number, and 3
l) ____________________
m) 15 subtracted from one-half a number
m) ___________________
n) The quotient of a number and 9
n)____________________
o) The sum of 3 times
p) The product of
x
and 4 times
y and z
y
subtracted from 10
q) Three times a number, minus 8
r) Twice the product of
x and y , decreased by 7
s) 7 less than one-third number, divided by 3
t) 9 times
x , subtracted from y
o) ___________________
p) ___________________
q) ___________________
r) ___________________
s) ___________________
t) ___________________
u) 8 more than three times a number
u) ___________________
v) 15 less than twice a number
v) ___________________
64
Word Problems – Number Statements
1)
Word Problem:
Solution:
Let
6 more than a number is 17. What is the number?
x
be the number.
(Step 1 – Identify the variables.)
Translate the words of the problem into an equation.
6 more than a number
is
x  6
The equation is:
Solving, we get:
x  6  17
x  17  6
x  11

17.
17
The word “is”
becomes “=”
in the
equation.
(Step 2 – Write the equation.)
(Step 3 – Solve the equation.)
The number is 11.
--------------------------------------------------------------------------------------Check the answer by substituting “11” for x in the equation to see if both
sides of the equation are equal.
(Step 4 – Check your solution.)
x  6  17
11  6  17
17  17
This is true. Thus, 11 is the correct answer.
--------------------------------------------------------------------------------------Another way to check. Go back to the problem itself and substitute “11” for “a
number”.
“6 more than a number is 17”.
“6 more than 11 is 17”. This statement is true. 11 is the correct answer.
65
2)
Word Problem:
4 less than twice a number is 36. What is the number?
Solution:
Let
x
be the number.
(Step 1 – Identify the variables.)
Translate the words of the problem into an equation.
4 less than twice a number is 36.
2x  4
Recall: "Less than"
appears backwards in the
algebraic expression!
The equation is:
Solving we get:

36
(Step 2 – Write the equation.)
(Step 3 – Solve the equation.)
2x  4
2x
2x
2x
2
x
 36
 36  4
 40
40

2
 20
The number is 20.
--------------------------------------------------------------------------------------------------------------
Check the answer in the equation: (Step 4 – Check your solution.)
2 x  4  36
2(20)  4  36
40  4  36
36  36
OR
This is true. Thus, 20 is the correct answer.
--------------------------------------------------------------------------------------------Check the answer by reading it back in the problem
“4 less than twice 20 is 36.”
Since twice 20 is 40, it is true that 4 less than 40 is 36.
20 is the correct answer.
66
Solve each word problem using an equation.
a)
10 less than a number is 6. What is the number?
b)
5 more than twice a number is 35. What is the number?
c)
Three times a number decreased by 10 is equal to 23. What is
the number?
67
d)
10 less than four times a number is 22. What is the number?
e)
4 less than one-half a number is 5. What is the number?
f)
6 less than one-third a number is equal to 8 more than the number.
What is the number?
68
Word Problems – Consecutive Integers
Consecutive Integers follow in order, with each integer being 1
more than the previous one.
Examples:
Two consecutive integers
7, 8
Three consecutive integers 24, 25, 26
Four consecutive integers
7, 8, 9, 10
Using algebra, we will call the first integer x .
The second integer is 1 more than the first. We will call it
x 1.
The third integer is 1 more than the second. We will call it x  2.
Consecutive Even Integers follow in order, with each integer being
2 more than the previous one.
Examples
Two consecutive even integers:
6, 8
Three consecutive even integers: 26, 28, 30
Four consecutive even integers:
14, 16, 18, 20
x.
Using algebra, we will call the first even integer
x  2.
The third integer is 2 more than the second. We will call it x  4.
The second integer is 2 more than the first. We will call it
Consecutive Odd Integers follow in order, with each integer being
2 more than the previous one.
Examples
Two consecutive odd integers:
11, 13
Three consecutive odd integers:
5, 7, 9
Four consecutive odd integers:
13, 15, 17, 19
Using algebra, we will call the first odd integer
x.
x  2.
The third integer is 2 more than the second. We will call it x  4.
The second integer is 2 more than the first. We will call it
69
1)
The sum of three consecutive integers is 87. Find the integers.
Solution: Take note we are looking for three integers.
Let the integers be x, x  1, and x  2.
x   x  1   x  2   87
(Step 1 – Identify the variables.)
(Step 2 – Write the equation.)
3x  3  87
3x  87  3
3x  84
3x
84

3
3
x  28
x  1  29
x  2  30
(Step 3 – Solve the equation.)
State your final answer !!
The numbers are 28, 29, and 30.
------------------------------------------Check: Are the numbers 28, 29, and 30 consecutive integers? Yes
Is their sum 87? Yes. (28 + 29 + 30 = 87) (Step 4 – Check your solution.)
2)
The sum of two consecutive even integers is 98. Find the integers.
Solution:
Take note we are looking for two integers.
Let the integers be x and x  2.
x   x  2   98
(Step 1 – Identify the variables.)
(Step 2 – Write the equation.)
2 x  2  98
2x
2x
2x
2
x
x2
 98  2
 96
96

2
 48
 50
(Step 3 – Solve the equation.)
The numbers are 48 and 50.
------------------------------------------Check: Are the numbers 48 and 50 consecutive even integers? Yes
Is their sum 98? Yes. (48 + 50 = 98)
(Step 4 – Check your solution.)
70
3)
The sum of three consecutive odd integers is 177. Find the integers.
Solution: Take note we are looking for three integers.
Let the integers be x , x  2, and x  4
x   x  2   ( x  4)  177
(Step 1 – Identify the variables.)
(Step 2 – Write the equation.)
3 x  6  177
3 x  177  6
3 x  171
3x
171

