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Linear Inequalities and Absolute Value Inequalities

Use interval notation

Using Interval Notation

Indicating the solution to an inequality such as $x\ge 4$ can be achieved in several ways.

We can use a number line as shown in . The blue ray begins at $x=4$ and, as indicated by the arrowhead, continues to infinity, which illustrates that the solution set includes all real numbers greater than or equal to 4.

We can use set-builder notation: $\{x|x\ge 4\},$ which translates to “all real numbers x such that x is greater than or equal to 4.” Notice that braces are used to indicate a set.

The third method is interval notation, in which solution sets are indicated with parentheses or brackets. The solutions to $x\ge 4$ are represented as $[4,\infty ).$ This is perhaps the most useful method, as it applies to concepts studied later in this course and to other higher-level math courses.

The main concept to remember is that parentheses represent solutions greater or less than the number, and brackets represent solutions that are greater than or equal to or less than or equal to the number. Use parentheses to represent infinity or negative infinity, since positive and negative infinity are not numbers in the usual sense of the word and, therefore, cannot be “equaled.” A few examples of an interval, or a set of numbers in which a solution falls, are $[-2,6),$ or all numbers between $-2$ and $6,$ including $-2,$ but not including $6;$ $(-1,0),$ all real numbers between, but not including $-1$ and $0;$ and $(-\infty ,1],$ all real numbers less than and including $1.$ outlines the possibilities.

Set Indicated	Set-Builder Notation	Interval Notation
All real numbers between a and b, but not including a or b	\(\{x\|a	$(a,b)$
All real numbers greater than a, but not including a	$\{x\|x>a\}$	$(a,\infty )$
All real numbers less than b, but not including b	\(\{x\|x	$(-\infty ,b)$
All real numbers greater than a, including a	$\{x\|x\ge a\}$	$[a,\infty )$
All real numbers less than b, including b	$\{x\|x\le b\}$	$(-\infty ,b]$
All real numbers between a and b, including a	\(\{x\|a\le x	$[a,b)$
All real numbers between a and b, including b	\(\{x\|a	$(a,b]$
All real numbers between a and b, including a and b	$\{x\|a\le x\le b\}$	$[a,b]$
All real numbers less than a or greater than b	$\{x\|xb\}$	$(-\infty ,a)\cup (b,\infty )$
All real numbers	$\{x\|x\ \text{is all real numbers}\}$	$(-\infty ,\infty )$

Example

Try it.

Use interval notation to indicate all real numbers greater than or equal to $-2.$

Solution

Use a bracket on the left of $-2$ and parentheses after infinity: $[-2,\infty ).$ The bracket indicates that $-2$ is included in the set with all real numbers greater than $-2$ to infinity.

Condensed — the full section is in OpenStax College Algebra 2e.

Using the Properties of Inequalities

When we work with inequalities, we can usually treat them similarly to but not exactly as we treat equalities. We can use the addition property and the multiplication property to help us solve them. The one exception is when we multiply or divide by a negative number; doing so reverses the inequality symbol.

Example

Try it.

Illustrate the addition property for inequalities by solving each of the following:

ⓐ $x-15<4$
ⓑ $6\ge x-1$
ⓒ $x+7>9$

Solution

The addition property for inequalities states that if an inequality exists, adding or subtracting the same number on both sides does not change the inequality.

ⓐ
$\begin{array}{ll}\ x-15<4 & \\ x-15+15<4+15 & \ \text{Add 15 to both sides}. \\ x<19 & \end{array}$
ⓑ
$\begin{array}{ll}\ 6\ge x-1 & \\ 6+1\ge x-1+1 & \ \text{Add 1 to both sides}. \\ 7\ge x & \end{array}$
ⓒ
$\begin{array}{ll}\ x+7>9 & \\ x+7-7>9-7 & \ \text{Subtract 7 from both sides}. \\ x>2 & \end{array}$

Example

Try it.

