Download Course Learning Outcomes for Unit I Reading Assignment Unit

UNIT I STUDY GUIDE
Introduction to Algebra
and Solving Equations
Course Learning Outcomes for Unit I
Upon completion of this unit, students should be able to:
1. Identify real numbers and their properties.
1.1 Apply the distributive law when multiplying a number and an expression.
1.2 Find the absolute value of a number and determine if a number is rational or irrational.
2. Perform mathematical computations involving addition, subtraction, multiplication, division, and
exponents on real numbers.
2.1 Translate phrases and sentences into algebraic expressions and equations.
2.2 Evaluate algebraic expressions and combine like terms.
2.3 Perform multiplication and division on expressions.
2.4 Apply the order of operations when simplifying an expression.
6. Perform mathematical computations to solve formulas for a specific variable and to solve systems of
equations, inequalities, square roots, and radical expressions in equations.
6.1 Perform mathematical computations to solve equations and solve for a specific variable.
6.2 Use equations to solve application problems.
Reading Assignment
Chapter 1:
Introduction to Algebraic Expressions
Chapter 2:
Equations, Inequalities, and Problem Solving, sections 2.1 through 2.5, pp. 79-126
Unit Lesson
These lesson notes are to be used in conjunction with your textbook. They provide additional information and
examples for each chapter covered. Links to optional lesson videos and problem solution videos have also
been provided in case you wish for additional information or to see step-by-step examples of how certain
types of problems can be solved.
Chapter 1: Introduction to Algebraic Expressions
Many of you taking this class will use some of the concepts we learn in your career field. Others of you will
never see this math on a day-to-day basis. No matter what your circumstances, taking a math course will help
you. Mathematics helps to sharpen our ability to think critically and logically, even when we are not dealing
with numbers.
Section 1.1: Introduction to Algebra
OPTIONAL: Click this link to watch a video lesson for this section: Flash version of the video for Section 1.1:
Introduction to Algebra (15:34).
Evaluating an algebraic expression: To evaluate an algebraic expression, substitute the number in for the
variable and perform the mathematical operation.
MAT 1302, Algebra I
1
Translating to algebraic expressions and equations: Please refer to the chart
onx page
4 ofGUIDE
the textbook
UNIT
STUDY
for common phrases and their interpretations.
Title
Example: “Abby’s speed minus twice the wind speed.” Since we are not given what Abby’s speed is, we will
use a variable to represent this. The variable you choose can be any letter. For example, we can let Abby’s
speed be represented with an “a.”
We also have the wind speed. We will represent the wind speed with “w.”
Now, minus means that we are subtracting. “Twice the wind speed” means we are taking the wind speed and
multiplying it by 2. What we have is
Abby’s speed
twice the wind speed
−
a
OR
minus
2w
a - 2w
Example: “One third of the sum of two numbers.” When you see a phrase such as this, notice that we are
finding one third of the sum, not one third of a number. Therefore, the one-third will be on the outside of a set
of parentheses, like this:
1
3
()
“Sum” means we are adding, and since we are told “two numbers” rather than the same number twice, we will
1
( 𝑥 + 𝑦)
use two different variables to represent the numbers, like this:
3
Example: “Ten less than twice a number is fifty.” When we see the phrase “less than,” it is easy to know that
we are referring to subtraction. What is not sometimes easy to see is that the subtraction is the opposite of
how the wording is given. “Ten less than” means that the 10 is being subtracted from a number: “(some
number) minus 10,” or x – 10.
“Twice a number” means we are multiplying a number by 2, which we can represent as 2x.
The word “is” in a sentence can be represented by an equal sign. Therefore, we can write “ten less than twice
a number is fifty” as 2x - 10 = 50.
Section 1.2: The Commutative, Associative, and Distributive Laws
OPTIONAL: Click this link to watch a video lesson for this section: Flash version of the lesson video for
Section 1.2: The Commutative, Associative, and Distributive Laws (17:01).
While understanding the commutative and associative laws can be helpful, in these lesson notes, we will
focus on the distributive law since it will have the most application for future purposes.
Distributive Law: For any numbers a, b, and c, a(b + c) = ab + ac
In other words, multiply the number outside the parentheses by the numbers inside the parentheses. Be sure
to view the examples given in the lesson videos as well as the textbook.
Section 1.3: Fraction Notation
OPTIONAL: Click this link to watch a video lesson for this section: Flash version of the lesson video for
Section 1.3: Fraction Notation (21:18).
