Lectures for College Algebra

Instructor: Mr. Rivera, Adjunct Lecturer Department of Mathematics, The City College of New York

Welcome to College Algebra

This took way too long to make, and I don't get paid enough (at all) for this.

As a brief introduction, I am an adjunct lecturer and tutor at CCNY at the time this letter was written. When I was 19, I took my first job as an educator here, tutoring in the Marshak Building. In my first month, I fell flat on my face many times trying to help students understand and solve problems. However, with time and dedication, I was able to improve.

I expect people who read my lecture notes to struggle to learn concepts just a little bit as I have when I first took these classes and when I first tried to explain these concepts. Rest assured, that is perfectly normal. Through time and practice, you too can improve and use math and its logical tools in many situations.

Math is not just a bunch of equations or numbers. Math is a language, a thought process, and a form of understanding our world. The purpose of these lecture notes is to show you my thinking when I see problems or face a new concept I have not seen before. They are also intended to act as commentary as you read them when you wish. It is also a script for myself to walk you guys through the concepts, so expect the notes and my lessons to be very similar to one another.

They will be based on textbooks, which will be referenced on the syllabus. However, I know it is not fun for me to repeat a book. You can safely assume I will have my own twists on what we discuss.

If you wish to point out mistakes or seek clarity, or if you believe I can better approach a topic, do not hesitate to reach out to me at jrivera5@ccny.cuny.edu. I believe in improving and updating lectures to make the experience more entertaining for my students, and even for those who read my lectures outside of my class.

Good luck this semester and I do hope you enjoy your time at my lectures.

- Mr. Rivera, Adjunct Lecturer

About this Notebook

This is NOT a textbook. It is intended to be presented to students as lecture material with examples and clarity. As such, you should not expect homework problems to appear in these notes, as those are provided in separate workbooks and MYOPENMATH.

While this is a condensed version of a textbook used in this class, it should not be considered a substitute by any means. Your textbooks are still important resources, and their approaches may offer a style of teaching better suited than mine.

Likewise, this material is not a substitute for in-person lectures. Students, you will learn best through a lecturer, along with your classmates or peers, and missing these lectures will rob you of the best experience possible.

I encourage students to reach out to me or to their peers whenever you need help with a problem or a concept. I also apologize in advance for any errors that may remain in these notes. Please be wary for typos, as these notes were prepared under time constraints.

I would like to thank the students I have personally tutored at the Math and Physics Tutoring Center in the Marshak Building for helping me develop my skills as a communicator. This material has been shaped with you in mind as some of my most valuable references.

Lecture 1: June 1st, 2026

Syllabus and Introducing Ourselves

The Introduction to the course and the policies can be found on the syllabus. This lecture will be slightly shorter because of that. Good luck this semester and do not be afraid to ask questions. This is a team effort for everyone to lift each other up to the next level. It is entirely reasonable to ask a question and answer incorrectly. I do not bite and you will not lose points by doing this, but hopefully gain a better understanding.

The Language of Algebra (1.1)

We will begin with the Natural Numbers, denoted as $\mathbb{N}$ . Natural Numbers are defined similarly to counting numbers, which are always positive.

How many Natural Numbers do you think exist? The answer is infinitely many; in other words, the number is currently growing and will always continue to grow. The reason why is because if you can think of a number infinitely large, I am always able to think of a number larger than it by some random margin.

Within this collection of numbers, we can separate these into three separate categories: unit numbers, prime numbers, and composite numbers.

Unit Number: There is only a single unit number. That number is the number $1$ . The number $1$ only has $1$ divisor, unlike primes and composites.
Prime Numbers: We define prime numbers to be numbers that only have $2$ divisors: $1$ and the number itself. For example, $3$ is a prime number, since it can only be divided by $1$ and $3$ . There are infinitely many prime numbers, and there is only $1$ even prime: The number $2$ .
Composite Numbers: Composite numbers are numbers that can be divided by $3$ or more numbers. Take the number $4$ for example. It can be divided by $1$ and $4$ , but it can also be divided by $2$ , meaning $4$ is not a prime. There are infinitely many composite numbers.

Say we have three natural numbers $a$ , $b$ , and $c$ .

