Skip to main content

Section 2.2 Linear Inequalities

In the previous section, we talked about linear equations, which are mathematical statements that indicate that two expressions are equal. But what if we would like to compare expressions that are different "sizes"?

Subsection 2.2.1 Linear Inequalities

The symbol \(\gt\) is called an inequality symbol , and the statement \(a\gt b\) is called an inequality . There are four inequality symbols:
\begin{align*} \amp\gt \amp\amp\text{is greater than} \\ \amp\lt \amp\amp\text{is less than} \\ \amp\ge \amp\amp\text{is greater than or equal to} \\ \amp\le \amp\amp\text{is less than or equal to} \end{align*}
Inequalities that include the symbols \(\gt\) or \(\lt\) are called strict inequalities ; those that include \(\ge\) or \(\le\) are called nonstrict .
If an inequality has one or more variables in it, a solution of that inequality is any set of values that can replace the variables to produce a true statement. For example, \(3\) is a solution for \(2x>2\) since \(2\cdot3>2\) is a true statement. However, \(3\) is not the only solution! In fact, any number greater than \(1\) is a solution. While linear equations have exactly one, zero, or infinitely many solutions, linear inequalities can have much more complex solution sets.
Solving linear inequalities is very similar to solving linear equalities. The main difference is in multiplying and dividing: if we multiply or divide both sides of an inequality by a negative number, the direction of the inequality must be reversed. For example, if we multiply both sides of the inequality
\begin{equation*} 2\lt 5 \end{equation*}
by \(-3\text{,}\) we get
\begin{align*} \alert{-3}(2) \amp\gt \alert{-3}(5)\amp\amp \text{Change inequality symbol from }\lt \text{ to }\gt. \\ -6 \amp\gt -15. \end{align*}
Because of this property, the rules for solving linear equations must be revised slightly for solving linear inequalities.

To Solve a Linear Inequality.

  1. We may add the same number to both sides of an inequality or subtract the same number from both sides of an inequality without changing its solutions.
  2. We may multiply or divide both sides of an inequality by a positive number.
  3. If we multiply or divide both sides of an inequality by a negative number, we must reverse the direction of the inequality symbol.

Example 2.11.

Solve the inequality \(4 - 3x \ge -17\text{.}\)
Solution.
Use the rules above to isolate \(x\) on one side of the inequality.
\begin{align*} 4 - 3x \amp\ge -17\amp\amp\text{Subtract 4 from both sides.} \\ -3x \amp\ge -21\amp\amp\text{Divide both sides by }-3. \\ x \amp\le 7 \end{align*}
Notice that we reversed the direction of the inequality when we divided by \(-3\text{.}\) Any number less than or equal to \(7\) is a solution of the inequality.
A compound inequality involves two inequality symbols. To solve a compound inequality, we use the same steps as before, applying the operations on all three "sides" of the inequality symbols.

Example 2.12.

Solve \(4 \le 3x + 10 \le 16\text{.}\)
Solution.
We isolate \(x\) by performing the same operations on all three sides of the inequality.
\begin{alignat*}{3} 4 \amp{}\le{} 3x\amp {}+{}\amp 10 \amp{}\le{} 16\hphantom{blank} \amp\text{Subtract }10\text{ from each piece.} \\ -6 \amp{}\le{} \amp 3x\amp {}{}\amp {}\le{} 6\hphantom{blank} \amp\text{Divide each piece by }3. \\ -2 \amp{}\le{} \amp x\amp {}{}\amp{}\le{} 2\hphantom{blank}\amp \end{alignat*}
The solutions are all numbers between \(-2\) and \(2\text{,}\) inclusive.

Caution 2.13.

Notice in the previous example that in a compound inequality, both of the inequality symbols are in the same "direction". You typically would not see something like \(4\lt x \geq 8\) or \(-3\geq x \leq 2\text{.}\)

Subsection 2.2.2 Interval Notation

A common way to represent solutions of an inequality is with interval notation. An interval is a set that consists of all the real numbers between two numbers \(a\) and \(b\text{.}\)
The set \(-2 \le x \le 2\) includes its endpoints \(-2\) and \(2\text{,}\) so we call it a closed interval , and we denote it by \([-2, 2]\) (see FigureΒ 2.14a). The square brackets tell us that the endpoints are included in the interval. An interval that does not include its endpoints, such as \(-2 \lt x \lt 2\text{,}\) is called an open interval , and we denote it with round brackets, \((-2, 2)\) (see FigureΒ 2.14b).
described in detail following the image
closed interval and open interval
Figure 2.14.

Caution 2.15.

Do not confuse the open interval \((-2, 2)\) with the point \((-2, 2)\text{!}\) The notation is the same, so you must decide from the context whether an interval or a point is being discussed.
We can also discuss infinite intervals , such as \(x\lt 3\) and \(x\ge -1\text{,}\) shown in FigureΒ 2.16. We denote the interval \(x\lt 3\) by \((-\infty, 3)\text{,}\) and the interval \(x\ge -1\) by \([-1, \infty)\text{.}\) The symbol \(\infty\text{,}\) for infinity, does not represent a specific real number but rather indicates that the interval continues forever along the real line. We always use round brackets next to \(\pm\infty\) in infinite intervals.
described in detail following the image
infinite intervals
Figure 2.16.
Finally, we can combine two or more intervals into a larger set. For example, the set consisting of \(x\lt -1\) or \(x\gt 2\text{,}\) shown in FigureΒ 2.17, is the union of two intervals and is denoted by \((-\infty,-2) \cup (2,\infty)\text{.}\)
described in detail following the image
number line with two disjoint infinite intervals
Figure 2.17.
Many solutions of inequalities are intervals or unions of intervals.

