Students will:

(1.) Define inequalities.
(2.) Discuss inequality in the world.
(3.) Discuss inequality in mathematics.
(4.) Solve linear inequalities.
(5.) Check the solution(s) of linear inequalities.
(6.) Graph the solution(s) of linear inequalities.
(7.) Solve quadratic inequalities.
(8.) Check the solution(s) of quadratic inequalities.
(9.) Graph the solution(s) of quadratic inequalities.
(10.) Solve polynomial inequalities.
(11.) Check the solution(s) of polynomial inequalities.
(12.) Graph the solution(s) of polynomial inequalities.
(13.) Solve absolute value inequalities.
(14.) Check the solution(s) of absolute value inequalities.
(15.) Graph the solution(s) of absolute value inequalities.
(16.) Solve rational inequalities.
(17.) Check the solution(s) of rational inequalities.
(18.) Graph the solution(s) of rational inequalities.

Skills Measured/Acquired

(1.) Use of prior knowledge
(2.) Critical Thinking
(3.) Interdisciplinary connections/applications
(4.) Technology
(5.) Active participation through direct questioning
(6.) Research

Please note:
added to
subtracted from
multiplied by
divided by

Check for prior knowledge. Ask students about these terms.

Bring it to English: equal, unequal, equality, inequality, more, less, greater, smaller, lesser, most, least, at most, at least, exactly, just, not

Bring it to Mathematics: inequality, greater than, less than, greater than or equal to, less than or equal to, at least, at most, exactly, equal, equals, equal to, not equal to

(1.) At least $5$ means $\ge 5$
It means that the minimum should be $5$

(2.) At most $5$ means $\le 5$
It means that the maximum should be $5$

(3.) Just $5$ means $= 5$
It means exactly $5$ or equal to $5$

(4.) More than $5$ means $\gt 5$

(5.) Less than $5$ means $\lt 5$

(6.) No more than $5$ means $\le 5$
It should not be more than $5$
It means $5$ or less

(7.) No less than $5$ means $\ge 5$
It should not be less than $5$
It means $5$ or more

Three Notable Notes When Solving Inequalities

There are three major things to note when solving Inequalities:
Let us first illustrate why equality (equation) is much better than inequality.

Equation
$3 = 3$
$-3 = -3$
Whether you multiply or divide both sides of an equation by $-1$, the equation is still the same.

Inequality
$3 \lt 4$
Multiply both sides by $-1$
$-1 * 3 \gt -1 * 4$
$-3 \gt -4$
Teacher: Did you notice what happened?
Student: The inequality sign is reversed.
Teacher: That's right.
(1.) Whenever you multiply or divide both sides of an inequality by a negative number, the inequality is reversed.


Equation
$2 + 3 = 5$
$5 = 2 + 3$
The $LHS$ is equal to the $RHS$
The $RHS$ is equal to the $LHS$
You can swap: make the $LHS$ to be the $RHS$ and the $RHS$ to be the $LHS$ and there is no problem.

Inequality
$(2 + 3) \lt 6$
$6 \gt (2 + 3)$
Teacher: Did you notice what happened?
Student: The inequality sign is reversed.
Teacher: That's right.
(2.) Whenever you swap (make the $LHS$ to be the $RHS$, and the $RHS$ to be the $LHS$); the inequality is reversed.


Equation
$3 = 3$

$\dfrac{1}{3} = \dfrac{1}{3}$

Whether you take the reciprocal of both sides of an equation, the equation remains the same.

Inequality
$3 \lt 4$
Take the reciprocal of both sides
$ \dfrac{1}{3} \gt \dfrac{1}{4} \\[5ex] $ Teacher: Did you notice what happened?
Student: The inequality sign is reversed.
Teacher: That's right.
(3.) Whenever you take the reciprocal of both sides of an inequality, the inequality is reversed.

Two Notable Notes After Solving Inequalities

(1.) For all inequalities besides the compound inequalities involving "AND"; express your solution such that the variable is always on the $LHS$.
Example: Write $x \lt 3$ rather than $3 \gt x$
For the compound inequalities involving "AND", write the variable in the middle.
Example: Write $3 \lt x \lt 7$

(2.) Always check the solution(s) of any inequality.
Yes, it is possible that it checks out right even though the solution might be wrong (unlike in the case of an equation). However, it is always a good practice to check!

Teacher: And always check with the ....
Student: main inequality or the original inequality
Teacher: Why is that ...(speaking like an American lol)
Student: Because any modified inequality might be incorrect.
Teacher: Good answer!

Symbols and Meanings

Common Mistakes

(1.) Forgot to change(reverse) the inequality after dividing by a negative number.
(2.) Forgot to change(reverse) the inequality after swapping.
(3.) Closed negative infinity.
(4.) Closed positive infinity.