We define $c$ as being a sum of $a$ and $b$ if added together: $a + b = c$
The same can be said for a difference between two natural numbers: $c - a = b$
When it comes to multiplication, we define $c$ being a multiple of $a$ if there is a number $b$ that if multiplied with a, we receive $c$ as the product: $a \times b = c \quad \text{or} \quad a \cdot b = c \quad \text{or} \quad ab = c$

In turn, we can define $a$ and $b$ as divisors of $c$ , if $c$ is a multiple of $a$ and $b$ . Another name for our divisors can be factors. We will be seeing the word factor a lot throughout this semester so it is important we understand its definition.

Factor: A number or expression that is a divisor for a larger number or expression without a remainder.

Divisibility Tricks

There are a few neat tricks to help us see what is a factor of a larger number:
1. A number is divisible by $2$ if the last digit is even.
2. A number is divisible by $3$ if the sum of digits is also divisible by $3$ .
3. A number is divisible by $5$ if the last digit is $5$ or $0$ .
4. A number is divisible by $6$ if divisible by both $2$ and $3$ .
5. A number is divisible by $10$ if the last digit is $0$ .

With these tricks in mind, we can write composite numbers as a product of prime numbers. This process is called prime factorization. Take for example $6$ . $6$ has $4$ divisors: $1, 2, 3,$ and $6$ . $2$ and $3$ are prime numbers and $2 \times 3 = 6$ , meaning $2$ and $3$ are the prime factors of $6$ .

We can do this with larger numbers as well. Let's try it with a number like $18$ . We know $18 = 2 \times 9$ and we know $9 = 3 \times 3$ . Therefore, the prime factorization of $18$ is: $2 \times 3 \times 3 = 2 \cdot 3^2$

With prime factoring, we can now go into the Least Common Multiple (LCM).

Least Common Multiple (LCM): The smallest possible number that is a multiple of two separate numbers.

We can find the LCM through prime factorization. Let us utilize $6$ and $9$ and find the LCM between those two: 1. Primarily factor both: $6 = 2 \times 3$ and $9 = 3 \times 3$ . 2. We then take the highest power of each prime and multiply them all together. (We will talk more about powers very shortly this lecture to better understand what I mean). 3. Doing this, we result in $2 \cdot 3 \cdot 3$ . The product of these factors is $18$ . Therefore, the LCM of $6$ and $9$ is $18$ .

$\begin{aligned} 6 &= 2 \cdot 3 \\ 9 &= 3 \cdot 3 \end{aligned}$

Remember: Take each unique prime into consideration. If two numbers have a common prime factor, only use the highest power representation. $\text{LCM}(6,9) = 2 \cdot 3 \cdot 3 = 18$

Arithmetic Operations & Expressions

Now that we have covered some basic properties of Natural Numbers, we can look into their operations to get different numbers:
* Addition: We can add two numbers $a$ and $b$ to get their sum $c$ :
$a + b = c$ * Subtraction: We can subtract $a$ from $c$ to get their difference $b$ :
$c - a = b$ * Multiplication: We can multiply $x$ and $y$ to get their product $z$ :
$x \times y = xy = x(y) = z$ * Division: We can divide $z$ (the dividend) and $y$ (the divisor) to get the quotient $x$ :
$z \div y = x$

With these operations, we can manipulate numbers and expressions once we touch on equations and inequalities. Before we go further into what they are, we should go into what expressions we can have. An expression will always consist of at least one of these components: a constant and a variable.

Constant: A number that will always remain fixed. It can never change. It can sometimes be a letter to store a value like $C$ for example to denote a constant.
Variable: A number whose value can change. This can be defined as a letter $x$ or $y$ , which are the two standard letters in Algebra.

Equalities and Inequalities

The symbol "=" is known as the equal sign. This simply tells us that one expression (or number) is equal to another ( $x$ is $y$ ).

For inequalities, there are a total of $5$ symbols we need to go through:
* $x \neq y$ ( $x$ is not equal to $y$ )
* $x > y$ ( $x$ is strictly greater than $y$ )
* $x < y$ ( $x$ is strictly less than $y$ )
* $x \geq y$ ( $x$ is inclusively greater than or equal to $y$ )
* $x \leq y$ ( $x$ is inclusively less than or equal to $y$ )

Equation: Expressions or numbers connected by " $=$ ". With this, we can also state the inequalities from earlier can serve the same purpose as an " $=$ " with our desired expressions.

