Opening the book…
- Express an interval of real numbers using set-builder notation, interval notation, and a number line
- Solve a linear inequality in one variable
- Determine whether a linear inequality has a solution or not
- Use the notation and properties of linear inequalities to investigate real-world scenarios
If we combine all rational and irrational numbers into one collection, we have a set of numbers called the set of real numbers. The set of real numbers is denoted by the symbol \(\mathbb{R}\).
Every point on the number line is associated with a unique real number. Conversely, every real number is associated with a unique position on the number line. In lieu of this correspondence, the number line is commonly referred to as the real line.
Ordering the Real Numbers
The real numbers are ordered on the real line in a manner identical to how we order integers on a number line.
Suppose that \(a\) and \(b\) are real numbers positioned on the real line as shown below.
- Because \(a\) lies to the “left of” \(b\), we say that \(a\) is “less than” \(b\), or in mathematical symbols, \(a<b\). The inequality symbol \(<\) is read “less than.”
- Alternately, \(b\) lies to the “right of” \(a\), so we can also say that \(b\) is “greater than” \(a\), or in mathematical symbols, \(b>a\). The inequality symbol \(>\) is read “greater than.”
Here are two more inequality symbols that we will use in this section.
If we want to say that \(a\) lies to the “left of” \(b\), or shares the same position as \(b\), then we say that \(a\) is “less than or equal to” \(b\) and write \(a ≤ b\). The inequality symbol \(≤\) is read “less than or equal to.”
If we want to say that \(b\) lies to the “right of” \(a\), or shares the same position as \(a\), then we say that \(b\) is “greater than or equal to" \(a\) and write \(b≥ a\). The inequality symbol \(≥\) is read “greater than or equal to.”
Set-Builder Notation
Mathematicians use a construct called set-builder notation to describe sets or collections of numbers. The general form of set-builder notation looks as follows: \[\{x : \text { some statement about } x\} \nonumber \]For example, suppose that we want to describe the set of “all real numbers that are less than \(2\).” We could use the following notation: \[A=\{x : x<2\} \nonumber \]
This is read aloud as follows: “\(A\) equals the set of all \(x\) such that \(x\) is less than \(2\).” Some prefer to use a vertical bar instead of a colon. \[A=\{x \, | \, x<2\} \nonumber \] In this text we use the colon in set-builder notation, but feel free to use the vertical bar instead. They mean the same thing. One might still object that the notation \[\{x : x<2\} \nonumber \] is a bit vague. One objection could be “What type of numbers \(x\) are you referring to? Do you want the integers that are less than two or do you want the real numbers that are less than two?” As you can see, this is a valid objection. One way of addressing this objection is to write: \[A=\{x \in \mathbb{R} : x<2\} \quad \text { or } \quad A=\{x \in \mathbb{N} : x<2\} \nonumber \] The first is read “\(A\) is the set of all \(x\) in the Real Numbers (\(\mathbb{R}\)) that are less than two,” while the second is read “\(A\) is the set of all \(x\) in the Natural Numbers (\(\mathbb{N}\)) that are less than two.”
In this text, unless there is a specific reference to the set of numbers desired, we will assume that the notation \(\{x : x<2\}\) is asking for the set of all real numbers less than \(2\).
Here, we’ve shaded the set of real numbers \(\{x : x<2\}\).
“less than” is the same as saying “left of,” we’ve shaded (in red) all points on the real line that lie to the left of the number two. Note that there is an “empty circle” at the number two. The point representing the number two is not shaded because we were only asked to shade the numbers that are strictly less than two.
While the shading in the above is perfectly valid, much of the information provided is unnecessary (and perhaps distracting). We only need to label the endpoint and shade the real numbers to the left of two, as we’ve done in here.
For contrast, suppose instead that we’re asked to shade the set of real numbers \(\{x : x\leq2\}\). This means that we must shade all the real numbers that are “less than or equal to \(2\)” or “left of and including \(2\).”
