# Algebra/Chapter 4/Inequalities
## Definition of Inequalities

As opposed to an equation, an inequality is an expression that states that two quantities are unequal or not equivalent to one another. In real life we often use inequalities more than equations (i.e. this shirt costs $2 more than that one).
Given that a and b are real numbers, there are four basic inequalities:
* a < b --> a is "less than" b
    Example: 2 < 4 ; -3 < 0; etc.
* a > b --> a is "greater than" b
    Example: -2 > -4 ; 3 > 0 ; etc.
* $a\leq b$ --> a is "less than or equal to" b
    Example: If we know that $x\leq 7$, then we can conclude that x is equal to any value less than 7, including 7 itself.
* $a\geq b$ --> a is "greater than or equal to" b
    Example: Conversely, if $x\geq 7$, then x is equal to any value greater than 7, including 7 itself.

## Possible Relationships

A number on the number line is always greater than any number on its left and less than any number on its right. The symbol "$<$" is used to represent "is less than", and "$>$" to represent "is greater than".
Consider this number line:
From the number line, we can easily tell that 3 is greater than -2, because 3 is on the right side of -2 (or -2 is on the left of 3). We write it as $3>-2$. A teacher in elementary school told me to think of the greater sign as a "greedy mouth". The mouth always tries to eat the bigger number. When I look at the number line I can see that the greater and lesser signs could also represent the arrows at the ends of the number line. In this case the inequality signs represent arrows that point which direction I have to go to get to the first number from the second number. If the second number is smaller than the first than I have to move to the right on the number line to go from the second number to the first (which is why the "mouth" points toward the first number). Similarly if the second number is larger than the first then I have to move to the left on the number line (and the "mouth" now points toward the second number). I could never remember my teacher's rule of thumb for the "greedy mouth", but when I thought of the sign as the arrow heads at the end of the line the direction of the operator made sense to me. How are you going to remember that "$<$" is used to represent "is less than", and "$>$" to represent "is greater than"?
Consider a number $a$ and a constant $C$. One and only one of the following statements can be true:
* $a>C$
* $a=C$, or
* $a<C$
This is the Law of Trichotomy. Another way of describing it is that given a fixed number and a variable the variable can represent numbers smaller than the number, at the same point as the number, or that are larger than the number. When we create inequalities from words using symbols we know that only one of these statements can be true.
Word Problems

### Properties

There are four important properties with inequalities: 1. Transitive property: For any three numbers $x$, $y$ and $z$, if $x>y$ and $y>z$, then $x>z$.
2. In an inequality, we can add or subtract the same value from both sides, without changing the sign (i.e. "$>$" or "$<$"). That is to say, for any three numbers $x$, $y$ and $p$:
* if $x>y$, then $x+p>y+p$ and $x-p>y-p$.
3. We can multiply or divide both sides by a positive number without changing the sign. For example, if we have any two numbers $x$ and $y$, and another positive number $p$:
* if $x>y$, then $x\times p>y\times p$ and ${\frac {x}{p}}>{\frac {y}{p}}$.
4. When we multiply or divide both sides by a negative number, we have to change the sign of the inequality (i.e., "$>$" changes to "$<$" and vice versa). So if we are given two numbers $x$ and $y$, and another negative number $p$
    if $x>y$, $x\times p<y\times p$ and ${\frac {x}{p}}<{\frac {y}{p}}$.
Now we can go on to solve any linear inequalities.

## Solving Inequalities

Solving inequalities is almost the same as solving linear equations. Let's consider an example: $x+4<13$. All we have to do is subtract 4 on both sides. We will then get $x<9$, and that is the answer! Note, however, that what you get is not a single answer, but a set of solutions. Any number that satisfies the condition $x<9$ (any number that is less than 9) is a solution to the inequality. It is very convenient to represent the solution using the number line:
Note: the circle(o) shows that the value 9 is not included. Later on, when we deal with less than or equal to and greater than or equal to (≤ and ≥), we use "*" to show that the value is included in the solution set.
Let's try another more complicated question: $3x-2>2(x-3)$. First, you may want to expand the right hand side: $3x-2>2x-6$. Then we can simply rearrange so that all the unknowns are on one side (usually the left): $3x-2x>-6+2$. Hence, we can easily get the answer: $x>-4$.
Here's an example where the direction of the inequality changes when finding the solution: solve $4-6x>22$.
* First subtract 4 from both sides: $-6x>18$.
* Now divide through by -6, changing the direction of the inequality: ${\frac {-6x}{-6}}<{\frac {18}{-6}}$.
So the solution to the inequality is $x<-3$.
Inequalities, unlike equations, commonly have an infinite number of solutions.
$x>A\!\$ means that x is greater than A
$x<A\!\$ means that x is less than A