Illustrate the multiplication property for inequalities by solving each of the following:

ⓐ $3x<6$
ⓑ $-2x-1\ge 5$
ⓒ $5-x>10$

Solution

ⓐ
$\begin{array}{l}\ 3x<6 \\ \frac{1}{3}(3x)<(6)\frac{1}{3} \\ x<2\end{array}$
ⓑ
$\begin{array}{ll}\ -2x-1\ge 5 & \\ -2x\ge 6 & \\ (-\frac{1}{2})(-2x)\ge (6)(-\frac{1}{2}) & \ \text{Multiply by }-\frac{1}{2}. \\ x\le -3 & \ \text{Reverse the inequality}.\end{array}$
ⓒ
$\begin{array}{ll}\ 5-x>10 & \\ -x>5 & \\ (-1)(-x)>(5)(-1) & \ \text{Multiply by }-1. \\ x<-5 & \ \text{Reverse the inequality}.\end{array}$

Solving Inequalities in One Variable Algebraically

As the examples have shown, we can perform the same operations on both sides of an inequality, just as we do with equations; we combine like terms and perform operations. To solve, we isolate the variable.

Example

Try it.

Solve the inequality: $13-7x\ge 10x-4.$

Solution

Solving this inequality is similar to solving an equation up until the last step.

\[\begin{array}{ll}\ 13-7x\ge 10x-4 & \\ 13-17x\ge -4 & \ \text{Move variable terms to one side of the inequality}. \\ -17x\ge -17 & \ \text{Isolate the variable term}. \\ x\le 1 & \ \text{Dividing both sides by }-17\ \text{reverses the inequality}.\end{array}\]

The solution set is given by the interval $(-\infty ,1],$ or all real numbers less than and including 1.

Example

Try it.

Solve the following inequality and write the answer in interval notation: $-\frac{3}{4}x\ge -\frac{5}{8}+\frac{2}{3}x.$

Solution

We begin solving in the same way we do when solving an equation.

\[\begin{array}{ll}\ -\frac{3}{4}x\ge -\frac{5}{8}+\frac{2}{3}x & \\ -\frac{3}{4}x-\frac{2}{3}x\ge -\frac{5}{8} & \ \text{Put variable terms on one side}. \\ -\frac{9}{12}x-\frac{8}{12}x\ge -\frac{5}{8} & \ \text{Write fractions with common denominator}. \\ -\frac{17}{12}x\ge -\frac{5}{8} & \\ x\le -\frac{5}{8}(-\frac{12}{17}) & \ \text{Multiplying by a negative number reverses the inequality}. \\ x\le \frac{15}{34} & \end{array}\]

The solution set is the interval $(-\infty ,\frac{15}{34}].$

Understanding Compound Inequalities

A compound inequality includes two inequalities in one statement. A statement such as \(4

Example

Try it.

Solve the compound inequality: $3\le 2x+2<6.$

Solution

The first method is to write two separate inequalities: $3\le 2x+2$ and $2x+2<6.$ We solve them independently.

\[\begin{array}{lll}3\le 2x+2 & \ \text{and}\ & 2x+2<6 \\ 1\le 2x & & \ 2x<4 \\ \frac{1}{2}\le x & & \ x<2\end{array}\]

Then, we can rewrite the solution as a compound inequality, the same way the problem began.

\[\frac{1}{2}\le x<2\]

In interval notation, the solution is written as $[\frac{1}{2},2).$

The second method is to leave the compound inequality intact, and perform solving procedures on the three parts at the same time.

\[\begin{array}{ll}3\le 2x+2<6 & \\ 1\le 2x<4 & \ \text{Isolate the variable term, and subtract 2 from all three parts}. \\ \frac{1}{2}\le x<2 & \ \text{Divide through all three parts by 2}.\end{array}\]

We get the same solution: $[\frac{1}{2},2).$

Example

Try it.

Solve the compound inequality with variables in all three parts: $3+x>7x-2>5x-10.$

Solution

Let's try the first method. Write two inequalities:

\[\begin{array}{lll}3+x>7x-2 & \ \text{and}\ & 7x-2>5x-10 \\ 3>6x-2 & & 2x-2>-10 \\ 5>6x & & \ 2x>-8 \\ \frac{5}{6}>x & & \ x>-4 \\ x<\frac{5}{6} & & \ -4The solution set is \(-4.

Solving Absolute Value Inequalities

As we know, the absolute value of a quantity is a positive number or zero. From the origin, a point located at $(-x,0)$ has an absolute value of $x,$ as it is x units away. Consider absolute value as the distance from one point to another point. Regardless of direction, positive or negative, the distance between the two points is represented as a positive number or zero.

