0.6 Basic Inequalities in One Variable

Vanessa Coffelt

0.6 Basic Inequalities in One Variable

0.6.1 Linear Inequalities

We now turn our attention to linear inequalities. Unlike linear equations which admit at most one solution, the solutions to linear inequalities are generally intervals of real numbers. While the solution strategy for solving linear inequalities is the same as with solving linear equations, we need to remind ourselves that, should we decide to multiply or divide both sides of an inequality by a negative number, we need to reverse the direction of the inequality. In the example below, we work not only some `simple’ linear inequalities in the sense there is only one inequality present, but also some `compound’ linear inequalities which require us to revisit the notions of intersection and union.

Example 0.6.1

Example 0.6.1.1

Solve the following inequalities for the indicated variable.

Solve for $x$ : $\dfrac{7-8x}{2} \geq 4x + 1$

Solution:

Solve for $x$ : $\dfrac{7-8x}{2} \geq 4x + 1$ .

We begin by clearing denominators. Then we gather all of the terms containing $x$ to one side of the inequality and put the remaining terms on the other.

$\begin{array}{rclr} \dfrac{7-8x}{2} & \geq & 4x + 1 & \\ [8pt] 2\left(\dfrac{7-8x}{2}\right) & \geq & 2(4x + 1) & \text{Multiply by $2$} \\ [10pt] \dfrac{\cancel{2}(7-8x)}{\cancel{2}} & \geq & 2(4x) + 2(1) & \text{Distribute} \\ [3pt] 7 - 8x & \geq & 8x + 2 & \\ (7-8x) + 8x-2 & \geq & 8x+2 + 8x -2 & \text{Add $8x$, subtract $2$} \\ 7 - 2 - 8x + 8x & \geq & 8x + 8x + 2 - 2 & \text{Rearrange terms} \\ 5 & \geq & 16x & \text{$8x + 8x = (8+8)x = 16x$} \\ [3pt] \dfrac{5}{16} & \geq & \dfrac{16x}{16} & \text{Divide by the coefficient of $x$} \\[8pt] \dfrac{5}{16} & \geq & x & \\ \end{array}$

We get $\frac{5}{16} \geq x$ or, said differently, $x \leq \frac{5}{16}$ . We express this set^[1] of real numbers as $\left(-\infty, \frac{5}{16}\right]$ .

Though not required to do so, we could partially check our answer by substituting $x = \frac{5}{16}$ and a few other values in our solution set ( $x =0$ , for instance) to make sure the inequality holds. (It also isn’t a bad idea to choose an $x > \frac{5}{16}$ , say $x = 1$ , to see that the inequality doesn’t hold there.) The only real way to actually show that our answer works for all values in our solution set is to start with $x \leq \frac{5}{16}$ and reverse all of the steps in our solution procedure to prove it is equivalent to our original inequality.

Example 0.6.1.2

Solve the following inequalities for the indicated variable.

Solve for $y$ : $\dfrac{3}{4} \leq \dfrac{7-y}{2} < 6$

Solution:

Solve for $y$ : $\dfrac{3}{4} \leq \dfrac{7-y}{2} < 6$ .

We have our first example of a `compound’ inequality. The solutions to

$\dfrac{3}{4} \leq \dfrac{7-y}{2} < 6$

must satisfy

$\dfrac{3}{4} \leq \dfrac{7-y}{2} \qquad \text{\underline{and}} \qquad \dfrac{7-y}{2} < 6$

One approach is to solve each of these inequalities separately, then intersect their solution sets. While this method works (and will be used later for more complicated problems), our variable $y$ appears only in the middle expression so we can proceed by working both inequalities at once:

$\begin{array}{rclr} \dfrac{3}{4} \leq & \dfrac{7-y}{2} & < 6 & \\ [10pt] 4\left(\dfrac{3}{4} \right) \leq & 4\left( \dfrac{7-y}{2}\right) & < 4(6) & \text{Multiply by $4$} \\ [12pt] \dfrac{\cancel{4} \cdot 3}{\cancel{4}} \leq & \dfrac{\cancelto{2}{4}(7-y)}{\cancel{2}} & < 24 & \\ [5pt] 3 \leq & 2(7-y) & < 24 & \\ 3 \leq & 2(7)-2y & < 24 & \text{Distrbute}\\ 3 \leq & 14-2y & < 24 & \\ 3 -14 \leq & (14-2y) - 14 & < 24 - 14 & \text{Subtract $14$}\\ -11 \leq & -2y & < 10 & \\ [3pt] \dfrac{-11}{-2} \geq & \dfrac{-2y}{-2} & > \dfrac{10}{-2} & \text{Divide by the coefficient of $y$} \\ [-5pt] & & & \text{Reverse inequalities} \\ [-3pt] \dfrac{11}{2} \geq & y & > -5 & \\ \end{array}$

Our final answer is $\frac{11}{2} \geq y > -5$ , or, said differently, $-5 < y \leq \frac{11}{2}$ . In interval notation, this is $\left( -5, \frac{11}{2} \right]$ .

We could check the reasonableness of our answer as before, and the reader is encouraged to do so.

Example 0.6.1.3

Solve the following inequalities for the indicated variable.

Solve for $t$ : $2t-1 \leq 4-t < 6t+1$

Solution:

Solve for $t$ : $2t-1 \leq 4-t < 6t+1$ .

We have another compound inequality and what distinguishes this one from our previous example is that $t$ appears on both sides of both inequalities. In this case, we need to create two separate inequalities and find all of the real numbers $t$ which satisfy both

$2t-1 \leq 4-t \textit{ and } 4-t < 6t + 1.$

The first inequality,

$\begin{array}{rcl} 2t-1 &\leq& 4-t \\ 3t &\leq& 5 \\ t \leq \frac{5}{3} \end{array}$

The second inequality,

$\begin{array}{rcl} 4-t &<& 6t+1 \\ 3 &<& 7t \\t &>& \frac{3}{7} \end{array}$

Thus our solution is all real numbers $t$ with $t \leq \frac{5}{3}$ and $t > \frac{3}{7}$ , or, writing this as a compound inequality, $\frac{3}{7} < t \leq \frac{5}{3}$ . Using interval notation,^[2] we express our solution as $\left( \frac{3}{7}, \frac{5}{3} \right]$ .

Example 0.6.1.4

Solve the following inequalities for the indicated variable.

Solve for $x$ : $5 + \sqrt{7} x \leq 4x + 1 \leq 8$

Solution:

Solve for $x$ : $5 + \sqrt{7} x \leq 4x + 1 \leq 8$ .

As before, with this inequality we have no choice but to solve each inequality individually and intersect the solution sets. Starting with the leftmost inequality, we first note that in the term $\sqrt{7} x$ , the vinculum of the square root extends over the $7$ only, meaning the $x$ is not part of the radicand. In order to avoid confusion, we will write $\sqrt{7} x$ as $x \sqrt{7}$ .

$\begin{array}{rclr} 5 + x\sqrt{7} & \leq & 4x+1 & \\ (5 + x\sqrt{7} ) -4x - 5 & \leq& (4x + 1) - 4x - 5 & \text{Subtract $4x$ and $5$} \\ x\sqrt{7} - 4x + 5 - 5 & \leq & 4x - 4x + 1 - 5 & \text{Rearrange terms} \\ x(\sqrt{7} - 4) & \leq & -4 & \text{Factor} \\ \end{array}$

At this point, we need to exercise a bit of caution because the number $\sqrt{7} - 4$ is negative.^[3]

When we divide by it the inequality reverses:

$\begin{array}{rclr} x(\sqrt{7} - 4) & \leq & -4 & \\[3pt] \dfrac{x(\sqrt{7}-4)}{\sqrt{7}-4} & \geq & \dfrac{-4}{\sqrt{7} - 4} & \text{Divide by the coefficient of $x$} \\ [-8pt] & & & \text{Reverse inequalities} \\ [-3pt] x & \geq & \dfrac{-4}{\sqrt{7} - 4} & \\ [10pt] x & \geq & \dfrac{-4}{-(4 - \sqrt{7})} & \\ [12pt] x & \geq & \dfrac{4}{4 -\sqrt{7}} & \\ \end{array}$

We’re only half done because we still have the rightmost inequality to solve.