### Special Cases - Multiplying by -1

The rule for multiplying or dividing each side of an inequality by a negative number states that you also need to change the direction of the inequality. Look at the inequality -1 < 1. Let's rewrite this generically as -1 op 1, although we know that the operator should be <. Now multiply both sides of the inequality by -1. You get (-1)*(-1) op (1)*(-1). Simplifying you get 1 op -1. For this statement to be true the operator now needs to be >.
If we wanted to we could change multiplying by a negative number into two operations: [(-1)*(Number)][-1 < 1][(Number)*(-1)]. This may seem obvious since we know that all we have to do to get a number to turn into its opposite is multiply by -1. But, what is it about multiplying by -1 that causes us to have to change the direction of the inequality operator? If you represent multiplication by the number 1 on a number line you see that all you do is move x(the number you are multiplying) units to the right of zero.
TODO: Need Graphic
Similarly, if you represent multiplication by the number -1 on a number line you see that you need to move x(the number you are multiplying) units to the left of zero.
TODO: Need Graphic
Now think about the difference between an equality and an inequality. If you know that x = y is true then you also know that x-y = 0. If you know that x > y then you know two things: that x - y > 0 (bigger - smaller is positive) and also that y - x < 0 (smaller - bigger is negative). If we multiply x > y by 1 we still get x > y. If we move y to the left side of the equation by subtracting y from both sides we get x - y > 0. Which we know is true. On the other hand when we multiply x > y by -1 and don't change the direction of the inequality we get -x > -y. When we move the y to the left side of the equation by adding y to both sides we get y - x > 0. Which we know is not true. In the previous step when we needed to say -x < -y because even though both numbers are negative, -x is further to the left, and therefore smaller, than -y.
Working algebra problems can become automatic, but it is important for you to keep in mind why the rules are true so that you can catch yourself if you accidentally forget part of a rule.

### Special Cases - Inequalities with a variable in the denominator

For example, consider the inequality
    ${\frac {2}{x-1}}<2\,$
In this case one cannot multiply the right hand side by (x-1) because the value of x is unknown. Since x may be either positive or negative, you can't know whether to leave the inequality sign as <, or reverse it to >. The method for solving this kind of inequality involves four steps:
* Find out when the denominator is equal to 0. In this case it's when $x=1$.
* Pretend the inequality sign is an = sign and solve it as such: ${\frac {2}{x-1}}=2\,$, so $x=2$.
* Plot the points $x=1$ and $x=2$ on a number line with an unfilled circle because the original equation included a < sign (note that it would have been a filled circle if the original equation included <= or >=). You now have three regions which are separated by unfilled circles. These regions are: $x<1$, $1<x<2$, and $x>2$.
* Test each region independently. in this case test if the inequality is true for 1<x<2 by picking a point in this region (e.g. x=1.5) and trying it in the original inequation. For x=1.5, the original inequality doesn't hold. Now, attempt x>2 (e.g. x=3). In this case the original inequation holds, and so the solution for the original inequation is x>2.

### Special Cases - Going back to equality

So far, you have seen various kinds of inequalities. However, there are still two kinds of inequalities that we must discuss:
* $x\geq A$ means that x is greater than or equal to A;
* $x\leq A$ means that x is less than or equal to A.
These kinds of inequalities also include the possibility of $x=A$. In most cases, they can be obtained if we negate simple inequalities like $x>A$ or ${x<A}$. For example, consider the inequality
$-5x>3$
To solve this inequality, we have to divide both sides by -5 (thus negating the inequality and isolating x). Notice that the right hand side (RHS) of the inequality will be negative ($-{\frac {3}{5}}$) now. So we must flip the sign.
In essence, to flip the $>$ sign, just replace it in the inequality with $\leq$ or vice versa. On the other hand, in order to flip the $<$ sign, replace it with $\geq$ or vice versa. Keeping all this in mind, we can continue solving the inequality above:
$x>{-{\frac {3}{5}}}$. After dividing both sides by -5, notice that we have an isolated x and a negative RHS. Flipping the sign gives:
$x\leq -{\frac {3}{5}}$. Which is the final answer!
Anyway, if you encounter inequalities with negative numbers, try to divide the whole inequality by the number next to x (so as to isolate x), flip the sign, and simplify.
Of course there are harder examples, where there are inequalities with multiple special cases. So, here's a tip: if you have a negative ratio with x in the denominator, just write a little line below the sign (since it gets inverted twice, due to negation and inversion). Just make sure to apply the negative sign to both sides first and then invert both sides after.
Solving linear inequalities involves finding solutions to expressions where the quantities are not equal.
A number on the number line is always greater than any number on its left and smaller than any number on its right. The symbol "<" is used to represent "is less than", and ">" to represent "is greater than".
For example:
From the number line, we can easily tell that 3 is greater than -2, because 3 is on the right side of -2 (or -2 is on the left of 3). We write it as $3>-2$ (or as $-2<3$). We can also derive that any positive number is always greater than negative number.
Consider any two numbers, a and b. One and only one of the following statements can be true:
* $a>b$,
* $a=b$, or
* $a<b$
This is the Law of Trichotomy.
For an inequality with one unknown, there may be many (sometimes infinite) possible solutions.