3
3
x  57
x  2  59
x  4  61
(Step 3 – Solve the equation.)
The numbers are 57, 59 and 61.
------------------------------------------Check: Are the numbers 57, 59, and 61 consecutive odd integers? Yes
Is their sum 177? Yes. (57 + 59 + 61 = 177) (Step 4 – Check your solution.)
Solve each word problem using an equation. State your final answer.
a)
The sum of two consecutive integers is 37. Find the integers.
71
b)
The sum of three consecutive even integers is 90. Find the integers.
c)
The sum of three consecutive odd integers is 153. Find the integers.
d)
The sum of three consecutive integers is 6. Find the integers.
72
e)
Find two consecutive even integers such that the smaller added to
four times the larger gives a sum of 28.
f)
The sum of four consecutive odd integers is 304. Find the integers.
g)
The first of two consecutive integers is 33 less than twice the
Find the two integers.
second.
73
Word Problems – Age
Aisha’s mother is 2 years older than twice Aisha’s age. The sum of their ages is 74. Find Aisha’s
age and her mother’s age.
Solution:
Let
x
be Aisha’s age.
2x  2
(Step 1 – Identify the variables.)
is the mother’s age.
Aisha + Mother = 74
(Step 2 – Write the equation.)
x 
(2 x  2)  74
3 x  2  74
3 x  74  2
3 x  72
(Step 3 – Solve the equation.)
3x
72

3
3
x  24
Aisha’s age is 24. Her mother’s age is
State the final answer !!
Aisha is 24.
Her mother is 50.
2x  2 .
2( 24)  2
48  2
50
----------------------------------------------------------------------------------------------------(Step 4 – Check your solution.)
Check:
Is the sum of their ages 74?
Yes, 24 + 50 = 74
Is Aisha’s mother 2 years older than twice Aisha?
Yes. Twice Aisha’s age is 48. Two years more than 48 is 50.
74
Solve each word problem using an equation. State your final answer.
a) Jane’s age is 2 years less than her sister’s age. The sum of their ages is 16.
Find the age of each person.
b) Maria’s mother is 2 years older than three times Maria’s age. The sum of
their ages is 50. Find Maria’s age and her mother’s age.
c) Hassan’s father is 6 years more than 3 times Hassan’s age. The sum of
their ages is 66. Find the age of each.
75
d) Millie’s age is 2 years less than one-half of Sara’s age. If the sum of
their ages is 58, find the age of each person
e) Ali is twice as old as his sister Aya. If the sum of their ages is 27 years,
how old is Ali and his sister?
f) Laila’s age is 5 years less than 2 times her brother’s age. The sum of
their ages is 40 years. What are their ages?
76
Word Problems - Geometry
Length
Rectangle
Width
The values for the length and
the width of a rectangle are
called its dimensions.
The length of a rectangle is 1 cm less than 3 times the width. The perimeter is 54 cm. What are
the dimensions of the rectangle?
(Step 1 – Identify the variables.)
Solution:
3x  1 ( L)
Let x represent the width.
x
3 x  1 is the length.
x
(W )
(W )
3x  1 ( L)
The perimeter is the sum of all sides.
P  W  W  L L
P
 2W

2L
54  2x  2 3x 1
(Step 2 – Write the equation.)
(Switch the equation around)
2 x  2  3 x  1  54
2 x  6 x  2  54
8 x  54  2
8 x  56
(Step 3 – Solve the equation.)
8x
56