Fortunately, that one seems rather mundane:

$\begin{array}{rcl} 4x+1 &\leq& 8 \\ x &\leq& \frac{7}{4} \end{array}$

without too much incident.

Our solution is $x \geq \frac{4}{4-\sqrt{7}}$ and $x \leq \frac{7}{4}$ .

We may be tempted to write $\frac{4}{4-\sqrt{7}} \leq x \leq \frac{7}{4}$ and call it a day but that would be nonsense! To see why, notice that $\sqrt{7}$ is between $2$ and $3$ so $\frac{4}{4 - \sqrt{7}}$ is between $\frac{4}{4-2} = 2$ and $\frac{4}{4-3} = 4$ . In particular, we get $\frac{4}{4 - \sqrt{7}} > 2$ . On the other hand, $\frac{7}{4} < 2$ . This means that our `solutions’ have to be simultaneously greater than $2$ AND less than $2$ which is impossible.

Therefore, this compound inequality has no solution, which means we did all that work for nothing.

Example 0.6.1.5

Solve the following inequalities for the indicated variable.

Solve for $w$ : $2.1 - 0.01w \leq -3$ or $2.1-0.01w \geq 3$

Solution:

Solve for $w$ : $2.1 - 0.01w \leq -3$ or $2.1-0.01w \geq 3$ .

Our last example is yet another compound inequality but here, instead of the two inequalities being connected with the conjunction `and‘, they are connected with `or‘, which indicates that we need to find the union of the results of each.

Starting with $2.1 - 0.01w \leq -3$ , we get $-0.01 w \leq -5.1$ , which gives^[4] $w \geq 510$ .

The second inequality, $2.1-0.01w \geq 3$ , becomes $-0.01w \geq 0.9$ , which reduces to $w \leq -90$ .

Our solution set consists of all real numbers $w$ with $w \geq 510$ or $w \leq -90$ . In interval notation, this is $(-\infty, -90] \cup [510, \infty)$ .

0.6.2 Absolute Value Inequalities

We now turn our attention to solving some basic inequalities involving the absolute value. Suppose we wished to solve $|x| < 3$ . Geometrically, we are looking for all of the real numbers whose distance from 0 is less than 3 units. We get $-3 < x < 3$ , or in interval notation, $(-3,3)$ . Suppose we are asked to solve $|x| > 3$ instead. Now we want the distance between $x$ and $0$ to be greater than $3$ units. Moving in the positive direction, this means $x > 3$ . In the negative direction, this puts $x < -3$ . Our solutions would then satisfy $x < -3$ or $x > 3$ . In interval notation, we express this as $(-\infty, -3) \cup (3, \infty)$ .

A number line with -3, 0, and 3 labeled. Open circles at -3 and 3 and the distance from each to 0 marked. — The solution to |x|<3

A number line with -3, 0, and 3 labeled. Open circles at -3 and 3 with a line to the left of -3 and a line to the right of 3 marked. — The solution to |x|>3.

Generalizing this notion, we get the following:

Theorem 0.8 Inequalities Involving Absolute Value

Let $c$ be a real number.

If $c> 0$ , $|x| < c$ is equivalent to $-c<x<c$ .
If $c \leq 0$ , $|x| < c$ has no solution.
If $c > 0$ , $|x| > c$ is equivalent to $x < -c$ or $x > c$ .
If $c \leq 0$ , $|x| > c$ is true for all real numbers.

If the inequality we’re faced with involves ` $\leq$ ‘ or ` $\geq$ ,’ we can combine the results of Theorem 0.8 with Theorem 0.4 as needed.