## Properties

* Transitive property:
       For any three numbers $x$, $y$, $z$, if $x>y$ and $y>z$, then $x>z$.
* Additive property:
       In an inequality, we can add or subtract the same value from both sides, without changing the sign (i.e. ">" or "<"). That is to say, for any three numbers $x$, $y$ and $p$, if $x>y$, then $x+p>y+p$ and $x-p>y-p$.
* Multiplicative property
       We can multiply or divide both sides by a positive number without changing the sign. For example, if we have any two numbers $x$ and $y$, and another positive number $p$, then if $x>y$, then $x\cdot p>y\cdot p$ and ${\frac {x}{p}}>{\frac {y}{p}}$.
       When we multiply or divide both sides by a negative number, we have to change the sign of the inequality (i.e, ">" change to "<" and vice versa). So if we are given two numbers $x$ and $y$, and another negative number $p$, then if $x>y$, $x\cdot p<y\cdot p$ and ${\frac {x}{p}}<{\frac {y}{p}}$.
Now we can go on to solve any linear inequalities.

## Solving Inequalities

Solving inequalities is almost the same as solving linear equations. Let's consider an example: $x+4<13$. All we have to do is to subtract 4 on both sides. We will then get $x<9$, and that is the answer! Note, however, what you get is not a single answer, but a set of solutions, i.e., any number that satisfies the condition $x<9$ (any number that is less than 9) can be a solution to the inequality. It is very convenient to represent the solution using the number line:
(Note: the open circle ("o") shows that the value 9 is not included in the solution set, as the inequality of this equation is less than 9, not less than or equal to 9. When we deal with less (greater) than or equal to (≤ or ≥) later on, we use a closed circle ("●") to show that the value is included in the solution set.)
Let us try another more complicated question: $3x-2\geq 2(x-3)$. First, you may want to expand the right hand side: $3x-2\geq 2x-6$. Then we can simply rearrange the terms so that all the unknown variables are on one side of the equation, usually the left hand side: $3x-2x\geq -6+2$. Hence we can easily get the answer: $x\geq -4$. This solution is represented on the number line below. Note that the solution requires a closed circle ("●"), because the $x$ is greater than or equal to 4.

### Inequalities with a variable in the denominator

For example consider the inequality
    ${\frac {2}{x-1}}<2\,$
In this case one cannot multiply the right hand side by $(x-1)$ because the value of x is unknown. Since x may be either positive or negative, you can't know whether to leave the inequality sign as $<$ (ie less than), or reverse it to > (ie greater than). The method for solving this kind of inequality involves four steps:
* Find out when the denominator is equal to zero. In the above example the denominator equals zero when $x=1$.
* Pretend the inequality sign is an $=$ sign and solve it as such: ${\frac {2}{x-1}}=2\,$, so $x=2$.
* Plot the points $x=1$ and $x=2$ on a number line with an unfilled circle because the original equation included < (it would have been a filled circle if the original equation included $\leq$ or $\geq$). You now have three regions: $x<1$, $1<x<2$, and $x>2$.
* Test each region independently. in this case test if the inequality is true for $1<x<2$ by picking a point in this region (e.g. $x=1.5$) and trying it in the original inequation. For x=1.5 the original inequation doesn't hold. So then try for $1>x>2)$ (e.g. $x=3)$). In this case the original inequation holds, and so the solution for the original inequation is $1>x>2)$.

## Compound Inequalities

A compound inequality is a pair of inequalities related by the words and or or. In an and inequality, both inequalities must be satisfied. All possible solution values will be located between two defined numbers, and if this is impossible, the compound inequality simply has no solutions.
Consider this example: $x+6\geq 2$ and $x\leq 2$. First, solve the first inequality for x to get $x\geq -4$. All and inequalities can be rewritten as one inequality, like this: $-4\leq$$x\leq 2$ (write x between two ≤'s or <'s or both with the smaller number on the left and the larger number on the right). Now, we can graph this inequality on a number line as a line segment. Remember, all solutions to ≤ or ≥ must be graphed with closed circles. Interpret this graphic as "all numbers between -4 and 2, including -4 and 2."
Now, let us consider or inequalities. Or inequalities usually do not have a set of solutions that satisfies both. Instead, they usually have two sets of infinite numbers that are solutions to each one. Because of this, or graphs define which numbers satisfy either equation. For example: $x\;<1$ or $x-1\geq 2$. First, solve for x in the second inequality to get $x\geq 3$. Now, graph the two inequalities on the same number line. Remember to use open and closed circles accordingly.

## Solving Inequalities with Absolute Value

Since $'''|x|=|-x|'''$ A inequality involving absolute value will have to solved in two parts.
Solving $'''|x-6|<5'''$
The first part would be $'''x-6<5'''$ which gives $x<11$. The second part would be $'''-(x-6)<5'''$ which solved yields $x>1$.
So the answer to $|x-6|<5$ is $1<x<11$

## Graphing Linear Inequalities

The graphing of linear inequalities is very similar to the graphing of linear functions. A linear inequality is written in
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