You can group numbers and expressions using parentheses $(\,)$ , brackets $[\,]$ , etc., to denote which specific operation to carry out first.

Exponents

We now touch on exponents. Let us go back to prime factoring briefly. We know the prime factor of $4$ is $2 \times 2$ . Let's say now I want to find out what $2 \times 2 \times 2 \times 2$ is, but I am too lazy to do actual multiplying. We know $2$ is the only number being used and it is being multiplied by itself $4$ times. We can make an exponent with $2$ as the base and $4$ as its power: $2 \times 2 \times 2 \times 2 = 2^4 = 16$

Base: A number or expression in which we multiply it by its respective power.
Power: The number (or sometimes expression, but that's for Precalc) of times we multiply the base by itself.

Looking back, we know $9 = 3 \times 3$ , which we can now write as $3^2$ (3 squared). This can be said for any second power. Looking at $2^2 = 4$ , we also know $4 \times 2 = 8$ . But we also know $2 \times 2 = 4$ , meaning $2 \times 2 \times 2 = 2^3 = 8$ (2 cubed). This can be said for any third power. We will visit other exponents much later in the semester. However, this is sufficient to understand prime factorization for now.

Order of Operations (P.E.M.D.A.S.)

With all of this information, we can now go into the order of operations. The concept is simple. We will be using what is called P.E.M.D.A.S.:
* P (Parentheses): We work around parentheses or brackets as mentioned earlier. We prioritize what operations we do in these groups.
* E (Exponents): We then turn our focus to exponents. We need to evaluate them to simplify them into a number.
* M (Multiplication): Next we multiply numbers together, this step is interchangeable with division.
* D (Division): Next we divide numbers from each other, this step is interchangeable with multiplication.
* A (Addition): Finally, we add it all up to get our final sum, simplifying our expression or number as much as possible. This step is interchangeable with subtraction.
* S (Subtraction): Finally, we subtract to get our final difference, simplifying our result completely. This step is interchangeable with addition.

Evaluating Expressions and Combining Like Terms

Now that we have our building blocks of math, we can move on to evaluating expressions when a variable is defined as a number. Say I have an expression $2x + 8$ and I have my variable $x$ defined as $3$ . This would then simplify to: $2(3) + 8 = 6 + 8 = 14$

These expressions can have several terms with multiple coefficients. We should know what we are talking about:

Term: A constant that is alone or paired up with some number of variables. For example, $x^2$ and $xy$ are both one term, but they are not the same, since $y \neq x$ in this case.
Coefficient: A leading constant that multiplies terms.

When we want to combine like terms, we want to take terms that are the same and add or subtract them.

Example: Simplify $5x^2 + 3x - 2 + 4x^2 - x + 7$ $\begin{aligned} &= (5x^2 + 4x^2) + (3x - x) + (-2 + 7) \\ &= 9x^2 + 2x + 5 \end{aligned}$

Notice how I separated the variable numbers from the constant numbers. Also notice that not all terms are put together equally. Some with higher exponents will be put together.

What we can also do is evaluate this expression for any number $x$ that I so choose. For example, let us evaluate the previous simplified expression $9x^2 + 2x + 5$ when $x = 2$ : $9(2^2) + 2(2) + 5 = 9(4) + 4 + 5 = 36 + 4 + 5 = 45$

On the Topic of Integers (1.2)

Now that we have gone pretty far into the natural numbers, allow us to expand our horizons a bit. We are now going to implement the number $0$ and all of the negative numbers. These numbers along with all of the natural numbers brings us the entire list of integers, normally denoted as $\mathbb{Z}$ .

Every natural number has a negative side or opposite on the number line (i.e., the negative side of $n$ is $-n$ for some natural number $n$ ) and they are both the same distance to $0$ . Another name for this is the additive inverse.

Zero cannot be positive or negative. It is the only neutral number in the entire number line.

Absolute Value

Now let's go over the concept of distance mentioned just now. Say I have $4$ and its negative, $-4$ . The distance between both numbers to $0$ is $4$ . The way we can write out this relationship is to use what is called the absolute value, denoted as $|n|$ for some integer $n$ . The absolute value of an integer will never be negative. This means we have our first critical inequality: $|n| \geq 0$

This works because there is no such thing as a negative distance travelled. As a thought experiment on this, ask yourself: Does it make sense to say I have travelled negative 2 miles? It sounds ridiculous, right? This is how people who work around numbers often think to verify if things truly makes sense. (This should remind you of Philosophy a little bit but I digress).