Strategy for Solving Inequalities Involving Absolute Value

In order to solve an inequality involving the absolute value of a quantity $|X|$ :

Isolate the absolute value on one side of the inequality.
Apply Theorem 0.8.

Example 0.6.2

Example 0.6.2.1

Solve the following inequalities.

$\left|x-\sqrt[4]{5} \right| > 1$

Solution:

Solve $\left|x-\sqrt[4]{5} \right| > 1$ for $x$ .

From Theorem 0.8, $\left|x-\sqrt[4]{5} \right| > 1$ is equivalent to $x-\sqrt[4]{5} < -1$ or $x-\sqrt[4]{5} > 1$ .

Solving this compound inequality, we get $x < -1 + \sqrt[4]{5}$ or $x > 1 + \sqrt[4]{5}$ .

Our answer, in interval notation, is: $\left(-\infty,-1 + \sqrt[4]{5} \right) \cup \left(1 + \sqrt[4]{5}, \infty \right)$ .

As with linear inequalities, we can only partially check our answer by selecting values of $x$ both inside and outside of the solution intervals to see which values of $x$ satisfy the original inequality and which do not.

Example 0.6.2.2

Solve the following inequalities.

$\dfrac{4 - 2|2x+1|}{4} \geq -\sqrt{3}$

Solution:

Solve $\dfrac{4 - 2|2x+1|}{4} \geq -\sqrt{3}$ for $x$ .

Our first step in solving $\frac{4 - 2|2x+1|}{4} \geq -\sqrt{3}$ is to isolate the absolute value.

$\begin{array}{rclr} \dfrac{4 - 2|2x+1|}{4} & \geq & -\sqrt{3} & \\ [5pt] 4 - 2|2x+1| & \geq & -4\sqrt{3} & \text{Multiply by $4$} \\ - 2|2x+1| & \geq & -4-4\sqrt{3} & \text{Subtract $4$} \\ |2x+1| & \leq & \dfrac{-4-4\sqrt{3}}{-2} & \text{Divide by $-2$, reverse the inequality} \\ [8pt] |2x+1| & \leq & 2 + 2\sqrt{3} & \text{Reduce} \\ \end{array}$

Due to the fact that we’re dealing with ` $\leq$ ‘ instead of just ` $<$ ,’ we can combine Theorems 0.8 and 0.4 to rewrite this last inequality as:^[5] $-(2 + 2\sqrt{3}) \leq 2x+1 \leq 2+2\sqrt{3}$ .

Subtracting the ` $1$ ‘ across both inequalities gives $-3-2\sqrt{3} \leq 2x \leq 1 + 2\sqrt{3}$ , which reduces to $\frac{-3-2\sqrt{3}}{2} \leq x \leq \frac{1+2\sqrt{3}}{2}$

In interval notation this reads as $\left[\frac{-3-2\sqrt{3}}{2}, \frac{1+2\sqrt{3}}{2}\right]$ .

Example 0.6.2.3

Solve the following inequalities.

$|2x - 1| \leq 3|4 - 8x| - 10$

Solution:

Solve $|2x - 1| \leq 3|4 - 8x| - 10$ for $x$ .

There are two absolute values in $|2x - 1| \leq 3|4 - 8x| - 10$ , so we cannot directly apply Theorem 0.8 here. Notice, however, that $|4 - 8x| = |(-4)(2x-1)|$ . Using this, we get:

$\begin{array}{rclr} |2x - 1| & \leq & 3|4 - 8x| - 10 & \\ |2x - 1| & \leq & 3|(-4)(2x-1)| - 10 & \text{Factor}\\ |2x - 1| & \leq & 3|-4||2x-1| - 10 & \text{Product Rule}\\ |2x - 1| & \leq & 12|2x-1| - 10 & \\ -11|2x - 1| & \leq & - 10 & \text{Subtract $12|2x-1|$} \\ |2x - 1| & \geq & \dfrac{10}{11} & \text{Divide by $-11$ and reduce} \\ \end{array}$

Now we are allowed to invoke Theorems 0.4 and 0.8 and write the equivalent compound inequality: $2x - 1 \leq -\frac{10}{11}$ or $2x-1 \geq \frac{10}{11}$ .

