Algebra/Equalities and Inequalities

Algebra

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:

<--|-----|-----|-----|-----|-----|-----|-----|-----|-----|-----|----->
  -5    -4    -3    -2    -1     0     1     2     3     4     5

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:

1. $a>b$,
2. $a=b$, or
3. $a<b$

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:

<-------------------o
<-+-----+-----+-----+-----+-----+-->
  6     7     8     9     10    11

(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.

              ●------------------->
<-+-----+-----+-----+-----+-----+-->
 -6    -5    -4    -3    -2    -1

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:

1. Find out when the denominator is equal to zero. In the above example the denominator equals zero when $x=1$.
2. Pretend the inequality sign is an $=$ sign and solve it as such: ${\frac {2}{x-1}}=2\,$, so $x=2$.
3. 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 $x>2$.
4. 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."

        ●-----------------●
<-+-----+-----+-----+-----+-----+-->
 -6    -4    -2     0     2     4

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.

<-------------o           ●-------->
<-+-----+-----+-----+-----+-----+-->
 -1     0     1     2     3     4

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 $1<x<11$

        ●----------------------------><-----------------------------●
<-+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+-->
  0     1     2     3     4     5     6     7     8     9     10    11    12

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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