Prime Factorization: Think of this as a number’s DNA. Each number has a unique prime factorization the
same way that each person has a unique DNA.
Example: Find the prime factorization of 136. The textbook shows one way to work prime factorization. Here
is another way we can perform prime factorization. First, write the number down—in our example, the number
MAT 1302, Algebra I
2
is 136. Then, to the left of the number, write its smallest prime factor—in this example,
2. Then,
divide the
UNIT x STUDY
GUIDE
number (136) by the prime factor (2) and write the quotient (68) above the number:
Title
68
2)136
Now, divide 68 by its smallest prime factor, which is also 2:
34
2)68
2)136
Keep doing this until our quotient on the top is a prime number:
17
2)34
2)68
2)136
The numbers in red make up the prime factorization. Therefore, the prime factorization of 136 is 2 ∙ 2 ∙ 2 ∙ 17
which can also be written as 23 ∙ 17.
Adding and Subtracting Fractions: When we add or subtract fractions, we must always have a common
denominator.
7
5
5
Example: Add. 8 + 12 + 16
Now, you might remember how to find the common denominator, but here is another method you can use
when the common denominator is not obvious. This method is called the cake method: Write down all three
denominators. Out to the left, write any number that will divide into ALL of the denominators. Divide each
denominator by that outside number and write the quotient above. Since we are dealing with 8, 12, and 16,
the number 2 will divide evenly into all of them:
4 6 8
2) 8 12 16
Now, do the same process for 4, 6, and 8. Again, since they are all even, 2 will divide all:
2 3 4
2)4 6 8
2) 8 12 16
Notice that nothing will divide all of the numbers, which are now 2, 3, and 4. Now, look at any two numbers.
Are there two numbers that can be divided by the same number? Yes! Two will divide itself and 4. Since 2
does not divide 3, just bring that 3 up:
1 3 2
2)2 3 4
2)4 6 8
2) 8 12 16
Since nothing else will divide any two of the numbers we now have—1, 3, and 2—we are done. Now, our
common denominator is the product of all of the numbers in red:
LCD = 2 * 2 * 2 * 1 * 3 * 2 = 48
Next, we rewrite our fractions using 48 as the denominator. Do this by multiplying the top and bottom of each
fraction by the same number—whatever number is needed to make the denominator equal 48:
7 6 42
∙ =
8 6 48
5 4 20
∙ =
12 4 48
5 3 15
∙ =
16 3 48
MAT 1302, Algebra I
3
UNIT x STUDY GUIDE
Our problem has now become:
Title
42 20 15
+
+
48 48 48
When we add up the numerators, keeping the denominator the same (48), we get:
42
20
15
+ 48 + 48 =
48
77
29
or 1
48
48
Multiplying and Dividing Fractions: When multiplying or dividing fractions, we do not need a common
denominator. When you multiply two fractions, cross simplify, then multiply straight across. When dividing,
use the catch phrase “keep change flip” to help you remember to keep the first fraction the same, change to
multiplication, and flip the second fraction.
Example: Divide.
3
7
6
This problem can also be written as
3
÷6
7
If we were to write 6 as a fraction, we simply write it over 1:
3 6
÷
7 1
“Keep change flip” and you will get
3 1
∙
7 6
Cross simplify: Notice that 3 can divide evenly into both 3 and 6 and reduce them to 1 and 2. Now, our
problem becomes
3 1 1 1
1
∙ = ∙ =
7 6 7 2 14
Section 1.4: Positive and Negative Real Numbers
OPTIONAL: Click this link to watch a video lesson for this section: Flash version of the lesson video for
Section 1.4: Positive and Negative Real Numbers (13:57).
Whole Numbers: {0, 1, 2, 3, …}
Integers: {…, -2, -1, 0, 1, 2, …}
**Notice that integers do NOT include decimals, fractions, or radicals**
Rational vs. Irrational Numbers: Both rational and irrational numbers are real numbers.
Rational numbers can be written in the form
𝑎
𝑏
where a and b are integers and b does not equal 0. This
includes all repeating decimals and decimals that terminate (ex:
1
7
= 0. 3̅ and 8 = 0.875).
3
Irrational numbers are numbers whose decimal representation has no rhyme or reason about it. In other
words, it does not repeat and does not terminate. Some common irrational numbers are 𝜋, 𝑒, and √2.
For absolute value, refer to page 34 in your textbook for a definition and examples.