We get $x \leq \frac{1}{22}$ or $x \geq \frac{21}{22}$ , which when written with interval notation becomes $\left(-\infty, \frac{1}{22}\right] \cup \left[\frac{21}{22}, \infty\right)$ .

Example 0.6.2.4

Solve the following inequalities.

$|2x - 1| \leq 3|4 - 8x| + 10$

Solution:

Solve $|2x - 1| \leq 3|4 - 8x| + 10$ for $x$ .

The inequality $|2x - 1| \leq 3|4 - 8x| + 10$ differs from the previous example in exactly one respect: on the right side of the inequality, we have ` $+10$ ‘ instead of ` $-10$ .’

$\begin{array}{rclr} |2x - 1| & \leq & 3|4 - 8x| + 10 & \\ |2x - 1| & \leq & 3|(-4)(2x-1)| + 10 & \text{Factor}\\ |2x - 1| & \leq & 3|-4||2x-1| + 10 & \text{Product Rule}\\ |2x - 1| & \leq & 12|2x-1| + 10 & \\ -11|2x - 1| & \leq & + 10 & \text{Subtract $12|2x-1|$} \\ |2x - 1| & \geq & -\dfrac{10}{11} & \text{Divide by $-11$ and reduce} \\ \end{array}$

The resulting inequality is always true. (Absolute value is, by definition, a distance and hence always $0$ or greater.)

Thus our solution to this inequality is all real numbers.

Example 0.6.2.5

Solve the following inequalities.

$2 < |x-1| \leq 5$

Solution:

Solve $2 < |x-1| \leq 5$ for $x$ .

To solve $2 < |x-1| \leq 5$ , we rewrite it as the compound inequality: $2 < |x-1|$ and $|x-1| \leq 5$ .

The first inequality, $2 < |x-1|$ , can be re-written as $|x-1|>2$ so it is equivalent to $x-1 < -2$ or $x-1 > 2$ .

Thus the solution to $2 < |x-1|$ is $x<-1$ or $x>3$ , which in interval notation is $(-\infty, -1) \cup (3, \infty)$ .

For $|x-1| \leq 5$ , we combine the results of Theorems 0.4 and 0.8 to get $-5 \leq x-1 \leq 5$ so that $-4 \leq x \leq 6$ , or $[-4,6]$ .

Our solution to $2 < |x-1| \leq 5$ is comprised of values of $x$ which satisfy both parts of the inequality, so we intersect $(-\infty, -1) \cup (3, \infty)$ with $[-4,6]$ to get our final answer $[-4,-1) \cup (3,6]$ .

Example 0.6.2.6

Solve the following inequalities.

$|10 x - 5| +|10 - 5x| \leq 0$

Solution:

Solve $|10 x - 5| +|10 - 5x| \leq 0$ for $x$ .

Our first hope when encountering $|10 x - 5| + |10 - 5x| \leq 0$ is that we can somehow combine the two absolute value quantities as we’d done in earlier examples. We leave it to the reader to show, however, that no matter what we try to factor out of the absolute value quantities, what remains inside the absolute values will always be different.

At this point, we take a step back and look at the equation in a more general way: we are adding two absolute values together and wanting the result to be less than or equal to $0$ . The absolute value of anything is always $0$ or greater, so there are no solutions to: $|10x - 5| + |10 - 5x| < 0$ .

Is it possible that $|10x - 5| + |10 - 5x| = 0$ ? Only if there is an $x$ where $|10x-5| = 0$ and $|10-5x| = 0$ at the same time.^[6]

The first equation holds only when $x = \frac{1}{2}$ , while the second holds only when $x = 2$ . Alas, we have no solution.

The astute reader will have noticed by now that the authors have done nothing in the way of explaining why anyone would ever need to know this stuff. These sections were designed to review skills and concepts that you’ve already learned. Thus, the deeper applications are in the main body of the text as opposed to here in Chapter 0.