An absolute value inequality is an equation of the form

\[|A|B,\ \text{or }\ |A|\ge B,\]

Where A, and sometimes B, represents an algebraic expression dependent on a variable x. Solving the inequality means finding the set of all $x$ -values that satisfy the problem. Usually this set will be an interval or the union of two intervals and will include a range of values.

There are two basic approaches to solving absolute value inequalities: graphical and algebraic. The advantage of the graphical approach is we can read the solution by interpreting the graphs of two equations. The advantage of the algebraic approach is that solutions are exact, as precise solutions are sometimes difficult to read from a graph.

Suppose we want to know all possible returns on an investment if we could earn some amount of money within $200 of $600. We can solve algebraically for the set of x-values such that the distance between $x$ and 600 is less than or equal to 200. We represent the distance between $x$ and 600 as $|x-600|,$ and therefore, $|x-600|\le 200$ or

\[\begin{array}{l}-200\le x-600\le 200 \\ -200+600\le x-600+600\le 200+600 \\ 400\le x\le 800\end{array}\]

This means our returns would be between $400 and $800.

To solve absolute value inequalities, just as with absolute value equations, we write two inequalities and then solve them independently.

Example

Try it.

Solve $|x-1|\le 3$ .

Solution

\[\begin{array}{l}|x-1|\le 3 \\ \\ -3\le x-1\le 3 \\ \\ -2\le x\le 4 \\ \\ [-2,4]\end{array}\]

Condensed — the full section is in OpenStax College Algebra 2e.

Key Concepts

Interval notation is a method to indicate the solution set to an inequality. Highly applicable in calculus, it is a system of parentheses and brackets that indicate what numbers are included in a set and whether the endpoints are included as well. See and .
Solving inequalities is similar to solving equations. The same algebraic rules apply, except for one: multiplying or dividing by a negative number reverses the inequality. See , , , and .
Compound inequalities often have three parts and can be rewritten as two independent inequalities. Solutions are given by boundary values, which are indicated as a beginning boundary or an ending boundary in the solutions to the two inequalities. See and .
Absolute value inequalities will produce two solution sets due to the nature of absolute value. We solve by writing two equations: one equal to a positive value and one equal to a negative value. See and .
Absolute value inequalities can also be solved by graphing. At least we can check the algebraic solutions by graphing, as we cannot depend on a visual for a precise solution. See .

Chapter Practice Test

For the following exercises, find the equation of the line with the given information.

For the following exercises, find the real solutions of each equation by factoring.

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Use interval notation to indicate all real numbers greater than or equal to $-2.$

答えを明らかにしろ

Use a bracket on the left of $-2$ and parentheses after infinity: $[-2,\infty ).$ The bracket indicates that $-2$ is included in the set with all real numbers greater than $-2$ to infinity.
Use interval notation to indicate all real numbers between and including $-3$ and $5.$

答えを明らかにしろ

$[-3,5]$
Write the interval expressing all real numbers less than or equal to $-1$ or greater than or equal to $1.$

答えを明らかにしろ

We have to write two intervals for this example. The first interval must indicate all real numbers less than or equal to 1. So, this interval begins at $-\infty$ and ends at $-1,$ which is written as $(-\infty ,-1].$
The second interval must show all real numbers greater than or equal to $1,$ which is written as $[1,\infty ).$ However, we want to combine these two sets. We accomplish this by inserting the union symbol, $\cup ,$ between the two intervals.
\[(-\infty ,-1]\cup [1,\infty )\]
Express all real numbers less than $-2$ or greater than or equal to 3 in interval notation.

答えを明らかにしろ

$(-\infty ,-2)\cup [3,\infty )$
Illustrate the addition property for inequalities by solving each of the following:
1. ⓐ $x-15<4$
2. ⓑ $6\ge x-1$
3. ⓒ $x+7>9$
答えを明らかにしろ
The addition property for inequalities states that if an inequality exists, adding or subtracting the same number on both sides does not change the inequality.
1. ⓐ
  $\begin{array}{ll}\ x-15<4 & \\ x-15+15<4+15 & \ \text{Add 15 to both sides}. \\ x<19 & \end{array}$
2. ⓑ
  $\begin{array}{ll}\ 6\ge x-1 & \\ 6+1\ge x-1+1 & \ \text{Add 1 to both sides}. \\ 7\ge x & \end{array}$
3. ⓒ
  $\begin{array}{ll}\ x+7>9 & \\ x+7-7>9-7 & \ \text{Subtract 7 from both sides}. \\ x>2 & \end{array}$
Solve: $3x-2<1.$