Note the difference between the two figures. In the first, we’re shading the set \(\{x : x<2\}\), so the number \(2\) is left unshaded (an empty dot). In the second, we’re shading the set \(\{x : x\leq2\}\), so the number \(2\) is shaded (a filled-in dot).
Interval Notation
There is another mathematical symbolism, called interval notation, that can be used to describe sets of real numbers. Consider the first set of numbers \(\{x : x \geq-3\}\).
Sweeping our eyes “from left to right”, we use \([-3, \infty)\) to describe this set of real numbers. Some comments are in order:
- The bracket at the left end means that \(-3\) is included in the set.
- As you move toward the right end of the real line, the numbers grow without bound. Hence, the \(\infty\) symbol (positive infinity) is used to indicate that we are including all real numbers to the right of \(-3\). However, \(\infty\) is not really a number, so we use parentheses to indicate we are “not including” this fictional point.
The set of numbers \(\{x : x<-1\}\) has the number line:
Sweeping our eyes “from left to right”, this set of real numbers is described with \((-\infty,-1)\). Again, comments are in order:
- The number \(-1\) is not included in this set. To indicate that it is not included, we use a parenthesis.
- As you move toward the left end of the real line, the numbers decrease without bound. Hence, the \(-\infty\) symbol (negative infinity) is used to indicate that we are including all real numbers to the left of \(-1\). Again, \(-\infty\) is not an actual number, so we use a parenthesis to indicate that we are not including this “fictional” point.
If you would like to ensure that you correctly use interval notation, place the numbers in your interval notation in the same order as they are encountered as you sweep your eyes from “left to right” on the real line.
A nice summary of set-builder and interval notation is presented in a table at the end of the section.
Equivalent Inequalities
Like equations, two inequalities are equivalent if they have the same solution sets.
Let \(a\) and \(b\) be real numbers with \[a < b \nonumber \]
If \(c\) is any real number, then \[a + c<b+ c \nonumber \]and\[a−c<b−c \nonumber \]That is, adding or subtracting the same amount from both sides of an inequality produces an equivalent inequality (does not change the solution).
Solve for \(x: \quad x-2 ≤ 7\). Sketch the solution on the real line, then use set-builder and interval notation to describe your solution.
Solution
To “undo” subtracting \(2\), we add \(2\) to both sides of the inequality.
\[\begin{aligned} x-2 &\leq 7 & & \color{blue} \text { Original inequality. } \\ x-2+2 &\leq 7+2 & & \color{blue} \text { Add } 2 \text { to both sides. } \\ x &\leq 9 & & \color{blue} \text { Simplify both sides. } \end{aligned} \nonumber \]
To shade the real numbers less than or equal to \(9\), we shade the number \(9\) and all real numbers to the left of \(9\).
Using set-builder notation, the solution is \(\{x : x \leq 9\}\). Using interval notation, the solution is \((-\infty, 9]\).
If we multiply or divide both sides of an inequality by a positive number, we have an equivalent inequality.
Let \(a\) and \(b\) be real numbers with \(a<b\). If \(c\) is a real positive number, then \[a c<b c \nonumber \]and\[\dfrac{a}{c}<\dfrac{b}{c} \nonumber \]
Solve for \(x : \quad 3 x \leq-9\) Sketch the solution on the real line, then use set-builder and interval notation to describe your solution.
Solution
To “undo” multiplying by \(3\), divide both sides of the inequality by \(3\). Because we are dividing both sides by a positive number, we do not reverse the inequality sign.
\[\begin{aligned} 3x & \leq -9 & & \color{blue} \text { Original inequality. } \\ \dfrac{3x}{3} & \leq \dfrac{-9}{3} & & \color{blue} \text { Divide both sides by } 3. \\ x & \leq -3 & & \color{blue} \text { Simplify both sides. } \end{aligned} \nonumber \]
Shade the real numbers less than or equal to \(-3\).