Therefore, $|-n| = |n| = n$ for any integer $n$ . Be careful however, some absolute values can be opposite. You can consider the absolute value similar to the P in PEMDAS.

Integer Operations

Remember those operations we did earlier with the natural numbers? Well, now they can be expanded to all of the integers, which can allow us to finally introduce the additive identity: Zero. Say I have some integer $a$ . Well, the additive identity allows us to preserve that integer: $a \pm 0 = a$ This was discovered by the Mayans and Indians separately, spreading across their respective hemispheres after further development.

I also wish to introduce a clear distinction when it comes to adding or subtracting integers. We already know the case of adding two positive numbers.

If we take $1 - 2$ , that will be equal to $-1$ . Another way of writing that would be: $1 + (-2) = -1$
We can even add and subtract negatives: $-1 + (-2) = -3 \quad \text{and} \quad -2 - (-1) = -2 + 1 = 1 - 2 = -1$

The question you likely think to yourself is: Why are we able to switch around signs of numbers like that when adding or subtracting? The reason why is because not only can we add and subtract negative numbers, but we can also multiply and divide them.

For example: $(-1) \times (-2) = 2$ . The reason why this works is I can factor out $(-1)$ from $2$ , since $(-1) \times (-1) = 1$ and $1 \times 2 = 2$ . And we know $1 \times (-1) = -1$ , so $4 \times (-2) = -8$ , and so on. This can be thought of similarly for division, say $8 \div (-4) = -2$ as a random example.

We can come up with a simple rule when multiplying and dividing: If the signs are the same, the result is positive, otherwise it is negative.

With this, we can also conclude: to find the opposite of an integer, simply multiply it by $-1$ . Like natural numbers mentioned earlier, we can evaluate expressions with integers now and perform order of operations (PEMDAS), giving us a lot more to consider.

To end our first lecture, I will leave you with a very inspirational quote:

"A NEGATIVE TIMES A NEGATIVE EQUALS A POSITIVE" > — Mr. Escalante, Stand and Deliver

Lecture 2: June 2nd, 2026

Now that we have covered integers, we can now expand our knowledge of the number line even further, by looking at numbers in between integers.

On the Topic of Fractions (1.3)

Say I have two integers, $p$ and $q$ and $q \neq 0$ . Now we are going to assume that $q$ is not able to divide $p$ , meaning there will be a remainder. If necessary, revisit long division of integers while considering a remainder as this will be much more important later on in the semester.

$\textbf{Example I: Divide } 17 \div 5$

$5 \times 3 = 15$

$17 - 15 = 2$

$17 \div 5 = 3 \text{ remainder } 2$

$\boxed{17 = 5(3) + 2}$

$\textbf{Example II: Divide } 19 \div 3$

$3 \times 6 = 18$

$19 - 18 = 1$

$19 \div 3 = 3 \text{ remainder } 1$

$\boxed{19 = 6(3) + 1}$

We can either write out the quotient as a remainder, which can be tedious work. Or we can leave $\frac{p}{q}$ in its simplest form. These numbers are called rational numbers, denoted as $\mathbb{Q}$ . This way of writing $p$ divided by $q$ is called a fraction. $p$ is known as the numerator or dividend, while $q$ is denoted as the denominator, or divisor. A fraction will be simplified if $p$ and $q$ have no more common factors amongst one another.

We can not allow the denominator to ever be zero, because we simply can't divide a number into zero groups. Think about that. I will be stressing this throughout the course. We call anything divided by 0 "undefined". If we have $\frac{0}{0}$ , the term will be "Indeterminate", a term you will learn more about in Precalc, so do not expect this often throughout my lectures.

$\frac{6}{8} = \frac{3}{4}, \quad \frac{-10}{15} = \frac{-2}{3}, \quad \frac{12}{4} = 3, \quad \frac{7}{1} = 7$

From the few equations we have from the previous page, each fraction is simplified so that numerator and denominator have no common factors. You will have a better idea on how we are able to simplify these fractions later.