Section 1.5: Addition of Real Numbers
OPTIONAL: Click this link to watch a video lesson for this section: Flash version of the video for Section 1.5:
Addition of Real Numbers (19:06).
MAT 1302, Algebra I
4
Combining like terms: The terms in an expression are only considered “like”UNIT
if theyx have
theGUIDE
same variable
STUDY
part. We combine them by adding or subtracting the coefficient, or number part,
attached to the variable.
Title
Example: −7 + 10.5𝑦 + 13 + (−11.5𝑦)
First, we will group according to like terms:
−7 + 13 + 10.5𝑦 + (−11.5𝑦)
We can combine the constants, -7 and 13, because they have no variable attached to them. The terms 10.5y
and -11.5y can be combined because they both have identical variables, parts of y. When we combine 10.5y
and -11.5y, the variable part is left alone. We only add or subtract the coefficients:
−7 + 13 = 6
10.5𝑦 + (−11.5𝑦) = −1𝑦
After combining like terms, we can more easily find our answer, which is
6 + (−𝑦).
Note that -y is the same as -1y.
Section 1.6: Subtraction of Real Numbers
OPTIONAL: Click this link to watch a video lesson for this section: Flash version of the video for Section 1.6:
Subtraction of Real Numbers (16:07).
Be sure to watch the video and/or read the textbook for examples of subtraction of real numbers.
When you are dealing with word problems involving subtraction of negative numbers, be careful not to
confuse subtraction in a problem with the sign of a negative number.
Example: In just 12 hours on February 21, 1918, the temperature in Granville, North Dakota, rose from −33℉
to 50℉. By how much did the temperature change?
In this problem, we need to calculate the difference to find how much the temperature changed. We will take
the second temperature and subtract the initial temperature (which was negative):
50 − (−33)
= 50 + 33
= 83
The temperature changed by 83℉.
Section 1.7: Multiplication and Division of Real Numbers
OPTIONAL: Click this link to watch a video lesson for this section: Flash version of the video for Section 1.7:
Multiplication and Division of Real Numbers (13:59).
Please refer to the video and/or the textbook, pages 51-52, for the rules and examples of multiplication and
division of positive and negative numbers.
Division involving zero: Here is a little tip to help you remember division with zero:
We cannot divide by 0. The answer is undefined: “NO or n/0.” It is “OK, or 0 / k,” to divide zero by a number.
The answer is 0.
Section 1.8: Exponential Notation and Order of Operations
OPTIONAL: Click this link to watch a video lesson for this section: Flash version of the video for Section 1.8:
Exponential Notation and Order of Operations (27:26).
One commonly confused issue in this section is the difference between −𝑎2 and (−𝑎)2 . The difference is the
use of the parentheses. Let’s see the difference through an example:
vs.
(−3)2
−32
MAT 1302, Algebra I
5
−(3 ∙ 3)
−9
(−3)(−3)UNIT x STUDY GUIDE
9 Title
As we can see, when we do not enclose the negative sign in parentheses, our answer will be negative, no
matter what our exponent may be. When we enclose the negative sign in parentheses and our exponent is
even, our answer will be positive.
Order of operations: A phrase you can use to remember the order of operations is “Please Excuse My Dear
Aunt Sally.”
Parentheses Exponents Multiplication Division Addition Subtraction
One thing to remember though, is that multiplication and division are done together left to right and addition
and subtraction are done together left to right.
Be sure to look at the examples in the textbook. Here are a couple more examples to view:
Example: 32 ÷ (−2)2 ∙ 4
Since there are no parentheses, we will go directly to the exponents:
32 ÷ (−2)2 ∙ 4
32 ÷ (−2)(−2) ∙ 4
32 ÷ 4 ∙ 4
We must be careful with the next step. Even though the word multiplication comes first in the wording of our
order of operations, it is multiplication and division left to right. Therefore, the division comes first:
(32 ÷ 4) ∙ 4
8 ∙ 4
32
Example: −48 ÷ (7 − 9)3 − 2[1 − 5(2 − 6) + 32 ]
First, we work inside the parentheses:
−48 ÷ (−2)3 − 2[1 − 5(−4) + 32 ]
Next, we work the exponents:
−48 ÷ (−8) − 2[1 − 5(−4) + 9]
Now, we work inside the brackets to simplify. We begin by multiplying 5 and -4:
−48 ÷ (−8) − 2[1 − (−20) + 9]
−48 ÷ (−8) − 2[1 + 20 + 9]
−48 ÷ (−8) − 2[30]
Now, we divide -48 by -8, and multiply 2 and 30. Since these two operations are separated by a subtraction
sign, we can do them at the same time:
6 − 60
−54
Click here to see a video showing another example of simplifying using order of operations.