Student: What is that symbol?
Teacher: You mean the eight - upside down or eight - laying down?
Student: ☺☺☺...yes Sir
Teacher: It is ad infinitum (Latin word), the infinity symbol (English word)
Teacher: What is the least number on earth?
Student: Zero?
Teacher: Which one is better - Mr. A who does not have anything and does not owe anything OR Mr. B who has $\$10$ but is owing $\$50$
Student: I get it. Zero is greater than negative forty
Teacher: Correct.
So, what is the least number on earth?
Student: I do not know.
Teacher: Well, there is no least number on earth.
Neither is there a greatest number on earth.
Student: Okay,...so it goes to either negative infinity or positive infinity
Teacher: That is correct!
In that case ...

NOTE: We do not close negative infinity or positive infinity.
We do not ever put a bracket around negative infinity or positive infinity.
We do not close negative infinity because there is no least number on earth.
We do not close positive infinity because there is no greatest number on earth.

Rules of Inequalities

$c, d, e$ are real numbers

$ (1.)\:\:If\:\: c \lt d \:\:and\:\: d \lt e, \:\:then\:\: c \lt e ...Transitive\:\:Rule \\[3ex] (2.)\:\:If\:\: c \gt d \:\:and\:\: d \gt e, \:\:then\:\: c \gt e ...Transitive\:\:Rule \\[3ex] (3.)\:\:If\:\: c \lt d, \:\:then\:\: d \gt c \\[3ex] (4.)\:\:If\:\: c \gt d, \:\:then\:\: d \lt c \\[3ex] (5.)\:\:If\:\: c \lt d, \:\:then\:\: -c \gt -d \\[3ex] (6.)\:\:If\:\: c \gt d, \:\:then\:\: -c \lt -d \\[3ex] (7.)\:\:If\:\: c \lt d, \:\:then\:\: \dfrac{1}{c} \gt \dfrac{1}{d} \\[5ex] (8.)\:\:If\:\: c \gt d, \:\:then\:\: \dfrac{1}{c} \lt \dfrac{1}{d} \\[5ex] (9.)\:\:If\:\: c \lt d, \:\:then\:\: (c + e) \lt (d + e) \\[3ex] (10.)\:\:If\:\: c \gt d, \:\:then\:\: (c + e) \gt (d + e) \\[3ex] (11.)\:\:If\:\: c \lt d, \:\:then\:\: (c - e) \lt (d - e) \\[3ex] (12.)\:\:If\:\: c \gt d, \:\:then\:\: (c - e) \gt (d - e) \\[3ex] (13.)\:\:If\:\: c \lt d, \:\:and\:\: e \gt 0; \:\:then\:\: ce \lt de \\[3ex] (14.)\:\:If\:\: c \lt d, \:\:and\:\: e \lt 0; \:\:then\:\: ce \gt de \\[3ex] (15.)\:\:If\:\: c \gt d, \:\:and\:\: e \gt 0; \:\:then\:\: ce \gt de \\[3ex] (16.)\:\:If\:\: c \gt d, \:\:and\:\: e \lt 0; \:\:then\:\: ce \lt de \\[3ex] (17.)\:\:If\:\: c \lt d, \:\:and\:\: e \gt 0; \:\:then\:\: \dfrac{c}{e} \lt \dfrac{d}{e} \\[5ex] (18.)\:\:If\:\: c \gt d, \:\:and\:\: e \gt 0; \:\:then\:\: \dfrac{c}{e} \gt \dfrac{d}{e} \\[5ex] (19.)\:\:If\:\: c \lt d, \:\:and\:\: e \lt 0; \:\:then\:\: \dfrac{c}{e} \gt \dfrac{d}{e} \\[5ex] (20.)\:\:If\:\: c \gt d, \:\:and\:\: e \lt 0; \:\:then\:\: \dfrac{c}{e} \lt \dfrac{d}{e} $

A linear inequality is an inequality containing linear expressions.

Student: What is a linear expression?
Teacher: A linear expression is an expression in which the highest exponent of the independent variable in the expression is $1$
Please review the pre-requisite topic: Expressions and Equations

A polynomial inequality is an inequality that has a polynomial function.

You can also write that it is an inequality of the forms:
$ f(x) \lt 0 \\[3ex] OR\:\: f(x) \lt some\:\: value/function \\[3ex] OR\:\: f(x) \le 0 \\[3ex] OR\:\: f(x) \le some\:\: value/function \\[3ex] OR\:\: f(x) \gt 0 \\[3ex] OR\:\: f(x) \gt some\:\: value/function \\[3ex] OR\:\: f(x) \ge 0 \\[3ex] OR\:\: f(x) \ge some\:\: value/function \\[3ex] where\:\: f(x) = ax^n + bx^{n - 1} + cx^{n - 2} + ... + z $

Notable Notes for Solving Polynomial Inequalities

(1.) Make sure the $RHS$ (Right Hand Side) is always $0$ before you begin to solve.
This is because $0$ is the only number that "actually" reveals the inequality: greater than, $\gt$; and less than, $\lt$
It is the only number that differentiates positive numbers from negative numbers.
Positive numbers are greater than zero.
Negative numbers are less than zero.

(2.) For "original" (not modified) polynomial inequalities in which $0$ is already on the $RHS$, there is no need to check. Why?
For such polynomial inequalities; we check as we solve. That saves time. ☺☺☺
It is a "check-as-you-solve" kind of thing.