Using set-builder notation, the solution is \(\{x : x \leq-3\}\). Using interval notation, the solution is \((-\infty,-3]\).
Reversing the Inequality Sign
Up to this point, the technique for solving inequalities appears to be nearly identical to the technique used to solve equations. However, in this section, we will encounter one exception.
Suppose we start with the valid inequality \(-2 < 5\), then we multiply both sides by \(2\), \(3\), and \(4\).
\[\begin{array}{rrrr}{-2<5} & {-2<5} & {-2<5} \\ {2(-2)<2(5)} & {3(-2)<3(5)} & {4(-2)<4(5)} \\ {-4<10} & {-6<15} & {-8<20}\end{array} \nonumber \]
Note that in each case, the resulting inequality is still valid.
\(\color{Red} \text {Caution! We’re about to make an error!} \)
Start again with \(−2 < 5\), but this time multiply both sides by \(−2\), \(−3\), and \(−4\).
\[\begin{array}{rrrr}{-2<5} & {-2<5} & {-2<5} \\ {-2(-2)<-2(5)} & {\;-3(-2)<-3(5)} & {\;-4(-2)<-4(5)} \\ {4<-10} & {6<-15} & {8<-20}\end{array} \nonumber \]
In each of the resulting inequalities, the inequality symbol is pointing the wrong way!
When you multiply both sides of an inequality by a negative number, you must reverse the inequality sign. Starting with \(−2 < 5\), multiply both sides by \(−2\), \(−3\), and \(−4\), but reverse the inequality symbol.
Some readers might prefer a more formal explanation for why we reverse the inequality when we multiply both sides by a negative number. Suppose that \(a<b\). Then, subtracting \(b\) from both sides gives the result \(a−b<0\). This means that \(a−b\) is a negative number. Now, if \(c\) is a negative number, then the product \((a−b)c\) is positive. Then:
\[\begin{aligned}(a-b) c &>0 \\ a c-b c &>0 \\ a c-b c+b c &>0+b c \\ a c &>b c \end{aligned} \nonumber \]
Thus, if you start with \(a<b\) and \(c<0\), then \(ac > bc\).
Let \(a\) and \(b\) be real numbers with \(a<b\). If \(c\) is a real negative number, then \[a c>b c \nonumber \] and \[\dfrac{a}{c}>\dfrac{b}{c} \nonumber \]That is, when multiplying or dividing both sides of an inequality by a negative number, you must reverse the inequality sign.
Solve for \(x: \quad −2x<4\). Sketch the solution on the real line, then use set-builder and interval notation to describe your solution.
Solution
To “undo” multiplying by \(−2\), divide both sides by \(−2\). Because we are dividing both sides by a negative number, we reverse the inequality sign.
\[\begin{aligned} -2 x&< 4 & & \color{blue} \text { Original inequality. } \\ \dfrac{-2x}{-2}&> \dfrac{4}{-2} & & \color{blue} \text { Divide both sides by }-2 \\ x&> -2 & & \color{blue} \text { Reverse the inequality sign. } \\ x&> -2 & & \color{blue} \text { Simplify both sides. } \end{aligned} \nonumber \]
Shade the real numbers greater than \(−2\).
Using set-builder notation, the solution is \(\{x : x>-2\}\). Using interval notation, the solution is \((-2, \infty)\).
Multiple Steps
Sometimes you need to perform a sequence of steps to arrive at the solution.
Solve for \(x: \quad 3 −5x ≤ 2x + 17\). Sketch the solution on the real line, then use set-builder and interval notation to describe your solution.