答えを明らかにしろ

$x<1$
Illustrate the multiplication property for inequalities by solving each of the following:
1. ⓐ $3x<6$
2. ⓑ $-2x-1\ge 5$
3. ⓒ $5-x>10$
答えを明らかにしろ
1. ⓐ
  $\begin{array}{l}\ 3x<6 \\ \frac{1}{3}(3x)<(6)\frac{1}{3} \\ x<2\end{array}$
2. ⓑ
  $\begin{array}{ll}\ -2x-1\ge 5 & \\ -2x\ge 6 & \\ (-\frac{1}{2})(-2x)\ge (6)(-\frac{1}{2}) & \ \text{Multiply by }-\frac{1}{2}. \\ x\le -3 & \ \text{Reverse the inequality}.\end{array}$
3. ⓒ
  $\begin{array}{ll}\ 5-x>10 & \\ -x>5 & \\ (-1)(-x)>(5)(-1) & \ \text{Multiply by }-1. \\ x<-5 & \ \text{Reverse the inequality}.\end{array}$
Solve: $4x+7\ge 2x-3.$

答えを明らかにしろ

$x\ge -5$
Solve the inequality: $13-7x\ge 10x-4.$

答えを明らかにしろ

Solving this inequality is similar to solving an equation up until the last step.
\[\begin{array}{ll}\ 13-7x\ge 10x-4 & \\ 13-17x\ge -4 & \ \text{Move variable terms to one side of the inequality}. \\ -17x\ge -17 & \ \text{Isolate the variable term}. \\ x\le 1 & \ \text{Dividing both sides by }-17\ \text{reverses the inequality}.\end{array}\]
The solution set is given by the interval $(-\infty ,1],$ or all real numbers less than and including 1.
Solve the inequality and write the answer using interval notation: $-x+4<\frac{1}{2}x+1.$

答えを明らかにしろ

$(2,\infty )$
Solve the following inequality and write the answer in interval notation: $-\frac{3}{4}x\ge -\frac{5}{8}+\frac{2}{3}x.$

答えを明らかにしろ

We begin solving in the same way we do when solving an equation.
\[\begin{array}{ll}\ -\frac{3}{4}x\ge -\frac{5}{8}+\frac{2}{3}x & \\ -\frac{3}{4}x-\frac{2}{3}x\ge -\frac{5}{8} & \ \text{Put variable terms on one side}. \\ -\frac{9}{12}x-\frac{8}{12}x\ge -\frac{5}{8} & \ \text{Write fractions with common denominator}. \\ -\frac{17}{12}x\ge -\frac{5}{8} & \\ x\le -\frac{5}{8}(-\frac{12}{17}) & \ \text{Multiplying by a negative number reverses the inequality}. \\ x\le \frac{15}{34} & \end{array}\]
The solution set is the interval $(-\infty ,\frac{15}{34}].$
Solve the inequality and write the answer in interval notation: $-\frac{5}{6}x\le \frac{3}{4}+\frac{8}{3}x.$

答えを明らかにしろ

$[-\frac{3}{14},\infty )$
Solve the compound inequality: $3\le 2x+2<6.$

答えを明らかにしろ

The first method is to write two separate inequalities: $3\le 2x+2$ and $2x+2<6.$ We solve them independently.
\[\begin{array}{lll}3\le 2x+2 & \ \text{and}\ & 2x+2<6 \\ 1\le 2x & & \ 2x<4 \\ \frac{1}{2}\le x & & \ x<2\end{array}\]
Then, we can rewrite the solution as a compound inequality, the same way the problem began.
\[\frac{1}{2}\le x<2\]
In interval notation, the solution is written as $[\frac{1}{2},2).$
The second method is to leave the compound inequality intact, and perform solving procedures on the three parts at the same time.
\[\begin{array}{ll}3\le 2x+2<6 & \\ 1\le 2x<4 & \ \text{Isolate the variable term, and subtract 2 from all three parts}. \\ \frac{1}{2}\le x<2 & \ \text{Divide through all three parts by 2}.\end{array}\]
We get the same solution: $[\frac{1}{2},2).$
Solve the compound inequality: $4<2x-8\le 10.$