Simplifying expressions: When we simplify expressions, we first clear out the parentheses, then combine
like terms.
Example: 4𝑎 − 𝑏 − 4(5𝑎 − 7𝑏 + 8𝑐)
To solve the above example, we begin our simplification process by distributing the -4 inside the parentheses.
Notice that we are including the sign of the 4 when we multiply. Since the 4 is negative, all of the signs for the
numbers inside the parentheses will change:
MAT 1302, Algebra I
6
4𝑎 − 𝑏 − 20𝑎 + 28𝑏 − 32𝑐
UNIT x STUDY GUIDE
Title
Now we combine like terms. Remember, like terms have the same variable part. For instance, 4a and -20a
are like terms because they both contain “a.” We must remember to be careful of our signs:
−16𝑎 + 27𝑏 − 32𝑐
Chapter 2: Equations, Inequalities, and Problem Solving
Section 2.1: Solving Equations
OPTIONAL: Click this link to watch a video lesson for this section: Flash version of the video for Section 2.1:
Solving Equations (16:55).
Section 2.2: Using the Principles Together
OPTIONAL: Click this link to watch a video lesson for this section: Flash version of the video for Section 2.2:
Using the Principles Together (17:08).
In this section, we put the knowledge we learned from section 2.1 into practice as well as learn some new
tools to help some equations easier to solve. Below are some examples of the equations that make use of
clearing fractions and decimals. Additionally, two specific types of equations are addressed, ones with no
solution and ones that have an infinite number of solutions.
Clearing Fractions
2
3
3
Example: 5 𝑥 − 2 𝑥 = 4 𝑥 + 3
First, we find the common denominator for all of the fractions. In this case, we are looking at 5, 2, and 4. Refer
to earlier in this lesson if you need help finding a common denominator. The least common denominator for 5,
2, and 4 is 20.
Now, we multiply everything in your equation by 20:
2
3
3
20 ( 𝑥 − 𝑥 = 𝑥 + 3)
5
2
4
2
3
3
20 ( 𝑥) − 20 ( 𝑥) = 20 ( 𝑥) + 20(3)
5
2
4
Notice that all of the denominators will simplify with 20:
2
3
3
4 ( 𝑥) − 10 ( 𝑥) = 5 ( 𝑥) + 20(3)
1
1
1
Now, multiply:
8𝑥 − 30𝑥 = 15𝑥 + 60
Simplify the left side by combining like terms:
−22𝑥 = 15𝑥 + 60
Move all x variables to one side:
−22𝑥 − 15𝑥 = 15𝑥 + 60 − 15𝑥
−37𝑥 = 60
Divide by -37:
−37𝑥
60
=
−37
−37
𝑥=−
MAT 1302, Algebra I
60
37
7
Clearing Decimals
UNIT x STUDY GUIDE
Title
Example: 0.91 − 0.2𝑧 = 1.23 − 0.6𝑧
To find what power of 10 to multiply by, use the greatest number of digits behind the decimal for any one of
the decimals. In this problem, we have 0.91, -0.2, 1.23, and 0.6. The largest number of digits behind the
decimal is 2 (in the numbers 0.91 and 1.23). Therefore, we will multiply everything by 100 (2 digits behind the
decimal gives you 2 zeros after the 1):
100(0.91 − 0.2𝑧 = 1.23 − 0.6𝑧)
The easiest way to multiply by 100 is to move all decimals two places to the right:
91 − 20𝑧 = 123 − 60𝑧
Move the x variables to one side:
91 − 20𝑧 + 60𝑧 = 123 − 60𝑧 + 60𝑧
91 + 40𝑧 = 123
Move the numbers to the other side:
91 + 40𝑧 − 91 = 123 − 91
40𝑧 = 32
Divide by 40:
40𝑧 32
=
40
40
Simplify:
𝑧=
32 4
=
40 5
The following examples show what to do when there is no solution or an infinite number of solutions.