Solution
We need to isolate terms containing \(x\) on one side of the inequality. Start by subtracting \(2x\) from both sides of the inequality.
\[\begin{aligned} 3-5x &\leq 2x+17 & & \color{blue} \text { Original inequality. } \\ 3-5x-2x &\leq 2x+17-2x & & \color{blue} \text { Subtract } 2x \text { from both sides. } \\ 3-7x &\leq 17 & & \color{blue} \text { Simplify both sides. } \end{aligned} \nonumber \]
We continue to isolate terms containing \(x\) on one side of the inequality. Subtract \(3\) from both sides.
\[\begin{aligned} 3-7x-3 &\leq 17-3 & & \color{blue} \text { Subtract } 3 \text { from both sides. } \\ -7x &\leq 14 & & \color{blue} \text { Simplify both sides. } \end{aligned} \nonumber \]
To “undo” multiplying by \(−7\), divide both sides by \(−7\). Because we are dividing both sides by a negative number, we reverse the inequality sign.
\[\begin{aligned} \dfrac{-7x}{-7} &\geq \dfrac{14}{-7} & & \color{blue} \text { Multiply both sides by }-7. \text { Reverse the inequality sign. } \\ x &\geq-2 & & \color{blue} \text { Simplify both sides. } \end{aligned} \nonumber \]
Using set-builder notation, the solution is \(\{x : x \geq-2\}\). Using interval notation, the solution is \([-2, \infty)\).
We can also clear fractions from an inequality by multiplying both sides by the least common denominator.
Solve for \(x : \quad \quad \dfrac{3}{4}-\dfrac{x}{12}>\dfrac{1}{3}\).
Solution
First, clear the fractions from the inequality by multiplying both sides by the least common denominator, which in this case is \(12\).
\[\begin{aligned} \dfrac{3}{4}-\dfrac{x}{12} &>\dfrac{1}{3} & & \color{blue} \text { Original inequality. } \\ 12\left[\dfrac{3}{4}-\dfrac{x}{12}\right] &>\left[\dfrac{1}{3}\right] 12 & & \color{blue} \text { Multiply both sides by } 12.\\ 12\left[\dfrac{3}{4}\right]-12\left[\dfrac{x}{12}\right] &>\left[\dfrac{1}{3}\right] 12 & & \color{blue} \text { Distribute the } 12.\\ 9-x &>4 & & \color{blue} \text { Cancel and Multiply. } \end{aligned} \nonumber \]
To “undo” adding \(9\), subtract \(9\) from both sides.
\[\begin{aligned} 9-x-9 &> 4-9 & & \color{blue} \text { Subtract } 9 \text { from both sides. } \\ -x &> -5 & & \color{blue} \text { Simplify both sides. } \end{aligned} \nonumber \]
We could divide both sides by \(−1\), but multiplying both sides by \(−1\) will also do the job. Because we are multiplying both sides by a negative number, we reverse the inequality sign.
\[\begin{aligned} (-1)(-x) &< (-5)(-1) & & \color{blue} \text { Multiply both sides by }-1. \text { Reverse the inequality sign. } \\ x &< 5 & & \color{blue} \text { Simplify both sides. } \end{aligned} \nonumber \]
Shade the real numbers less than \(5\).
Using set-builder notation, the solution is \(\{x : x<5\}\). Using interval notation, the solution is \((-\infty, 5)\).
We clear decimals from an inequality in the usual manner, by multiplying both sides by the appropriate power of ten.
Solve for \(x : \quad 3.25-1.2 x>4.6\).
Solution
First, clear the decimals from the inequality by multiplying both sides by \(100\), which moves each decimal point two places to the right.
\[\begin{aligned} 3.25-1.2 x &>4.6 & & \color{blue} \text { Original inequality. }\\ 325-120 x &>460 & & \color{blue} \text { Multiply both sides by } 100.\\ 325-120 x-325 &>460-325 & & \color{blue} \text { Subtract } 325 \text { from both sides. }\\ -120 x &>135 & & \color{blue} \text { Simplify both sides. } \\ \dfrac{-120 x}{-120} &<\dfrac{135}{-120} & & \color{blue} \text { Divide both sides by } -120. \text { Reverse the inequality sign.}\\ x &<-\dfrac{27}{24} & & \color{blue} \text {Reduce to lowest terms.} \end{aligned} \nonumber \]
Shade the real numbers less than \(−27/24\).