8
8
x  7
The width is 7 . The length is 3 x  1 .
3(7)  1
21  1
width = 7 cm; length = 20 cm
20
--------------------------------------------------------------------------------------------------------(Step 4 – Check your solution.)
Check:
Is the length 1 less than 3 times the width?
Yes. 20 is 1 less than 3 times 7.
Is the Perimeter 54? Yes. 2(7) + 2(20) = 14 + 40 = 54
77
Use an equation to solve each word problem. State your final answer.
a) The length of a rectangle is 1 cm more than twice the width. If the perimeter of
the rectangle is 74 cm, find the dimensions (length and the width) of the
rectangle.
b) The length of a rectangular playing field is 5 metres less than twice the width.
The perimeter of the playing field is 230 metres. Find the dimensions of the
field.
78
c) The length of a garden, shaped like a rectangle, is 2 m more than three times
its width. The perimeter of the rectangle is 108 m. Find its dimensions.
d) The length of a sheet of newspaper, shaped like a rectangle, is 14 cm less
than three times its width. The perimeter of the sheet is 212 cm. Find its length
and width.
79
e) The length of a rectangular book cover is 13 cm more than one-half the width.
If the perimeter is 68 cm, find its dimensions.
f) The length and width of a rectangle are two consecutive even integers.
The perimeter is 1006 m more than the width. What are the dimensions?
80
Unit 1 – Algebra I
Topic 8: Graphing Linear Equations
5.1
Graphing linear equations
5.1.1
Graph linear equations by finding the coordinates of two points
5.1.2
Find the x and y intercepts of a straight line graph
5.1.3
Find the slope of a straight line graph
5.1.4
Interpret the intercepts and slope of a given straight line graph
5.1.5
Find the equation corresponding to a given straight line graph
Graphing A Line
• A line is determined by 2 points.
• If we have the coordinates of 2 points, we can draw the line through them.
• The points may be given in 2 forms:
o ordered pairs
o table of values
Graph the line through the points (4, −3) and (−2,6) .
(−2,6)
(4, −3)
81
Graph the line through the points
x
y .
−4 −2 2
3 5
Slope and y-intercept
• The y-intercept of a line is the point of intersection of the line and the
y-axis.
• To graph a line, we must always start with some point that we know is on
the line. This point may be the y-intercept.
• Once we have one point plotted, we use the slope in the rise over run form
to locate another point on the line.
• Remember, we only need two points to draw a line.
You may use a 3rd point to check your work.
82
Graph the line with slope
2
and y-intercept ( 0,2 ) .
3
Step 1: Locate 2 on the y-axis. This is your starting point.
Step 2: Use slope
=
rise 2
to locate a second point.
=
run 3
Step 3: 2 up and 3 to the right
Step 4: Draw the line through the 2 points.
Graph the line with slope −
( 0,2 )
4
and y-intercept ( 0, − 3) .
3
83
Slope and any point
• To graph a line, start with any point that you know is on the line.
• Once you have one point plotted, use the slope in the rise over run form to
locate another point on the line.
• Remember, you only need two points to draw a line.
You may use a 3rd point to check your work.
Graph the line with slope −
4
and through the point ( 2,5 ) .
3
Step 1: Locate the point ( 2,5 ) . This is your starting point.
rise
4
Step 2: Use slope =
= − to locate a second point.
3
run
Step 3: 4 down and 3 to the right
or
(4 up and 3 to the left)
( 2,5)
Step 4: Draw the line through the 2 points
Graph the line with slope
5
6
and through the point ( 3,0)
84
1. Graph the line through the points (−2, −3) and (4,6) . Label the line, A.
2. Graph the line through the points (−7,3) and (6, −2) . Label the line, B.
3. Graph the line with slope −
4. Graph the line with slope
2
and y-intercept ( 0,3) . Label the line, A.
5
3
and y-intercept ( 0, − 4 ) . Label the line, B.
2
85
5. Graph the line with slope 3 and through the point ( −4,5 ) . Label the line, A.
6. Graph the line with slope −
1
and through the point ( 3, −5 ). Label the line, B.
3
86
Two Intercept Method
x-intercept
• The point where the line intersects the x-axis
• Where y = 0
• In this graph, the x-intercept is located at ( 2,0 )
y-intercept
• The point where the line intersects the y-axis
• Where x = 0
• In this graph, the y-intercept is located at ( 0,2 )
12 using the Two Intercept Method.
Graph 2 x + 3 y =
Solution
Step 1: Find the x-intercept.
Let y = 0
2x + 3y =
12
2 x − 3( 0 ) =
12
2x
Step 2: Find the y-intercept.
Let x = 0
2x + 3y =
12
2 ( 0 ) + 3 y = 12
= 12
x =6
3 y = 12
y=4
x
6
0
y
0
4
(0,4)
Step 3: Plot the points.
Step 4: Draw the line through
the 2 points.
(6,0)
87
Graph x + 2y = 6 using the Two Intercept Method.
−12 using the Two Intercept Method.
Graph 4 x − 3 y =
88
Finding the Equation of a Line
Slope-intercept form
=
y mx + b
Where m is the slope and b is the y-intercept.
Find the equation of a line with slope 2 and y-intercept 3 .
We have m = 2 and b = 3 .
Substitute into the slope-intercept form
=
y mx + b
=
y 2x + 3
Find the equation of a line with slope 5 and y-intercept 2 .
Slope
y-intercept
Equation
89
Find the equation of a line with slope 2 , and passing through the point with
coordinates ( 3,4 ) .
In order to find the equation, we need to find the y-intercept ( b ).
y mx + b , with the given values of
To do that we use the slope-intercept form,=
x and y , ( 3,4 ) .
=
y mx + b
4 2 ( 3) + b
=
4 6 +b
=
4−6=
b
−2 =b
Now we use slope-intercept form again to write the equation.
Find the equation of a line with slope
coordinates ( −6,5 ) .
Slope
y mx + b
=
y 2x − 2
=
1
, and passing through the point with
2
y-intercept
Equation
90
Find the equation of a line through the points with coordinates (1, −1) and ( 3,5 ) .
This example is the same as the previous example with an extra step.
We must first find the slope of the line.
y − y1
.
To do that we use the slope formula, m = 2
x2 − x1
Slope
(1, −1) and ( 3, 5)
( x1, y1 ) and ( x2 , y2 )
m=
y2 − y1
x2 − x1
−
5− 1
3 −1
6
m=
2
m=3
m=
y-intercept
Equation
Use slope-intercept form
to find the y-intercept.
Use slope-intercept form
to write the equation.
=
y mx + b
=
y mx + b
=
5 3 ( 3) + b
=
y 3x − 4
=
5 9 +b
5−9 =
b
−4 =b
Find the equation of a line through the points with coordinates (1, 2 ) and ( 3, −2 ) .
Slope
y-intercept
Equation
91
Find the equation of the line in the graph.
Slope
The slope can be found
by using:
y-intercept
Equation
From the graph we know
the y-intercept is 3.
Use slope-intercept form
to write the equation.
rise −3 3
= =
run 4 −4
slop=e
m=−
b=3
=
y mx + b
3
=
y − x+3
4
3
4
Find the equation of the line shown in the graph.
Slope
y-intercept
Equation
92
1. Find the equation of the line that has slope 3 and y-intercept −5 .
2. Find the equation of the line that has a slope of −
5
and a y-intercept of 3 .
4
3. Determine the equation of the line that has slope
3
and passes through the
7
point (14, − 3) .
4. Find the equation of the line that passes through ( −5,6 ) and has a slope of 4 .
5. Determine the equation of the line which passes through points
( −3, − 4 ) and (8,2 ) .
93
6. Find the equation of the line which passes through ( −5,0 ) and ( 6, −2 ) .
7. For each graph, state the x-intercept, y-intercept, and the equation of the line.
a)
b)
c)
d)
94
Unit 1 – Algebra I
Topic 9: Systems of Linear Equations – 2 Variables
5.2
Solve a system of linear equations in two variables using the following
methods:
5.2.1 Graphing
5.2.2 Substitution
5.2.3 Addition-subtraction (Elimination)
5.2.4 Determinants (Cramer’s Rule)
• To solve a system of linear equations means to find the coordinates of all the
points that satisfy both equations
• We will study four methods for solving linear systems involving two variables:
o Graphing
o Substitution Method
o Multiplication/Addition Method (also called Elimination)
o Determinants (Cramer’s Rule)
Graphing
Solve the system of equations by graphing.
2
{4x x−+y2=y−=
16
Recall from topic 7, for these equations, we can use the two intercept method.
x− y=
−2
4x + 2 y =
16
x y
x y
How can you
4 0
−2 0
check your
0 8
0 2
answer?
( 2,4 )
=
and y 4 .
The solution to the system is ( 2,4 ) . It can also be written
as x 2=
95
Solve the system of equations by graphing.
9
{4−2x x−+3 yy =
=
−5
For these equations, we will use the slope-intercept method.
4x − 3y =
9
− 3y =
−4 x + 9
−3 y −4 x 9
=
+
−3
−3 −3
4
y = x−3
3
4
So m1 = and b1 = −3
3
−2 x + y =
−5
=
y 2x − 5
So m2 = 2 and b2 = −5
( 3,1)
So the solution is ( 3,1) or x = 3 and y = 1 .
Solve each system by graphing. Check your solutions.
1.
4
{3xx+−y2=
y=
2
96
2.
3.
0
{22xx −+ 33yy =
12
=
−2
{2x x−+y y==
2
97
4.
10
=
{3xx−−2yy =
0
5.
=
2
{24xx +− 33yy =
4
98
Substitution
• Solving a system by the Substitution Method involves replacing a variable in
one equation with an expression from another.
Solve the system by the Substitution Method.
8
{=3yx +22xy−=
3
To solve this system by the Substitution Method, we note that the second equation
is already solved for y . Since y = 2 x − 3 , we can substitute 2 x − 3 for y in the
first equation and solve for x .
3x + 2 y =
8
3 x + 2 ( 2 x − 3) =
8
3x + 4 x − 6 =
8
7 x= 8 + 6
7 x = 14
x=2
To find y , we substitute 2 for x in either equation of the system.
y 2 x − 3 is already solved for y , it is easier to substitute into this equation.
Since =
=
y 2x − 3
=
y 2 ( 2) − 3
y= 4 − 3
y =1
So the solution is ( 2,1) .
Solve the following system by the Substitution Method.
x + 2y =
−10
=
y 2x + 5
{
99
Solve the following system by the Substitution Method.
10
{32xx −+ 23yy =
=
−2
Solution
We need to solve one of the equations for one of the variables.
Since no term has a coefficient of 1 , let’s solve the 2nd equation for x .
2x + 3y =
−2
2x =
−3 y − 2
2 x −3 y 2
=
−
2
2
2
3
x=
− y −1
2
3
We now substitute − y − 1 for x in the 1st equation and solve for y .
2
3x − 2 y =
10
 3