Example: 2(𝑥 + 5) + 4𝑥 = 6(5 + 𝑥) − 20
First, simplify each side by distributing 2 and 6:
2𝑥 + 10 + 4𝑥 = 30 + 6𝑥 − 20
Simplify both sides by combining like terms:
6𝑥 + 10 = 6𝑥 + 10
At this point, you may notice that we have the same exact equation on the right and left sides. We could stop
at this point. When you see that you have the same expression on both sides, the answer is all real
numbers. Just to show what would happen if we did keep going, we would have the following:
6𝑥 + 10 − 6𝑥 = 6𝑥 + 10 − 6𝑥
10 = 10
10 − 10 = 10 − 10
0=0
Example: 𝑥 + 8 = 3 + 𝑥 + 7
Simplify by combining 3 and 7:
𝑥 + 8 = 𝑥 + 10
Move the x variables to one side:
𝑥 + 8 − 𝑥 = 𝑥 + 10 − 𝑥
8 = 10
MAT 1302, Algebra I
8
This is not a true statement. The number 8 will never be the same as (equal to)
10. Therefore,
there is no
UNIT
x STUDY GUIDE
solution.
Title
Click here to see a video showing another example of solving equations.
Section 2.3: Formulas
OPTIONAL: Click this link to watch a video lesson for this section: Flash version of the video for Section 2.3:
Formulas (06:27).
When solving a formula for a specific variable, it can sometimes be helpful to sub in numbers in each step,
just to help you see what step to take next. Please make sure, though, that once you know what step to take,
you replace your number with the original variable.
Section 2.4: Applications with Percent
OPTIONAL: Click this link to watch a video lesson for this section: Flash version of the video for Section 2.4:
Applications with Percent (08:06).
In this section of the lesson, you will be presented with an example of calculating an original amount after it
has been increased. This method can also work when the original number was decreased.
Example: From 2007 to 2011, the number of new international students increased by 37 percent to 215,000.
How many new international students were there in 2007?
Let’s think of the number of students in 2007 as 100%. Then, the percentage after the increase would be
100+37 = 137%. Using this piece of information, we can set up a proportion to solve. A helpful hint when
setting up proportions is that the units—what is being represented--will be the same in both numerators and
the same in both denominators. Here, we are dealing with student count in 2007 (our first unit) and student
count in 2011 (our second unit).
In the numerator of one side, we will have 100%, which will correspond with the number of students in 2007,
x, in the numerator of the other side. The denominator under 100% will be 137%, which will correspond to the
number of students after the increase, or 215,000:
% 𝑓𝑜𝑟 2007 𝑛𝑢𝑚𝑏𝑒𝑟 𝑐𝑜𝑢𝑛𝑡 𝑓𝑜𝑟 2007
=
% 𝑓𝑜𝑟 2011 𝑛𝑢𝑚𝑏𝑒𝑟 𝑐𝑜𝑢𝑛𝑡 𝑓𝑜𝑟 2011
100
𝑥
=
137 215000
Cross multiply:
137𝑥 = 100 ∗ 215000
137𝑥 = 21500000
Divide by 137:
137𝑥 21500000
=
137
137
𝑥 ≈ 157,000
Section 2.5: Problem Solving
OPTIONAL: Click this link to watch a video lesson for this section: Flash version of the video for Section 2.5:
Problem Solving (22:16).
Refer to page 112 of your textbook for the five steps for problem solving in algebra. While this is not a magical
method to help you arrive at the right answer, it is a good method to help you understand and solve the
problem efficiently and accurately. Let’s look at two more examples of problem solving.
MAT 1302, Algebra I
9
Example: Through “extreme couponing,” Marie saved 85% of her grocery bill.UNIT
If shexpaid
$21,GUIDE
what was the
STUDY
original amount of the bill?
Title
For this problem, we will use a proportion to solve. The numerators will represent the original amount of the
bill. The denominators will represent what she actually paid.
Now, notice that 85% corresponds to how much she saved, not how much she spent. Since the total bill can
be considered 100%, she spent 100 - 85 = 15% of the total bill. Therefore, 15% of the total bill is $21:
% 𝑠𝑝𝑒𝑛𝑡
𝑎𝑚𝑜𝑢𝑛𝑡 𝑠𝑝𝑒𝑛𝑡
=
% 𝑡𝑜𝑡𝑎𝑙 𝑎𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝑡𝑜𝑡𝑎𝑙
15
21
=
100
𝑥
Cross multiply:
15𝑥 = 21 ∙ 100
15𝑥 = 2100
Divide by 15:
15𝑥 2100
=
15
15
𝑥 = $140
Example: A standard rectangular highway billboard sign has a perimeter of 124 feet. The length is 6 feet
more than three times the width. Find the dimensions of the sign.