Using set-builder notation, the solution is \(\{x : x<-27 / 24\}\). Using interval notation, the solution is \((-\infty,-27 / 24)\).
Just like some equations are identities with infinite solutions and some are contradictions with no solutions, inequalities may be identities or contradictions, too. We recognize these forms when we are left with only constants as we solve the inequality. If the result is a true statement, we have an identity. If the result is a false statement, we have a contradiction.
Solve the inequality \(8x−2(5−x)<4(x+9)+6x\), graph the solution on the number line, and write the solution in interval notation.
Solution
| Simplify each side as much as possible. | 8x−2(5−x)<4(x+9)+6x |
| Distribute. | 8x−10+2x<4x+36+6x |
| Combine like terms. | 10x−10<10x+36 |
| Subtract 10x from both sides to collect the variables on the left. | 10x−10−10x<10x+36−10x |
| Simplify. | −10<36 |
| The x’s are gone, and we have a true statement. | The inequality is an identity. The solution is all real numbers. |
| Graph the solution on the number line. |  |
| Write the solution in interval notation. | \((-\infty, \infty)\) |
Solve the inequality \(\frac{1}{3}a - \frac{1}{8}a > \frac{5}{24}a + \frac{3}{4}\), graph the solution on the number line, and write the solution in interval notation.
Solution
| Multiply both sides by the LCD, 24, to clear the fractions. |  |
| Simplify. |  |
| Combine like terms. |  |
| Subtract 5a from both sides to collect the variables on the left. |  |
| Simplify. |  |
| The statement is false! | The inequality is a contradiction. |
| There is no solution. | |
| Graph the solution on the number line. |  |
| Write the solution in interval notation. | There is no solution. |
Translate to an Inequality and Solve
To translate English sentences into inequalities, we must recognize the phrases indicating inequality. Some words are easy, like ‘more than’ and ‘less than’, but others are not as obvious.
Think about the phrase ‘at least’ – what does it mean to be ‘at least 21 years old’? It means 21 or more. The phrase ‘at least’ is the same as ‘greater than or equal to’.
This table shows some common phrases that indicate inequalities.
| > | \(\geq\) | < | \(\leq\) |
|---|---|---|---|
| is greater than | is greater than or equal to | is less than | is less than or equal to |
| is more than | is at least | is smaller than | is at most |
| is larger than | is no less than | has fewer than | is no more than |
| exceeds | is the minimum | is lower than | is the maximum |
Solve Applications with Linear Inequalities
Many real-life situations require us to solve inequalities. The method we will use to solve applications involving linear inequalities is very similar to the one we used when solving applications with equations.
We will read the problem and make sure all the words are understood. Next, we will identify what we are looking for and assign a variable to represent it. We will restate the problem in a single sentence to facilitate its translation into an inequality. Then, we will solve the inequality.
Sometimes an application requires the solution to be a whole number, but the algebraic solution to the inequality is not a whole number. In that case, we must round the algebraic solution to a whole number. The context of the application will determine whether we round up or down.
Dawn won a mini-grant of $4,000 to buy tablet computers for her classroom. The tablets she would like to buy cost $254.12 each, including tax and delivery. What is the maximum number of tablets Dawn can buy?