3  − y − 1 − 2 y =
10
 2

9
− y − 3 − 2y =
10
Multiplied by 2 to eliminate the fraction.
2
− 9y − 6 − 4y =
20
−13 y =20 + 6
−13 y =
26
y = −2
To find x , we substitute −2 for y in either equation of the system.
3
− y − 1 is already solved for x , it is easier to substitute into this equation.
Since x =
2
3
− y −1
x=
2
3
x=
− ( −2 ) − 1
2
So the solution is ( 2, −2 ) .
Would this have been easier
with a different method?
x= 3 − 1
x=2
100
Solve each system of equations using the Substitution Method.
1.
{2xx−+y y==4−4
2.
2
{4x x−+y2=y−=
16
3.
7
{3xx++y4=
24
y=
4.
−20
{64xx −+ 22 yy =
=
−10
101
5.
{3xx−+yy==2−14
6.
7.
1
{3xx−+y2=
−12
y=
−5
8. 5 x − 2 y =
3x + y =
−14
=
10
{3xx−−2yy =
0
{
102
Multiplication/Addition Method (Elimination)
• To solve a system of linear equations by the Multiplication/Addition Method,
we eliminate one of the variables, x or y . This method can also be called
Elimination.
Solve the system using by the Multiplication/Addition Method.
=
23
{34xx −+ 55yy =
−9
The coefficients of y have opposite signs ( -5 and +5).
We combine the two equations using addition.
4x + 5 y =
23
+ 3x − 5 y =
−9
7x
= 14
x = 2
’s have been eliminated.
Substitute into the 1st equation
4x + 5 y =
23
4 ( 2) + 5 y =
23
8 + 5y =
23
5=
y 23 − 8
5 y = 15
y =3
Could we have substituted into
the 2nd equation?
So,the solution is ( 2, 3) .
Solve the following system by the Multiplication/Addition Method.
2x + 3y =
4
3x − 3 y =
6
{
103
Solve the system by the Multiplication/Addition Method.
30
10 → × 3 → 6 x − 15 y =
2 x − 5 y =
3 x − 2 y =
−7 →× ( −2 ) → −6 x + 4 y =
14

− 11 y =
44
y=−4
10
=
{32xx −− 25 yy =
−7
Here we have to multiply
both equations to make
opposite coefficients for one
of the variables.
Substitute into the 1st equation.
2x − 5 y =
10
2 x − 5 ( −4 ) =
10
10
2 x + 20 =
= 10 − 20
2x
2 x = − 10
x = −5
How could we have eliminated
the y’s?
So the solution is ( −5, −4 ) .
Solve the following system by the Multiplication/Addition Method.
3x + 4 y =
−17
4x − 3y =
−6
{
104
Solve the systems by the Multiplication/Addition Method.(Elimination)
1.
−20
{64xx −+ 22 yy =
=
−10
{2x x−+y y==4−4
2.
{
29
4. 3 x + 4 y =
2x − 5 y =
−19
−2
3. 5 x + 8 y =
4x + 6 y =
−2
{
105
{
5.
2
=
{7xx++86y y=
−4
5
6. 3 x − y =
2x + 3y =
10
7.
8
{xy ++ 2xy==
5
8.
{4(y +x 2+xy=) 10= 42+ 4−yy
106
Determinants
• A matrix is a rectangular array of numbers enclosed by square brackets
• Matrices are usually named using upper case letters.
• Some examples of matrices are:
 5 −2 
B=