Like integers, we can perform PEMDAS on rational numbers. Say I have 2 rational numbers $\frac{a}{b}$ and $\frac{c}{d}$ already simplified and all four of the numbers are unique integers. When I multiply, I simply take the product of the numerators and divide it by the product of the denominators. $\frac{a}{b} * \frac{c}{d} = \frac{ac}{bd}$ Note that $\frac{ac}{bd}$ can be simplified if ac and bd have common factors. More on this later in the lecture.

To briefly get this out of the way, $a * \frac{b}{c} = \frac{a}{1} * \frac{b}{c} = \frac{ab}{c}$ This is to simply deprive ourselves of any confusion we may run into as we go on in the lectures.

Now we can introduce ourselves to the multiplicative identity: 1. This can be obtained by multiplying a rational number $\frac{a}{b}$ by its multiplicative inverse $\frac{b}{a}$ . We can easily see the result be $\frac{a}{b} * \frac{b}{a} = \frac{ab}{ba} = 1$

We will explore $ab = ba$ later this lecture, so for now take my word for it. Now we will looking at adding and subtracting fractions, as they involve the same steps. Take my two rational numbers from earlier and say I want to add or subtract.

There are 3 types of fractions we want to observe. Those with the same denominator, those with different denominators, those with similar denominators.

Fractions with different denominators:

Say I have $a,b,c,d$ as integers, $b,d \neq 0$ such that we have the expression

$\frac{a}{b} \pm \frac{c}{d}$

The requirement is for the denominators to be exactly the same when we take the sum or difference. In other words, we need to find the least common denominator (LCD). The process is the same as the least common multiple as discussed in the previous lecture The only difference is us looking at the denominator instead of the numerator.

$\text{LCD}(\frac{1}{3},\frac{1}{2}) = \text{LCM}(2,3) = 6$

The strategy is to multiply one fraction by the other denominator in the form of 1. In worther words, we need an equivalent fraction, which now is a very good time to show you what it is. For example, $1=\frac{d}{d}$ and therefore $\frac{a}{b} = 1*\frac{a}{b} = \frac{d}{d}*\frac{a}{b} = \frac{ad}{bd}$ This is an equivalent fraction. Following this process, I can rewrite $\frac{c}{d}$ as $\frac{bc}{bd}$ . Now I can either add or subtract ad with bc, giving me the result $\frac{a}{b} \pm \frac{c}{d} = \frac{ad \pm bc}{bd}$

$\frac{2}{3} + \frac{1}{4} = \frac{2*4}{3*4} + \frac{1*3}{4*3} = \frac{8}{12} + \frac{3}{12} = \frac{11}{12}$

Fractions with the same denominator c

Now say I have $a,b,c$ as integers and $c \neq 0$ . With this, I have $\frac{a}{c} \pm \frac{b}{c}$ . Then I can just operate directly across the numerators since I already have the same denominator. $\frac{a \pm b}{c}$

$\frac{3}{7} + \frac{2}{7} = \frac{3+2}{7} = \frac{5}{7}$

Fractions with similar denominators:

Now say I have $a,b,c,n$ as integers, $c, n \neq 0$ as before but now I have $\frac{a}{c} \pm \frac{b}{nc}$ . Luckily we already know that nc is a multiple of c so we only need to edit one of the fractions. Using equivalent fractions, $\frac{a}{c} = 1*\frac{a}{c} = \frac{n}{n}*\frac{a}{c} = \frac{na}{nc}$ This prompts me to finish the job. $\frac{a}{c} \pm \frac{b}{nc} = \frac{na}{nc} \pm \frac{b}{nc} = \frac{na \pm b}{nc}$

$\frac{5}{6} - \frac{1}{12} = \frac{10}{12} - \frac{1}{12} = \frac{9}{12} = \frac{3}{4}$

All three of our results may be simplifiable, depending if $ab \pm bc$ has any common factors with $bd, c$ , or $nc$ to make an equivalent, simplified fraction.

Now we can finally move on to division. Take our two rational numbers from earlier and I make what is called a complex fraction. This is basically one fraction as a numerator, and another as a denominator.