答えを明らかにしろ

\(6
Solve the compound inequality with variables in all three parts: $3+x>7x-2>5x-10.$

答えを明らかにしろ

Let's try the first method. Write two inequalities:
\[\begin{array}{lll}3+x>7x-2 & \ \text{and}\ & 7x-2>5x-10 \\ 3>6x-2 & & 2x-2>-10 \\ 5>6x & & \ 2x>-8 \\ \frac{5}{6}>x & & \ x>-4 \\ x<\frac{5}{6} & & \ -4The solution set is \(-4.
Solve the compound inequality: $3y<4-5y<5+3y.$

答えを明らかにしろ

$(-\frac{1}{8},\frac{1}{2})$
Describe all values $x$ within a distance of 4 from the number 5.

答えを明らかにしろ

We want the distance between $x$ and 5 to be less than or equal to 4. We can draw a number line, such as in , to represent the condition to be satisfied.
The distance from $x$ to 5 can be represented using an absolute value symbol, $|x-5|.$ Write the values of $x$ that satisfy the condition as an absolute value inequality.
\[|x-5|\le 4\]
We need to write two inequalities as there are always two solutions to an absolute value equation.
\[\begin{array}{lll}x-5\le 4 & \ \text{and}\ & x-5\ge -4 \\ x\le 9 & & \ x\ge 1\end{array}\]
If the solution set is $x\le 9$ and $x\ge 1,$ then the solution set is an interval including all real numbers between and including 1 and 9.
So $|x-5|\le 4$ is equivalent to $[1,9]$ in interval notation.
Describe all x-values within a distance of 3 from the number 2.

答えを明らかにしろ

$|x-2|\le 3$
Solve $|x-1|\le 3$ .

答えを明らかにしろ

\[\begin{array}{l}|x-1|\le 3 \\ \\ -3\le x-1\le 3 \\ \\ -2\le x\le 4 \\ \\ [-2,4]\end{array}\]
Given the equation $y=-\frac{1}{2}|4x-5|+3,$ determine the x-values for which the y-values are negative.

答えを明らかにしろ

We are trying to determine where $y<0,$ which is when $-\frac{1}{2}|4x-5|+3<0.$ We begin by isolating the absolute value.
\[\begin{array}{ll}-\frac{1}{2}|4x-5|<-3 & \ \text{Multiply both sides by -2, and reverse the inequality}. \\ |4x-5|>6 & \end{array}\]
Next, we solve for the equality $|4x-5|=6.$
\[\begin{array}{lll}4x-5=6 & & 4x-5=-6 \\ 4x=11 & \ \text{or}\ & \ 4x=-1 \\ x=\frac{11}{4} & & \ x=-\frac{1}{4}\end{array}\]
Now, we can examine the graph to observe where the y-values are negative. We observe where the branches are below the x-axis. Notice that it is not important exactly what the graph looks like, as long as we know that it crosses the horizontal axis at $x=-\frac{1}{4}$ and $x=\frac{11}{4},$ and that the graph opens downward. See .
Solve $-2|k-4|\le -6.$

答えを明らかにしろ

$k\le 1$ or $k\ge 7;$ in interval notation, this would be $(-\infty ,1]\cup [7,\infty ).$
When solving an inequality, explain what happened from Step 1 to Step 2:
$\begin{array}{ll}\text{Step 1} & \ -2x>6 \\ \text{Step 2} & \ x<-3\end{array}$

答えを明らかにしろ

When we divide both sides by a negative it changes the sign of both sides so the sense of the inequality sign changes.
When solving an inequality, we arrive at:
\(\begin{array}{l}x+2
Explain what our solution set is.
When writing our solution in interval notation, how do we represent all the real numbers?

答えを明らかにしろ

$(-\infty ,\infty )$
When solving an inequality, we arrive at:
$\begin{array}{l}x+2>x+3 \\ 2>3\end{array}$
Explain what our solution set is.
Describe how to graph $y=|x-3|$

答えを明らかにしろ

We start by finding the x-intercept, or where the function = 0. Once we have that point, which is $(3,0),$ we graph to the right the straight line graph $y=x-3,$ and then when we draw it to the left we plot positive y values, taking the absolute value of them.
$4x-7\le 9$
$3x+2\ge 7x-1$