3 −7 
A=
[ − 4 3 −6 9 ]
 4 −1 6 
=
C  2 7 −4 


 0 1 −3
 5 −2 
In the following matrix X = 

 3 −7 
• The dimensions of matrix X are 2 rows by 2 columns. The dimensions are
written as 2 × 2 and read as “two by two”. Since the number of rows and the
number of columns are the same, this is also a square matrix.
• Associated with every square matrix is a value called the determinant. This
value for a 2 × 2 matrix is the number that results from the difference between
the products of the numbers in each diagonal of the matrix.
a b 
• If X = 
 represents any 2 × 2 matrix, then the determinant is the result of
c d 
ad – bc. The determinant of X can be represented as det [ X ] or as X .
Notation:
det [ X=
]
a b
= ad − bc
c d
Find the determinant of each matrix:
5 3 
1. G = 

 7 2
=
det G
( 5 )( 2 ) − ( 7 )( 3 )
= 10 − 21
= −11
 6 −4 

 2 −3 
2. H = 
( 6 )( −3 ) − ( 2 )( −4 )
det H=
= −18 − −8
= −10
 −3 x 8 x 
−4 
3. A = 
 −2
det A = ( −3 x )( −4 ) − ( −2 )( 8 x )
= 12 x − −16 x
= 4x
= 28x
107
1. Find the determinant of each matrix:
a)
 2 −3 
A=

 −4 −10 
b)
7 3
B=

 −4 −2
c)
5a −2a 
C=

 −3 2 
d)
 −8 x 5 
D=

 −4 x 3 
To solve systems of linear equations using determinants we use Cramer’s Rule.
c1
a1x + b1y =
c2
a2 x + b2 y =
Cramer’s Rule for solving a 2x2 system: 
c1
c
x= 2
a1
a2
b1
b2
b1
b2
a1 c1
a c2
y= 2
a1 b1
a2 b2
108
Solve each system of equations using Cramer’s Rule.
(a)
1
2x + y =

−11
5 x − 2 y =
109
(b)
3 x - 2 y = 4

2
 x + 3y =
(c)
17
 −6 x + 8 y =

−4
13 x − 2y =
110
Unit 1 – Algebra I
Topic 10: Systems of Linear Equations – 3 Variables
5.3 Solve a system of linear equations in three variables by determinants
Solving by Determinants – Option 1 (Expansion)
 The determinant of a 3  3 matrix is defined in terms of the determinants of
minors
 The determinants following a1, a2, and a3 are the minors for a1, a2, and a3,
respectively
 Writing the determinant of a 3  3 matrix in terms of minors is called expansion
by minors
 In the definition below, we expanded by minors about the first column
 However, you can expand using expansion by minors about any row or
column using the sign array that follows:



 






 
Find the determinant of the matrix.
1
 2

 0
1
3
5
2
4
6  1
0
7
9
4
6
7 9
  2  
3
4
7
3
5
7
9
5 
6

9 
0
3 5
4
6
 1  36   42   2   27  35   0  18   20 
 1  78  2   8   0
 78  16
 62
111
Solving by Determinants – Option 2 (“Loops”)
 To find the determinant of a 3×3 matrix, copy the first two columns of the
matrix to the right of the original matrix.
 Next, multiply the numbers on the three downward diagonals, and add these
products together.
 Multiply the numbers on the upward diagonals, and add these products
together.
 Then subtract the sum of the products of the upward diagonals from the sum
of the product of the downward diagonals (subtract the second number from
the first number):
 a1
a
 2
a3
b1
b2
b3
c1 
 a1 b1 c1  a1 b1

c2 
 a2 b2 c2  a2 b2



a3 b3 c3  a3 b3
c3 
Find the determinant of the matrix.
3
1
 2 4

 0 7
5 
3
1


6 
 2 4


 0 7
9 
1
 X    2
 0
3
4
7
5 
6

9 
5  1
3

6 2 4

9  0 7
det  X   1 4  9    3  6  0    5  2  7    0  4  5    7  6  1  9  2  3 
 34   96 
 62
112
Find the determinant of the matrix.
(a)
(b)
3
2
8
5
5
0
4
9
6
5
4
1
 2 6
8
7
1
1
7 0 0
(c)
2
4 5
1
4 2
113
Cramer’s Rule (3x3)
 A system of three linear equations in three variables can be solved by using
determinants and Cramer’s Rule
 Note that Dx , D y , and Dz are obtained from D by replacing the x , y , or z  column
with the constants d1, d 2 , and d3
x yz4
Use Cramer’s Rule to solve the system:
x  y  3
x  2y  z  0
 To calculate D , we use the expansion method
1
1
1
1 0
1 0
1 1
D  1 1 0  1 
 1
 1
2 1
1 1
1 2
1 2 1
 1 1  0   1  1  0   1  2   1 
 1 1 3
5
114
 To solve for Dx , D y , and Dz you can expand by minors using any row or
column of your choice for each of the three determinants
Use Cramer’s Rule to solve each system.
x  y  z  6