$\frac{\frac{a}{b}}{\frac{c}{d}}$

It is a very good thing I have introduced the multiplicative inverse earlier, otherwise we would have been stuck. We know that dividing an integer by itself gives us 1. Another way of saying this taking our integer and multiplying it by its multiplicative inverse. In math terms: $1 = \frac{a}{a} = a * \frac{1}{a}$ . Which means, we can take $\frac{a}{b}$ and instead multiply it by the multiplicative inverse of $\frac{c}{d}$ That would end up being $\frac{d}{c}$ Therefore, $\frac{\frac{a}{b}}{\frac{c}{d}} = \frac{a}{b} * \frac{d}{c}$ . We turn this into a multiplication of fractions, which we have covered earlier. This method is called the Keep Change Flip method (KCF) Method and as you can see it is very useful to solve problems. Just a heads up that more complex problems can have us adding or subtracting fractions in the numerator and/or denominators which can be a problem with PEMDAS. We would have to address that on the spot before we actually address the complex fraction itself.

Example of Complex Fractions with KCF: $\frac{\frac{2}{3}}{\frac{4}{5}} = \frac{2}{3} * \frac{5}{4} = \frac{10}{12} = \frac{5}{6}$

Last thing I briefly want to touch up on is exponents. Say I have $(\frac{a}{b})^2$ Well, we know from earlier that we can multiple numbers by themselves some number of times. With that said, $(\frac{a}{b})^2 = \frac{a}{b} * \frac{a}{b} = \frac{a^2}{b^2}$ . We will see this again later in the semester to over exponents specifically.

$\left(\frac{3}{4}\right)^2 = \frac{3}{4} * \frac{3}{4} = \frac{9}{16}$

Like integers, we can simplify and even evaluate expressions by utilizing rational numbers.

$\frac{2x}{4} = \frac{x}{2} \quad \text{and} \quad \frac{6y^2}{3y} = 2y$

Before moving on, I want to reiterate the use of multiplying by $-1$ . If I have $\frac{a}{b}, b \neq 0$ , where I multiply by $-1$ , the following holds true:

$-1 * \frac{a}{b} = \frac{-a}{b} = \frac{a}{-b} = -\frac{a}{b}$

On the Topic of Real Numbers (1.5)

At last, we get to the real numbers. Real Numbers, denoted as $\mathbb{R}$ , hold all of the natural numbers, integers and rational numbers. However, it also holds nonrational, or irrational numbers. These irrational numbers, denonted as $\mathbb{I}$ , will not be able to be presented as a clean fraction with only integers. Since that is the case, they can't be expressed as a decimal with a pattern either. To give a pair of examples, the number $\pi$ or pi (sounds like pie) is an irrational number (3.14...). Another irrational number is Euler's number e (2.714...) Like rational numbers, irrational numbers can be operated under PEMDAS, meaning all of the real numbers can be operated under PEMDAS as well. Now that we have gone over all possible numbers on the number line, we can now go into their properties.

The commutative property for addition is a property of real numbers such that no matter what order of the numbers you put, you will always get the same result. Say I have three real numbers, $a,b$ and $c$ , such that $a + b = c$ , but we also know $c = b + a$ . Therefore, $a + b = b + a$ Next, the commutative property for multiplication is like that of addition. Say I take $d,e$ and $f$ as real numbers such that $de = f = ed$ , meaning $ed = de$ . This covers that $ab = ba$ example we went over a little while earlier.

A fair warning that this will not always hold true for subtraction or division.

$2 - 1 = 1 \neq 1 - 2 = (-1)$ $\frac{3}{4} = 3 \div 4 \neq \frac{4}{3} = 4 \div 3$

Now for the associative property. I take 3 real numbers, $a,b$ and $c$ again and no matter where I start, similar to the communative property, I will always get the same result. $a + b + c = a + (b + c) = (a + b) + c$ . The same can be done for multiplication. $abc = a(bc) = (ab)c$

This leads to the additive identity 0 and the multiplicative identity 1, which I have gone over earlier.

Two small notes on multiplication regarding 0: 1: If you take a real number and multiply by 0, you will get 0. $0*a=0$ 2: You can not divide by 0 because 0 does not have a multiplicative inverse. I will leave you to think on why that is the case.

We now cover the distributive property which at times can allow you to break PEMDAS if that process is too tedious. Take our 3 real numbers from earlier such that the expression $a(b + c)$ is possible. Then we can result in $a(b + c) = ab + ac$ . Similarly, $(b + c)a = ba + ca$ and we can do some manipulation if we wish thanks to the communative property.