Solution
| Step 1. Read the problem. | |
| Step 2. Identify what you are looking for. | the maximum number of tablets Dawn can buy |
| Step 3. Name what you are looking for. Choose a variable to represent that quantity. | Let \(n\) = the number of tablets. |
| Step 4. Translate Write a sentence that gives the information to find it. | $254.12 times the number of tablets is no more than $4,000. |
| \( \qquad \quad \) Translate into an inequality. | \( 254.12n\leq 4000 \) |
| Step 5. Solve the inequality. | \( n\leq 15.74 \) but \(n\) must be a whole number of tablets, so round down to 15 since 16 would cost too much. \( n\leq 15 \) tablets |
| Step 6. Check the answer in the problem and make sure it makes sense. | Rounding down the price to $250, 15 tablets would cost $3,750, while 16 tablets would be $4,000. So a maximum of 15 tablets at $254.12 seems reasonable. |
| Step 7. Write a sentence that answers the question. | Dawn can buy a maximum of 15 tablets. |
Felicity has a calligraphy business. She charges $2.50 per wedding invitation. Her monthly expenses are $650. How many invitations must she write to earn a profit of at least $2,800 per month?
Solution
| Step 1. Read the problem. | |
| Step 2. Identify what you are looking for. | the number of invitations Felicity needs to write |
| Step 3. Name what you are looking for. Choose a variable to represent that quantity. | Let \(j\) = the number of invitations. |
| Step 4. Translate Write a sentence that gives the information to find it. | $2.50 times the number of invitations minus $650 is at least $2,800. |
| \( \qquad \quad \) Translate into an inequality. | \( 2.50j−650\geq 2,800 \) |
| Step 5. Solve the inequality. | \( 2.5j\geq 3,450 \) \( j\geq 1,380 \) invitations |
| Step 6. Check the answer in the problem and make sure it makes sense. | If Felicity wrote 1400 invitations, her profit would be \(2.50(1400)−650\), or $2,850. This is more than $2800. |
| Step 7. Write a sentence that answers the question. | Felicity must write at least 1,380 invitations. |
In many situations, several quantities contribute to the total expense. When we solve problems like this, we must account for all the individual expenses.
Malik is planning a six-day summer vacation trip. He has $840 in savings, and he earns $45 per hour for tutoring. The trip will cost him $525 for airfare, $780 for food and sightseeing, and $95 per night for the hotel. How many hours must he tutor to have enough money to pay for the trip?
Solution
| Step 1. Read the problem. | |
| Step 2. Identify what you are looking for. | the number of hours Malik must tutor |
| Step 3. Name what you are looking for. Choose a variable to represent that quantity. | Let \(h\) = the number of hours. |
| Step 4. Translate Write a sentence that gives the information to find it. | The expenses must be less than or equal to the income. The cost of airfare plus the cost of food and sightseeing and the hotel bill must be less than the savings plus the amount earned tutoring. |
| \( \qquad \quad \) Translate into an inequality. | \( 525+780+95(6)\leq 840+45h \) |
| Step 5. Solve the inequality. | \( 1,875\leq 840+45h \) \( 1,035\leq 45h \) \( h\geq 23 \) hours |
| Step 6. Check the answer in the problem and make sure it makes sense. | We substitute 23 into the inequality. \( 1,875\leq 840+45h \) \( 1,875\leq 840+45(23) \) \( 1,875\leq 1875 \) |
| Step 7. Write a sentence that answers the question. | Malik must tutor at least 23 hours. |
Summary Table of Set-Builder and Interval Notation
Here is a summary table of the set-builder and interval notation.
| Shading on the real line | Set-builder | Interval |
|---|---|---|
 |
\(\{x : x>-5\}\) | \((-5, \infty)\) |
 |
\(\{x : x \geq-5\}\) | \([-5, \infty)\) |
 |
\(\{x : x<-5\}\) | \((-\infty,-5)\) |
|
\(\{x : x \leq-5\}\) | \((-\infty,-5]\) |
This page titled 1.4: Linear Inequalities was last modified on Fri, 29 Aug 2025 03:30:38 GMT and is shared under a CC BY-NC 4.0 license and was authored, remixed, and/or curated by Math Department via source content that was edited to the style and standards of the LibreTexts platform.
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