(a)  x  y  z  2
2x  y  z  7

115
x  y  z  0

(b)  2y  2z  0
 3 x  y  1

(c) Solve for x only using determinants (Cramer’s Rule):
xy 8


 x  2z  0
x  y  z  1

116
(d)
(e)
Solve for y only using determinants (Cramer’s Rule):
2x  3y  z  4

 3 x  z  3
 x  2y  2z  5

Solve for z only using determinants (Cramer’s Rule):
x  y  z  3

x  z  2
 2 x  y  2z  3

117
Unit 1 – Algebra I
Topic 11: Word Problems – 2 & 3 Variables
5.4
Solve word problems involving systems of linear equations in two and three
variables
Problem-Solving Guide






Read the problem several times. Highlight or underline key phrases.
Draw a diagram, chart, or table to help you
Choose different variables to represent the unknowns.
Write equations involving each of the variables.
Solve the system using the most convenient method.
Write the answer using the words of the problem.
Two shirts and one thobe cost 400 QR. Three shirts and two thobes cost 650 QR.
Determine the prices of one shirt and one thobe.
Solution
Let x  the price of one shirt
Let y  the price of one thobe
Two shirts and one thobe cost 400 QR
2x

 400
y
This gives us the system
2 x  y  400

3 x  2 y  650
EQN # 1   2 
 4 x  2 y  800
3 x  2 y  650
x
 150
x 150
Solved using Elimination
Three shirts and two thobes cost 650 QR.
3x

2y
 650
2 x  y  400
 2 150   y  400

Substitute x  150 in EQN #1  300  y  400
y  400  300

y  100

Using the words of the problem.
The price of one shirt is 150 QR.
The price of one thobe is 100 QR.
118
Tickets for a concert cost 40 QR for adults and 30 QR for children. Nine hundred
tickets were sold for a total of 33 000 QR. How many adult tickets were sold and
how many children tickets were sold?
Solution
Tickets for a concert cost 40 QR for adults and 30 QR for children. Nine
hundred tickets were sold for a total of 33 000 QR. How many adult tickets
were sold and how many children tickets were sold?
Adult Children
Total
Number of tickets
x
y
900
Cost of tickets
40 x
30 y
33000
To see how different parts fit together
we can draw a table.
This gives us the system
x  y  900
40 x  30 y  33000
 EQN #1
 30   30 x  30 y  27000
40 x  30 y  33000
10 x
 6000
x  600
x  y

Substitute into EQN #1 600 
y 
 y 

 900
y  900
900  600
300
There were 600 adult tickets sold and 300
children tickets sold.
119
You need a 15% acid solution for a certain test, but your supplier only ships a 10%
solution and a 30% solution. You decide to mix some of the 10% solution with some
of the 30% solution to make your own 15% solution. You need 20 liters of the 15%
acid solution. How many liters of 10% solution and 30% solution should you use?
Solution
Let x represent the number of liters of 10% solution
Let y represent the number of liters of 30% solutions.
For “mixture” problems it is helpful to create a table to organize your thoughts
liters sol'n percent acid
total liters acid
10% sol'n
x
0.10
0.10x
30% sol'n
y
0.30
0.30y
mixture
x + y = 20
0.15
(0.15)(20) = 3.0
Since x + y = 20, then x = 20 – y. Using this, we can substitute for x in our grid, and
eliminate one of the variables: Copyright © Elizabeth Stapel 1999-2011 All Rights Reserved
liters sol'n percent acid Total liters acid
10% sol'n
20 – y
0.10
0.10(20 – y)
30% sol'n
y
0.30
0.30y
mixture
x + y = 20
0.15
(0.15)(20) = 3.0
When the problem is set up like this, you can usually use the last column to write
your equation: The liters of acid from the 10% solution, plus the liters of acid in the
30% solution, add up to the liters of acid in the 15% solution. Then:
0.10  20  y   0.30 y  3.0
2  0.10 y  0.30 y  3.0
2  0.20 y  3
0.20 y  1
y 5
 x  20  5  15
We need 5 L of the 30% acid solution, and 15 L of the 10% acid solution.
120
A mathematical model shown below compares the average summer temperature
change over a ten-year period between three cities London (x), Hong Kong (y), and
Cape Town (z). What was the temperature change of each city?
Solution
121
Solve each word problem using systems of equations.
1.
At a café, three coffees and two teas cost 135 QR. One coffee and three teas
cost 80 QR. What is the price of one tea? What is the price of one coffee?
2.
Adults can buy tickets to a football game for 20 QR. The price for children is
10 QR. If 350 tickets were sold for 4500 QR, then how many adult and
children’s tickets were sold?
122
3.
Nasser purchased 4 mobiles and 2 iPods for 5200 QR. Hassan purchased 2
mobiles and 3 iPods for 6600 QR. Find the cost of 1 mobile and 1 iPod.
4.
Your teacher is giving you a test worth 100 points containing 40 questions.
There are two‐point and four‐point questions on the test. How many of
each type of question are on the test?
123
5.
A 10% chlorine solution is to be mixed with a 25% chlorine solution to obtain
30 litres of 20% solution. How many litres of each must be used?
6.
Jassim wants to make a 100 ml of 5% alcohol solution by mixing a quantity
of a 2% alcohol solution with a 7% alcohol solution. What are the quantities
of each of the two solutions he has to use?
124
7.
A woman invests a total of $20,000 in two accounts. The first account
earns 5% interest and the second account earns 8% interest. If her interest
amount is 1 180 QR, how much did she invest in each account?
8.
In analyzing the forces on the bell-crank mechanism shown in the diagram
below, the following equations are found. What is the value of force A?
A  0.60F  80
B  0.80F  0
6.0 A  10F  0
125
9.
Sara wants to determine the weights of her two dogs: Fogo and Pluto.
However, neither dog will sit on the scale by herself. Sara, Fogo, and Pluto
altogether weigh 175 kg. Sarah and Fogo together weigh 143 kg. Sarah and
Pluto together weigh 139 kg. How much does Fogo weight?
126
Unit 1 Answers – Course Notes
Topic 1: Number Systems
Page 5, Practice:
#1
Number
Natural
Whole
Integer
Rational