And like rational numbers, we can evaluate and simplify real numbers that were variables as the situation calls for.

As a brief example, we have the expression $\frac{6x}{9x}$ , $x \neq 0$ , we can simplify by dividing numerator and denominator by $3x$ : $\frac{6x}{9x} = \frac{6x \div 3x}{9x \div 3x} = \frac{2}{3}$

How did I know to divide by 3 along with $x$ ? That will be discussed in a future lecture, but feel free to think and take a guess as to why.

Finally, for any real number $a$ , $0 * a = 0$ and for any real number $b \neq 0$ , $0 * \frac{1}{b} = \frac{0}{b} = 0$

Lecture 3: June 3rd, 2026

With everything we have covered with real numbers, we can now go over how to generally solve linear equations. We want to find solutions.

Strategizing Linear Equations (2.1)

Solution: A value that can be substituted for a variable to make a true statement in an equation.

To verify if a value we find is a solution to our equation, we simply substitute our variable for our value and evaluate both sides of the equation if necessary to see if the equation holds true.

As an example, we have: $2x + 3 = 11$

Suppose we want to verify if $x = 4$ is a solution. Substitute: $2(4) + 3 = 8 + 3 = 11$

Since both sides equal $11$ , the value $x = 4$ is a solution.

The primary focus for this section will be a linear equation. A linear equation is one that has a variable with some leading coefficient $a$ , added on with $b$ such that $y = ax + b$ . Where $a$ and $b$ are real numbers and $a \neq 0$ .

One of the best strategies for solving linear equations is to go backwards in the PEMDAS process if your equation is simple enough to do so.

Given the equation: $3x + 5 = 20$

Step 1: Undo addition by subtracting $5$ from both sides: $3x + 5 - 5 = 20 - 5 \implies 3x = 15$
Step 2: Undo multiplication by dividing both sides by $3$ : $\frac{3x}{3} = \frac{15}{3} \implies x = 5$

So, the solution is $x = 5$ .

Another strategy would be to first simplify both sides of your equation through PEMDAS. Then I would put all of my like terms on either side. For instance, I would put all of my variables on the left hand side of my equation and all of my constants on the right hand side, then utilize reverse PEMDAS. I can check my work by simply plugging in my result into my initial equation to verify if the solution is correct.

With arguably the best strategy in all of algebra, we now need to classify equations. There are three classifications for solutions of linear equations. They can be nonexistent, meaning there is no solution. Another name for this can be called a contradiction. Solutions can also be conditional. This means that there is a finite number of solutions for an equation and all other numbers are not solutions. Or they can be an identity, meaning every single number on the number line is a solution to our equation.

Examples of Classifications

Contradiction: $2x + 3 = 2x + 5$ Subtract $2x$ from both sides: $3 = 5 \quad \text{(False! No solution)}$
Conditional Solution: $3x - 7 = 2$ $3x = 2 + 7$ $3x = 9$ $x = 3 \quad \text{(One solution, } 1 < \infty \text{, conditional)}$
Identity: $4(x + 2) = 4x + 8$ $4x + 8 = 4x + 8$ $0 = 0 \quad \text{(True for all } x \text{, identity)}$

For the rest of the lecture, I will simply be going through a few problems of linear equations with fractions, in order for us to be a bit more comfortable with these types of problems. Another reason is to actually make sure we go back and address any issues any of you may have had over the first two lectures before the first quiz. If you were not present, you do have the text book and the work book to work off of. This is a chance to work a bit more together and be creative with what problems we want to create for ourselves and challenge ourselves, which is why I will only 1 example regarding fractions to give those of you absent an idea.

$0 = -4 + \frac{3x}{4} \implies 4 = \frac{3x}{4} \implies 16 = 3x \implies x = \frac{16}{3}$

Lecture 4: June 4th, 2026

For some reason, the website is not allowing me to put up certain images that I want you to see. All of what I paste in these lecture notes are personally coded onto a separate server. For now, the images will have to wait until the Brightspace is up and running. When that happens, I will upload the PDF version of my lecture notes and you will be able to see all of the images and the visual examples there. For now, I do apologize for the ugliness and lack of images. Hopefully this is the last lecture I have to do this with for the website.