We close this section with an example of how the properties in Theorem 0.3 are used in Calculus. Here, ` $\varepsilon$ ‘ is the Greek letter `epsilon’ and it represents a positive real number. Those of you who will be taking Calculus in the future should become very familiar with this type of algebraic manipulation.

$\begin{array}{rclr} \left| \dfrac{8-4x}{3} \right| & < & \varepsilon & \\ [12pt] \dfrac{|8 - 4x|}{|3|} & < & \varepsilon & \text{Quotient Rule}\\ [12pt] \dfrac{|-4(x-2)|}{3} & < & \varepsilon & \text{Factor} \\ [12pt] \dfrac{|-4| |x-2|}{3} & < & \varepsilon & \text{Product Rule} \\ [12pt] \dfrac{4 |x-2|}{3} & < & \varepsilon & \\ [12pt] \dfrac{3}{4} \cdot \dfrac{4 |x-2|}{3} & < & \dfrac{3}{4} \cdot \varepsilon & \text{Multiply by $\dfrac{3}{4}$} \\ [12pt] |x -2 | & < & \dfrac{3}{4} \varepsilon & \\ \end{array}$

0.6.3 Section Exercises

In Exercises 1 – 18, solve the given inequality. Write your answer using interval notation.

$3 - 4x \geq 0$
$2t - 1 < 3 - (4t-3)$
$\dfrac{7 -y}{4} \geq 3y + 1$
$0.05R + 1.2 > 0.8 - 0.25R$
$7 - (2-x) \leq x+3$
$\dfrac{10m+1}{5} \geq 2m - \dfrac{1}{2}$
$x \sqrt{12} - \sqrt{3} > \sqrt{3} x + \sqrt{27}$
$2t - 7 \leq \sqrt[3]{18} t$
$117y \geq y\sqrt{2} - 7y \sqrt[4]{8}$
$-\dfrac{1}{2} \leq 5x - 3 \leq \dfrac{1}{2}$
$-\dfrac{3}{2} \leq \dfrac{4 - 2t}{10} < \dfrac{7}{6}$
$-0.1 \leq \dfrac{5-x}{3} - 2 < 0.1$
$2y \leq 3-y < 7$
$3x \geq 4-x \geq 3$
$6-5t > \dfrac{4t}{3} \geq t - 2$
$2x+1 \leq -1$ or $2x+1 \geq 1$
$4-x \leq 0$ or $2x+7 < x$
$\dfrac{5-2x}{3} > x$ or $2x + 5 \geq 1$

In Exercises 19 – 30, solve the inequality. Write your answer using interval notation.

$|3x - 5| \leq 4$
$|7t + 2| > 10$
$|2w+1| - 5 < 0$
$|2-y| - 4 \geq -3$
$|3z+5| + 2 < 1$
$2|7-v| +4 > 1$
$3 - |x+\sqrt{5}| < -3$
$|5t| \leq |t|+3$
$|w-3| < |3-w|$
$2 \leq |4-y| < 7$
$1 < |2w - 9| \leq 3$
$3 > 2|\sqrt{3} - x| > 1$

Section 0.6 Exercise Answers can be found in the Appendix … Coming soon

Using set-builder notation, our `set' of solutions here is $\{ x \, | \, x \leq \frac{5}{16} \}$ . ↵
If we intersect the solution sets of the two individual inequalities, we get the answer, too: $\left(-\infty, \frac{5}{3}\right] \cap \left(\frac{3}{7}, \infty\right) = \left( \frac{3}{7}, \frac{5}{3} \right]$ . ↵
As a result of $4 < 7 < 9$ , it stands to reason that $\sqrt{4} < \sqrt{7} < \sqrt{9}$ and thus $2 < \sqrt{7} < 3$ . ↵
Don't forget to change the direction of the inequality! ↵
Note the use of parentheses: $-(2+2\sqrt{3})$ as opposed to $-2 + 2\sqrt{3}$ . ↵
Do you see why? ↵