 69
a)
b)

9

Irrational
Real
c)
48.676767…


d)
7
10


e)
17


f)
6


g)
8.75

#2
a) Q, R
#3
a) 55,
12
3
b) Q , R
b) 55,
12
3
c) I, Q, R
c) -4, 55,
d) Q, R
12
22 12
d) -4, 55,
, , 3.4
7 3
3
1

e) W, I, Q, R
e) 987.3856813
#4
Number
Natural
Whole
Integer
Rational
a)
8




b)
 8
c)
8
4
e)
6.125
g)
36
h)
i)
j)
Irrational












Real
Imaginary
Complex
a)
9  10i

b)
3i

c)
49
e)
 100
g)
4i
h)
2








2





Page 9, Practice:
Number


5
8
5e
22
7
Real
Natural
Whole
Integer
Rational
Irrational
Real





10
5

7 31
Imaginary
Complex
 4


10i


2e


Page 11, Practice:
Example
Property
 2   3   3   2
Commutative - Addition
Identity - Multiplication
7  1  1 7
6 a  b   6a  6b
Distributive
4x   4x   0   4x   4x
Inverse - Addition
 6  5   2  6   5  2
Associative - Multiplication
 3  7    3  6   39
3  x  3  2  3x  6
or
a) i)
ii)
3  7  6   39
or
3  x  2   3x  6
b) Distributive
3

Topic 2: Order of Operations
Page 12, Practice:
a) 4
b) -12
c) -4
d) -11
e) -9
f) -1
g) 5
h) 1
i) 1
a) -32
b) 32
c) -15
d) 28
e) 8
f) 60
g) 27
h) 36
i) -4
j) 8
k) -6
l) 6
a) 28
b) 41
c) 20
d) 21
e) 24
f) 41
g) 34
h) 5
i) -7
j) 20
k) -8
l) 106
m) -42
n) -26
Page 13, Practice:
Page 15, Practice:
Topic 3: Units of Measurement
Page 19, Practice:
1) 0.046 m
2) 0.0762 dam
3) 0.0046 km
4) 0.0458 dg
6) 457 mg
7) 3870000 ml
8) 957 mL
9) 6.62 L
5) 2cg
Page 22, Practice:
1) 3 L/hr
2) 137.52 km/hr
3) 0.2 m/s
4) 22 m/s
5) 3528000 cm/min2
6) 575000 kg/m3
7) 0.0000745 kg/mm2
8) 18792 km/hr2
9) 0.08595 mg/cm3
10) 129.6 kg/hr2
4
Topic 4: Laws of Exponents
Page 26, Practice:
a) 216
b) -36
c) 36
d) -81
e) 27
g) -49
h) -32
i) -343
a) x10
b) x 9
f) 3p2 x27
k) 30 p 6 r
f) 81
c) x34 y5
d) 40x5 y15
e) 60a 3 x5
g) 10x13 z12
h) 8x 23
i) 90 pr
j) 12ax 2
l) 15x10 y 6
m) xz
n) 18p2 r 2
o) 4axy 6
a) 81
b) x3
c) a 8b
d) x2 y12 z 5
e) 1
g) 3a 2
h)
Page 27, Practice:
Page 30, Practice:
2 p 4
3
i)
4 pq 13 pqr
j)
23
9
k) 3p 4 qr 4
f) 3
l) 10m4 np5
Page 32, Practice:
a) x16
d) 8a12
c) 8m12 n9
b) 729
e) x8
g) 81x16 y8 z12
h) 8a15b12 c 3
i) 4 p 6 q 4 r10 s 2
j) 16x12 y 24 z 20
k)  x28 y 21
l) 16x 4 y 4
Page 33, Extra Practice:
a) 15x 4 y 5
b) 3a8 x12
2qr 5
 w5
f)
g)
9
7
c) 8x9 y5 z 3
h) 125x6 y 9
d) 4x7 y 6 z 2
e) 4a 2b7 c 2
i) 16x8 y12 z16 j) 8a15b12 c 3
5
f)  x12
Page 36, Practice:
a)
16
y4
b)
64 x 6
9
c)
27 x15
8 y 30
d) 
m15 n15
125x6
e)
5
x2
c)
1
125
d) 
1
125
e) 9
f)
i)
2
4 4
x y
j)
x6
9 y3
k) 4a 4b
l)
y5
x3
y
x5
m4 n4
16
f) 
x5 y 5
32
Page 40, Practice:
a)
1
25
b)
g)
125 x3
27
h) 
m)
2qr
3 p3
n) a 3
8
27
4
5x 3
Page 42, Practice:
a) p18
b)
y3
27 x 3
c) 8 p3q3
d)
x4
4
e)
x6
16 y 2
f)
g) 9h 7 k 8
h)
8b 3
a 3c 6
i)
x9
4 y9
j)
x4
4 y14
k)
x6
4 y9
l) 
m) 
27m21n6
8
x15
n) 17
9y
16y 8
x4
o)
p)
16 x 4
y9
x 32
9 y 26
Topic 5: Algebraic Expressions
Page 46, Practice:
a) 6mn2
b) 9 p 2  7 p  2
c) 7 x2  6 xy
d) 5a 2  a
e) 4an 2
f) 7 p2  4 px  5 p  6
g) c 2  4cd  c  5
h) 3 p 2  2 p
i) 2 x 2  2 y
j) 2x  2z
k) 10 yz 2  13 yz
l) 9.7 p 2  6 p  5
m) 6y 2
n) 11m2 n  0.3mn  2
o) 4 x5 y 4  8x3 y 2  7
p) 4m2 n  3mn  6m2
6
Page 49, Practice:
#1
a) 5
b) 25
c) 29
d) 14
#2
a) 5
b) 43.4
c) 4.8
d) -4.8
e) 51.44
b) x  11
c) x  3
d) x  3
e) x  8
f) x  11
h) x  2
i) x  5
j) x  1.7
k) x  10
l) x  
c) x  6
d) x  22
e) x  16
f) x  4
d) x  14
e) x  2
f) x 
j) x  6
k) x  
f) 115.832
Topic 6: Linear Equations
Page 51, Practice:
a) x  10
g) x 
27
7
m) x  3
n) x 
73
9
32
5
Page 53, Practice:
a) x  23
g) x 
16
7
b) x  2
h) x  1
Page 56, Practice:
a) x  14
b) x  12
g) x  7
h) y  
5
2
c) x 
5
9
i) x  63
1
4
90
13
l) x  5
Page 59, Practice:
#4
a) t 
f) m 
j) h 
I
Pr
y b
x
3V
 r2
b) s 
d) l 
P  2w
2
e) B  180  A  C
A P
Pt
h) y 
6  2x
3
i) z 
2S
t2
l) x 
p
4
c) w 
g) r 
k) g 
V
lh
7
3 y  24
2
4y
3x
Topic 7: Word Problems – 1 Variable
Page 63, Practice:
e) 3( x  4)
f) 7  3x
i) 3x 16
j) 9x  6
k) 2x  29
l)
x
9
o) 3 x  4 y
p) 10  yz
q) 3x  8
r) 2( xy )  7
t) y  9 x
u) 3x  8
v) 2x 15
b) x  15
c) x  11
d) x  8
e) x  18
f) x  21
d) 1, 2, 3
e) 4, 6
b) 4 y
g) rs  2
h)
(r  s )
5
n)
m)
x
 15
2
x