Working with Formulas to Solve for a Variable (2.3)

When solving for variables, we can generally solve linear equations.
Say I have 3 real numbers, $a$ , $b$ , and $c$ and 2 variables $x$ and $y$ , such that

$ax + by = c.$

The general idea is to solve for $y$ in terms of $x$ :

$ax + by = c \implies by = c - ax \implies y = \frac{c - ax}{b}.$

This work can be applied to formulas you may have seen in high school geometry.

For example, the volume of a sphere is

$V = \frac{4}{3} \pi r^3,$

where $r$ is the radius.

The volume of a cylinder is

$V = \pi r^2 h,$

and we can solve for a height $h$ given a radius $r$ .

Its surface area $S$ can be defined as

$S = 2\pi(r^2 + rh).$

The area of a rectangle is

$A = w \cdot l,$

and we can solve for one variable given the other.

The area of a triangle is

$A = \frac{1}{2} b h,$

where $b$ is the base and $h$ is the height.

The area of a circle is

$A = \pi r^2,$

where $r$ is the radius.

We can also convert temperature between Celsius and Fahrenheit using

$C = \frac{5}{9}(F - 32),$

which can be treated as a linear equation.

Theorem

Theorem: A statement proven not by itself, but through reasoning and based on truths that already exist.
In other words, we deduce the statement using logic and known facts.

With that, I can now introduce one of the most popular theorems in all of mathematics.

The Pythagorean Theorem

The Pythagorean Theorem, discovered by Pythagoras, states that for any right triangle with legs $a$ and $b$ , the sum of their squares equals the square of the hypotenuse $c$ :

$a^2 + b^2 = c^2.$

The proof of this theorem is at the end of the book. When I prove things in that section, I show you why the statement is true.

Example

If a right triangle has hypotenuse $c = 5$ and height $b = 3$ , the base $a$ is

$a^2 + 3^2 = 5^2$

$a^2 = 16$

$a = \sqrt{16} = \pm 4.$

Negative solutions are mathematically valid, but in measurement, we use only positive values.

Perimeter Formulas

We primarily look at the perimeter for rectangles, triangles, and circles. Their derivations (where they come from) at this level are beyond the scope of this class, so I will simply present each of them.

Rectangle:

$P(\text{rectangle}) = 2l + 2w = 2(l + w)$

Triangle:

$P(\text{triangle}) = b + h + l$

Circle:

$P(\text{circle}) = 2\pi r$

The circumference is another name for the perimeter of a circle.

Solving Linear Inequalities (2.5)

Now that we have some introduction on what an equation is, we can now introduce ourselves to inequalities.

Like equations, we can write two expressions and compare them to each other. However, as the name suggests, the expressions will not always be equal.

Suppose we have 3 real numbers, $a$ , $b$ , and $c$ where $a \neq 0$ . We can write all of these inequalities as such:

$ax + b < c$

$ax + b \le c$

$ax + b > c$

$ax + b \ge c$

$ax + b \ne c$

As a friendly reminder, there are expressions that have the chance of being equal to another expression or number, which is why you see a little line under the alligator sign.

Inequalities have all of the properties of an equation, but an equation does not hold all properties of an inequality.

For example, we know that

$2 > 1.$

Multiply both sides by 2:

$4 > 2.$

Now divide both sides by $-2$ :

$-2 > -1.$

This statement is false. To make it true, we must reverse the inequality:

$-2 < -1.$

Rule: When we multiply or divide negative numbers on an inequality, we must always flip the alligator sign to preserve the statement.

Classifications of Inequalities

Like equations, inequalities can be conditional, an identity, or a contradiction.

Example of a Condition

$x + 3 > 5 \implies x > 2$

Example of an Identity

$x - x < 1 \implies 0 < 1$

Example of a Contradiction

$2x + 1 < 2x \implies 1 < 0$

Graphing Inequalities

Solve and graph:

$x + 2 < 1 \implies x < -1$

Solve and graph:

$x \ge 1$

I know this part is difficult without an image but try to visualize this as best you can. I promise when Brightspace comes up you will actually see what I am talking about if you did not come to my lecture today.

Open circle → endpoint not included
Closed circle → endpoint included

We will discuss this further when we study compound inequalities.