First, we need to know the formula for the perimeter of a rectangle:
𝑃 = 2𝐿 + 2𝑊
Now, we look at the problem one sentence at a time. The first sentence tells us that the perimeter is 124 feet,
so our formula now looks like
124 = 2𝐿 + 2𝑤
The second sentence tells us what the length is in terms of width. “Six feet more than” tells us that we are
adding 6. “Three times the width” can be written as 3w. We put the two parts together, which gives us 3w + 6.
Remember that this is the length, so we will substitute this in for L in our equation:
124 = 2(3𝑤 + 6) + 2𝑤
Now, distribute 2 to the parentheses:
124 = 6𝑤 + 12 + 2𝑤
Simplify by combining like terms:
124 = 8𝑤 + 12
Move 12 over to the left-hand side:
124 − 12 = 8𝑤 + 12 − 12
112 = 8𝑤
Divide by 8:
112 8𝑤
=
8
8
𝑤 = 14
Now, use w = 14 and find the length:
𝐿𝑒𝑛𝑔𝑡ℎ = 3𝑤 + 6
MAT 1302, Algebra I
10
𝐿𝑒𝑛𝑔𝑡ℎ = 3(14) + 6 = 42 + 6 = 48
UNIT x STUDY GUIDE
Title
Click here to see a video showing how to solve an angles of triangle word problem or here to see a video
showing how to solve a word problem involving consecutive even integers.
Learning Activities (Non-Graded)
Problem Solution Video Reruns
The following videos showing problem solutions are linked in the unit lesson. They are being included here for
easy accessibility in case you want to review them and the step-by-step solutions they demonstrate:
MAT1302 Section 1.8 Order of Operation https://www.youtube.com/watch?v=yRdg1yTWwUE
MAT1302 Section 2.2 Solving Equations https://www.youtube.com/watch?v=vdiB3Rx2qK4
MAT1302 Section 2.5 Angles of Triangle Word Problem https://www.youtube.com/watch?v=UWzEHH29PMk
MAT1302 Section 2.5 Consecutive Even Integers Word Problem
https://www.youtube.com/watch?v=U3VwC7-lDOk
Practice What You Have Learned
After reading Chapters 1 and 2, improve your mastery skills by working the odd problems from the Exercise
Sets in each section:













Section 1.1 exercises, pages 8-11
Section 1.2 exercises, pages 16-18
Section 1.3 exercises, pages 26-28
Section 1.4 exercises, pages 34-36
Section 1.5 exercises, pages 41-43
Section 1.6 exercises, pages 48-51
Section 1.7 exercises, pages 57-59
Section 1.8 exercises, pages 66-68
Section 2.1 exercises, pages 86-87
Section 2.2 exercises, pages 93-95
Section 2.3 exercises, pages 99-102
Section 2.4 exercises, pages 108-112
Section 2.5 exercises, pages 121-127
After completing the odd problems, find out how well you did by checking your answers in the back of the
book. These activities are non-graded, which means you do not have to submit them. If you experience
difficulty in mastering any of the concepts, contact your instructor for additional information and guidance.
Review What You Have Learned
Before attempting the homework and the unit assessments, study the chapter summaries, review the
concepts taught in the chapters, and work the odd-numbered problems in the review exercises:




Chapter 1 Mid-Chapter Review, page 37
Chapter 1 Review Exercises, pages 75-77
Chapter 1 Test (practice), page 78
Chapter 2 Mid-Chapter Review, page 103
Check your answers in the back of the book. NOTE: A Study Summary is provided at the end of each chapter
to help you review the concepts taught in the chapter.
If you experience difficulty in mastering any of the concepts, contact your instructor for additional information
and guidance.
MAT 1302, Algebra I
11
Other Resources and Activities
UNIT x STUDY GUIDE
Title
If you need additional guidance or information, access the Multimedia Library in MyMathLab. This library
provides Animations that show how to solve selected types of problems, as well as Video Lectures,
PowerPoints, and Interactive Figures for information about a variety of topics from this unit. The Multimedia
Textbook is an electronic version of your textbook.
To access the Multimedia Library, go to MyMathLab and access your course. Once in the course, click on
“Multimedia Library” in the toolbar on the left side of the screen. When the dialog box comes up, select the
chapter, section, and media types from the dropdown menus, then click “Find Now.” This will bring up a listing
of media assets that you can click on to view.
Non-Graded Learning Activities are provided to aid students in their course of study. You do not have to
submit them. If you have questions contact your instructor for further guidance and information
MAT 1302, Algebra I
12