(3.) For all other polynomial inequalities (polynomial inequalities in which we need to modify so that $0$ is on the $RHS$), we have to check our work. And as usual, we have to check with the original/main inequalities.

(4.) For all polynomial inequalities; to get the boundary points, the polynomials on the $LHS$ must be in factored form.
If the polynomial is not in factored form, then we need to factor it in order to get the boundary points.
For the test points, we can use the polynomial "as is" or we can use the factored form (much better).
However, to get the boundary points; it has to be in factored form.

An absolute value inequality is an inequality in which at least one of it's terms has an absolute value.

The absolute value of a term is the magnitude or modulus of that term regardless of sign.
The absolute value of a term say $x$ is denoted by $|x|$
Some resources/calculators represent it as $abs(x)$

Whenever we solve absolute value inequalities, we have to consider two cases.
First Case: The term inside the absolute value is positive.
Second Case: The term inside the absolute value is negative.

And of course, with Mr. C; you have to always Check your solutions

Recall the definitions:

A rational number is any number that can be written as a fraction where the denominator is not equal to zero.
You can also say that a rational number is a ratio of two integers where the denominator is not equal to zero.
A rational number is a number that can be written as: $$\dfrac{c}{d}$$ where $c, d$ are integers and $d \neq 0$
A rational number can be an integer.
It can be a terminating decimal. Why?
It can be a repeating decimal. Why?
It cannot be a non-repeating decimal. Why?
Ask students to tell you what happens if the denominator is zero.

A rational function is a function of the form: $$\dfrac{c(x)}{d(x)}$$ where $c(x), d(x)$ are functions and $d(x) \neq 0$

Based on the prior definitions, ask students to suggest the definition(s) of a rational inequality.

A rational inequality is an inequality containing a rational function.
You can also say that a rational inequality is an inequality of the forms: $$ \dfrac{c(x)}{d(x)} \lt 0 \\[5ex] OR\:\:\: \dfrac{c(x)}{d(x)} \lt some\:\: value/function \\[5ex] OR\:\:\: \dfrac{c(x)}{d(x)} \gt 0 \\[5ex] OR\:\:\: \dfrac{c(x)}{d(x)} \gt some\:\: value/function \\[5ex] OR\:\:\: \dfrac{c(x)}{d(x)} \le 0 \\[5ex] OR\:\:\: \dfrac{c(x)}{d(x)} \le some\:\: value/function \\[5ex] OR\:\:\: \dfrac{c(x)}{d(x)} \ge 0 \\[5ex] OR\:\:\: \dfrac{c(x)}{d(x)} \ge some\:\: value/function \\[5ex] where\:\: c(x), d(x) \:\:are\:\: polynomial\:\: functions\:\: and\:\: d(x) \neq 0 $$

Notable Notes for Solving Rational Inequalities

(1.) Make sure the $RHS$ (Right Hand Side) is always $0$ before you begin to solve.
This is because $0$ is the only number that "actually" reveals the inequality: greater than, $\gt$; and less than, $\lt$
It is the only number that differentiates positive numbers from negative numbers.
Positive numbers are greater than zero.
Negative numbers are less than zero.

(2.) For "original" (not modified) rational inequalities in which $0$ is already on the $RHS$, there is no need to check. Why?
For such rational inequalities; we check as we solve. That saves time. ☺☺☺
It is a "check-as-you-solve" kind of thing.

(3.) For all other rational inequalities (rational inequalities in which we need to modify so that $0$ is on the $RHS$), we have to check our work. And as usual, we have to check with the original/main inequalities.

(4.) For rational inequalities that involves greater than or equal to, $\ge$; and less than or equal to, $\le$; it is important we note the domain of that inequality before writing our solution.

(5.) For all rational inequalities that we solve, we should have:
Only one fraction/rational function on the $LHS$ and
$0$ on the $RHS$
If we have several fractions/rational functions, we should bring all of them to the $LHS$, and we should simplify them to be only one fraction/rational fraction.
We should solve them as fractions. We should never attempt to remove the fractions.
Here is another major problem caused by Inequalities. I informed you earlier that Inequality does not only cause problem in the real-world. It causes problems in Mathematics.
NOTE: Please do not solve rational inequalities by multiplying each term by the $LCD$ (like we did in Rational Equations).
We have to solve the Rational Inequalities as fractions. We will not get rid of the fractions. We just have to like fractions...and we just have to promote equality in our world!!!
One of the easiest ways of solving Rational Equations is to get rid of the fractions by multiplying each term by the $LCD$.
Well, with Rational Inequality...not so. We have to solve them as fractions.
We need to get all the fractions on the $LHS$, and simplify them as one fraction.
We need to have only $0$ on the $RHS$
Then, after solving; we need to check with the original inequality.

(6.) For all rational inequalities; to get the boundary points:
the numerator polynomial (as applicable) and denominator polynomial on the $LHS$ must be in factored form.
For the test points, we can use the rational inequality "as is" or the factored form (much better).
However, to get the boundary points; we have to put both the numerator and the denominator in factored form.