答えを明らかにしろ

$(-\infty ,\frac{3}{4}]$
$-2x+3>x-5$
$4(x+3)\ge 2x-1$

答えを明らかにしろ

$[-\frac{13}{2},\infty )$
$-\frac{1}{2}x\le -\frac{5}{4}+\frac{2}{5}x$
$-5(x-1)+3>3x-4-4x$

答えを明らかにしろ

$(-\infty ,3)$
$-3(2x+1)>-2(x+4)$
$\frac{x+3}{8}-\frac{x+5}{5}\ge \frac{3}{10}$

答えを明らかにしろ

$(-\infty ,-\frac{37}{3}]$
$\frac{x-1}{3}+\frac{x+2}{5}\le \frac{3}{5}$
$|x+9|\ge -6$

答えを明らかにしろ

All real numbers $(-\infty ,\infty )$
$|3x-1|>11$

答えを明らかにしろ

$(-\infty ,-\frac{10}{3})\cup (4,\infty )$
$|2x+1|+1\le 6$
$|x-2|+4\ge 10$

答えを明らかにしろ

$(-\infty ,-4]\cup [8,+\infty )$
$|-2x+7|\le 13$

Symbols used here

\infty

infinity
Not a number: "grows without bound" in limits and intervals.

A \cup B,\ A \cap B,\ A \setminus B

union, intersection, difference
In either; in both; in A but not B.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\sup,\ \inf

supremum, infimum
Least upper bound, greatest lower bound.

\pm

plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.

\neq

not equal
The two sides are different.

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

|x|

absolute value / modulus
Distance from zero: |−3| = 3. For a complex number, distance from the origin.

imaginary unit
i² = −1.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\mathbb{N},\ \mathbb{Z},\ \mathbb{Q},\ \mathbb{R},\ \mathbb{C}

number sets
Naturals, integers, rationals, reals, complex numbers.

How to: Linear Inequalities and Absolute Value Inequalities

Use interval notation
Use properties of inequalities.
Solve inequalities in one variable algebraically.
Solve absolute value inequalities.
Interval notation is a method to indicate the solution set to an inequality. Highly applicable in calculus, it is a system of parentheses and brackets that indicate what numbers are included in a set and whether the endpoints are included as well. See
Solving inequalities is similar to solving equations. The same algebraic rules apply, except for one: multiplying or dividing by a negative number reverses the inequality. See
Compound inequalities often have three parts and can be rewritten as two independent inequalities. Solutions are given by boundary values, which are indicated as a beginning boundary or an ending boundary in the solutions to the two inequalities. See
Absolute value inequalities will produce two solution sets due to the nature of absolute value. We solve by writing two equations: one equal to a positive value and one equal to a negative value. See

Questions people ask

What does it mean to solve an equation?

To find every value of the unknown that makes both sides equal. Each step is an operation applied to both sides that keeps the solution set the same, until the unknown stands alone.

Why do I sometimes get two answers?

A quadratic can cross the axis twice, so it can have two solutions. A degree-n polynomial has up to n. The graph shows where each one comes from.

How do I know whether to factor or use the quadratic formula?

Try factoring for a few seconds: look for two numbers that multiply to a·c and add to b. If nothing obvious appears, the discriminant b² − 4ac tells you how many real roots there are, and the formula finds them without guessing.

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Parts of this page are adapted from OpenStax College Algebra 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

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Set Indicated	Set-Builder Notation	Interval Notation
All real numbers between a and b, but not including a or b	\(\{x\|a	\((a,b)\)
All real numbers greater than a, but not including a	\(\{x\|x>a\}\)	\((a,\infty )\)
All real numbers less than b, but not including b	\(\{x\|x	\((-\infty ,b)\)
All real numbers greater than a, including a	\(\{x\|x\ge a\}\)	\([a,\infty )\)
All real numbers less than b, including b	\(\{x\|x\le b\}\)	\((-\infty ,b]\)
All real numbers between a and b, including a	\(\{x\|a\le x	\([a,b)\)
All real numbers between a and b, including b	\(\{x\|a	\((a,b]\)
All real numbers between a and b, including a and b	\(\{x\|a\le x\le b\}\)	\([a,b]\)
All real numbers less than a or greater than b	\(\{x\|xb\}\)	\((-\infty ,a)\cup (b,\infty )\)
All real numbers	\(\{x\|x\ \text{is all real numbers}\}\)	\((-\infty ,\infty )\)