  7
3

s) 
3
7
b
d) 5 y  8
a) x  8
c)
7x
3
Page 67, Practice:
a) x  16
Page 71, Practice:
a) 18, 19
b) 28, 30, 32
f) 73, 75, 77, 79
c) 49, 51, 53
g) 31, 32
Page 75, Practice:
a) Jane is 7 years old and her sister is 9.
b) Maria is 12 years old and her mom is 38.
c) Hassan is 15 years old and his dad is 51. d) Sara is 40 years old and Millie is 18.
e) Aya is 9 years old and Ali is 18.
f) Laila is 25 years old and her brother is 15.
Page 78, Practice:
a) Width = 12cm, Length = 25cm
b) Width = 40m, Length = 75m
c) Width = 13m, Length = 41m
d) Width = 30cm, Length = 76cm
e) Width = 14cm, Length = 20cm
f) Width = 334m, Length = 336m
8
Topic 8: Graphing Linear Equations
Page 82, Practice:
Page 83, Practice:
Page 84, Practice:
Page 85, Practice:
#1 and #2
Page 86, Practice:
#5 and #6
Page 85, Practice:
#3 and #4
9
Page 88, Practice #1:
Page 88, Practice #2:
Page 89, Practice:
y  5x  2
Page 90, Practice:
y
Page 91, Practice:
y  2 x  4
Page 92, Practice:
3
y   x 3
2
1
x8
2
Page 93, Practice:
1) y  3x  5
6
26
5) y  x 
11
11
x  int  6
y  int  3
y
1
x3
2
x  int  4
7c) y  int  4
y  x  4
5
2) y   x  3
4
3) y 
2
10
6) y   x 
11
11
x  int  3
7d) y  int  2
2
y  x2
3
10
3
x9
7
4) y  4 x  26
x  int  3
7a) y  int  3
y  x3
7b)
Topic 9: Systems of Linear Equations - 2 Variables
Page 96, Practice:
#1 Solution is  2,2
#2 Solution is  3, 2 
#4 Solution is  2, 6 
#5 Solution is 1,0 
Page 99, Practice:
#3 Solution is  0,2
(-4,-3)
Page 101, Practice:
1)  0, 4 
2)  2,4
3)  4,3 
4)  3,1
5)  3, 5
6)  2, 6 
7)  2, 3
8)  3, 5
Page 103, Practice:
 2,0
Page 104, Practice:
 3, 2
11
Page 105, Practice:
1)  0, 4 
2)  3,1
3)  2,1
4)  3,5 
 4 3
5)  ,  
5 5
 25 20 
6)  , 
 11 11 
7) (2 , 3)
8)  8, 2 
b) -2
c) 4a
d) 4x
Page 108, Practice:
a) -32
Page 109, Practice:
a)  1,3
 16 2 
, 
 11 11 
 1 197 
c)  ,

 46 92 
b) 
Topic 10: Systems of Linear Equations – 3 Variables
Page 113, Practice:
a) -550
b) 122
c) 84
Page 115, Practice:
x 1
a) y  2
z 3
x0
b) y  1
c) x  6
d) y  2
e) z  3
z  1
Topic 11: Word Problems – 2 & 3 Variables
Page 122, Practice:
1.
2.
3.
4.
5.
6.
7.
8.
9.
One coffee costs 35QR and one tea costs 15QR.
100 adult tickets and 250 children tickets were sold.
One mobile costs 300QR and one iPod costs 2000QR.
There are 30 questions worth two points each and 10 questions worth 4 points each.
You need 10L of the 10% solution and 20L of the 25% solution.
Jassim needs 40mL of the 2% solution and 60mL of the 7% solution.
The woman invested $14,000 in the 5% account and $6,000 in the 8% account.
The value of force A is 125N.
Fogo weighs 36kg.
12