Example 2.18.

Write each of the solution sets with interval notation and graph the solution set on a number line.
  1. \(\displaystyle 3 \le x \lt 6\)
  2. \(\displaystyle x \ge -9\)
  3. \(\displaystyle x\le 1 ~\text{ or }~ x\gt 4\)
  4. \(\displaystyle -8 \lt x \le -5 ~\text{ or }~ -1 \le x \lt 3\)
Solution.
  1. \([3, 6)\text{.}\) This is called a half-open or half-closed interval. (See FigureΒ 2.19.)
    described in detail following the image
    half-open interval
    Figure 2.19.
  2. \([-9,\infty)\text{.}\) We always use round brackets next to the symbol \(\infty\) because \(\infty\) is not a specific number and is not included in the set. (See FigureΒ 2.20.)
    described in detail following the image
    half-closed infinite interval
    Figure 2.20.
  3. \((-\infty, 1] \cup (4, \infty)\text{.}\) The word or describes the union of two sets. (See FigureΒ 2.21.)
    described in detail following the image
    disjoint infinite intervals
    Figure 2.21.
  4. \((-8,-5] \cup [-1, 3)\text{.}\) (See FigureΒ 2.22.)
    described in detail following the image
    disjoint finite intervals
    Figure 2.22.

Exercises 2.2.3 Exercises

1. Interval Notation.

This is similar to Problem 1 from your Section 2.2 worksheet.
Write each inequality in interval notation. If an interval uses infinity, type "INF".
a) \(x\lt 20\)
b) \(x\geq -6\)
c) \(-11 \lt x \leq 11\)
d) \(4 \leq x \leq 4\)
Hint.
Pay special attention to the direction of the inequalities, as well as whether they are strict or non-strict.

2. Solving Inequalities.

This is similar to Problem 3 from your Section 2.2 worksheet.
Solve the following inequality. Write your solution in interval notation. If an interval uses infinity, type "INF".
\(1 x + 4 \leq 4\)

3. Solving Inequalities.

This is similar to Problem 3 from your Section 2.2 worksheet.
Solve the following inequality. Write your solution in interval notation. If an interval uses infinity, type "INF".
\(-\dfrac{2}{5} x \lt -15\)

4. Solving Inequalities.

This is similar to Problem 3 from your Section 2.2 worksheet.
Solve the following inequality. Write your solution in interval notation. If an interval uses infinity, type "INF".
\(-2(m+5)+8 \geq -9(m-4)\)

5. Solving Inequalities.

This is similar to Problem 3 from your Section 2.2 worksheet.
Solve the following inequality. Write your solution in interval notation. If an interval uses infinity, type "INF".
\(5 x - \dfrac{3}{2}(x+4)>-\dfrac{3}{4}(x+6)+4\)

6. Solving Compound Inequalities.

This is similar to Problem 3 from your Section 2.2 worksheet.
Solve the following inequality. Write your solution in interval notation. If an interval uses infinity, type "INF".
\(-11\lt y-15\lt 13\)
Hint.
When solving compound inequalities, make sure you’re doing the same thing to all three "pieces" (such as adding the same number to all three pieces).

7. Solving Compound Inequalities.

This is similar to Problem 3 from your Section 2.2 worksheet.
Solve the following inequality. Write your solution in interval notation. If an interval uses infinity, type "INF".
\(-5 \leq \dfrac{4 q- 1}{1} \lt 0\)
Hint.
When solving compound inequalities, make sure you’re doing the same thing to all three "pieces" (such as adding the same number to all three pieces). If you multiply or divide by a negative number, be sure to reverse all inequality symbols.

8. Positive, Negative, and Zero.

This is similar to Problem 4 from your Section 2.2 worksheet.
By setting up an appropriate inequality and solving it, determine for what values of \(x\) the expression \(8 x - 3\) will be (a) positive, (b) negative, and (c) zero. Write your solution in interval notation. If an interval uses infinity, type "INF".
a) \(8 x - 3\) will be POSITIVE for values of \(x\) in the interval
b) \(8 x - 3\) will be NEGATIVE for values of \(x\) in the interval
c) \(8 x - 3\) will be ZERO for values of \(x\) in the interval
Hint.
Recall that an expression is positive if it is greater than zero and it is negative if it is less than zero.

9. Car Rentals.

This is similar to Problem 5 from your Section 2.2 worksheet.
A car rental company offers two options. Option A is $\(19\) per day plus $\(0.09\) per mile. Option B is $\(8\) per day plus $\(0.21\) per mile. If a car is rented for one day, how many miles can you drive so that Option A is cheaper?
Option A is cheaper for one day if you drive
miles.
Hint.
Try writing an expression that represents how much it costs to drive \(m\) miles with each option, assuming you rent the car for one day.
Option A being cheaper means that the cost of Option A is less than the cost of Option B.