What is an Absolute Value Inequality?

Absolute value inequalities might sound complex, but they are quite manageable once you get the hang of them. Let’s break it down step by step.

What is Absolute Value?

Before diving into inequalities, it’s crucial to understand what absolute value means. The absolute value of a number is its distance from zero on a number line, regardless of direction. For example, the absolute value of both -3 and 3 is 3. We denote absolute value with vertical bars: $|3| = 3$ and $|-3| = 3$

Absolute Value Inequality Basics

An absolute value inequality is an inequality that contains an absolute value expression. It looks something like this:

$|x| < a$ or $|x| > a$

Here, $a$ is a positive number. Let’s explore what these mean.

$|x| < a$

This inequality states that the distance of $x$ from zero is less than $a$. This translates to:

$-a < x < a$

For example, if we have $|x| < 4$, it means $-4 < x < 4$. So, $x$ can be any number between -4 and 4.

$|x| > a$

This inequality means that the distance of $x$ from zero is greater than $a$. This translates to:

$x < -a$ or $x > a$

For instance, if we have $|x| > 4$, it means $x < -4$ or $x > 4$. So, $x$ can be any number less than -4 or greater than 4.

Solving Absolute Value Inequalities

Let’s solve a few examples to make this clearer.

Example 1: $|x – 2| < 5$

  1. Start by writing the inequality without the absolute value:

$-5 < x – 2 < 5$

  1. Solve for $x$ by adding 2 to all parts of the inequality:

$-5 + 2 < x – 2 + 2 < 5 + 2$

$-3 < x < 7$

So, the solution is $-3 < x < 7$

Example 2: $|2x + 1| > 3$

  1. Write the inequality without the absolute value:

$2x + 1 < -3$ or $2x + 1 > 3$

  1. Solve each part separately:

For $2x + 1 < -3$:

$2x < -4$

$x < -2$

For $2x + 1 > 3$:

$2x > 2$

$x > 1$

So, the solution is $x < -2$ or $x > 1$

Graphing Absolute Value Inequalities

Graphing these inequalities can help you visualize the solutions. Let’s graph the solutions to the examples above.

Graphing Example 1: $-3 < x < 7$

On a number line, you would shade the region between -3 and 7, not including -3 and 7 themselves (since the inequality is strict, using < and >).

Graphing Example 2: $x < -2$ or $x > 1$

On a number line, you would shade all regions to the left of -2 and to the right of 1. Again, -2 and 1 are not included.

Real-World Applications

Absolute value inequalities are not just abstract concepts; they have real-world applications too. For instance, they can be used in error tolerances in engineering, where a measurement must fall within a specific range.

Example: Manufacturing Tolerances

Suppose a machine part must be within 0.02 inches of 2 inches in diameter. This can be represented as:

$|x – 2| leq 0.02$

Which translates to:

$-0.02 leq x – 2 leq 0.02$

Solving for $x$:

$1.98 leq x leq 2.02$

So, the diameter must be between 1.98 and 2.02 inches.

Conclusion

Understanding absolute value inequalities is an essential skill in algebra. They help us solve problems where we need to consider distances from a point. By breaking down the problem into manageable steps and practicing with examples, you can master this concept.

Citations

  1. 1. Khan Academy – Absolute Value Inequalities
  2. 2. Purplemath – Solving Absolute Value Inequalities
  3. 3. Math is Fun – Absolute Value Inequalities

Related

(2) O3 + H → O2 + OH k2 = 1.78×10^-11 cm^3 s^-1 (3) O + OH → O2 + H k3 = 4.40×10^-11 cm^3 s^-1 (5) O + HO2 → O2 + OH k5 = 3.50×10^-11 cm^3 s^-1 (6) H + HO2 → O2 + H2 k6 = 5.40×10^-12 cm^3 s^-1 (9) OH + HO2 → O2 + H2O2 k9 = 4.00×10^-11 cm^3 s^-1 (10) HO2 + HO2 → O2 + H2O2 k10 = 2.50×10^-12 cm s^-1 (11) O + O2 + M → O3 + M k11 = 1.05×10^-34 cm^6 s^-1 (14) H + O2 + M → HO2 + M k14 = 8.08×10^-32 cm^6 s^-1 (15) H + H + M → H2O + M k15 = 3.31×10^-27 cm^6 s^-1 (16) O2 + hv → 2 O k16 = (1.26×10^-8 s^-1) φ (17) H2O + hv → H + OH k17 = (3.4×10^-6 s^-1) φ (18) O3 + hv → O2 + O k18 = (7.10×10^-5 s^-1) φ

Table 1 Reactions, rate constants and activation energies used in the model* No. Reaction kopt (M⁻¹ s⁻¹) 1 OH + H₂ → H + H₂O 3.74 x 10⁷ 2 OH + HO₂ → HO₂ + OH⁻ 5 x 10⁹ 3 OH + H₂O₂ → HO₂ + H₂O 3.8 x 10⁷ 4 OH + O₂ → O₂ + OH 9.96 x 10⁹ 5 OH + HO₂ → O₂ + H₂O 7.1 x 10⁹ 6 OH + OH → H₂O₂ 5.3 x 10⁹ 7 OH + e⁻aq → OH⁻ 3 x 10¹⁰ 8 H + O₂ → HO₂ 2.0 x 10¹⁰ 9 H + HO₂ → H₂O₂ 2.0 x 10¹⁰ 10 H + H₂O₂ → OH + H₂O 3.44 x 10⁷ 11 H + OH → H₂O 1.4 x 10¹⁰ 12 H + H → H₂ 1.94 x 10¹⁰ 13 e⁻aq + O₂ → O₂⁻ 1.9 x 10¹⁰ 14 e⁻aq + O₂ → HO₂⁻ + OH⁻ 1.3 x 10¹⁰ 15 e⁻aq + HO₂ 2.0 x 10¹⁰ 16 e⁻aq + H₂O₂ 1.1 x 10¹⁰ 17 e⁻aq + HO₂ → OH + OH⁻ 1.3 x 10¹⁰ 18 e⁻aq + H⁺ → H 2.3 x 10¹⁰ 19 e⁻aq + e⁻aq → H₂ + OH⁻ + OH⁻ 2.5 x 10⁹ 20 HO₂ + O₂ → O₂ + HO₂ 1.3 x 10⁹ 21 HO₂ + HO₂ → O₂ + H₂O₂ 8.3 x 10⁵ 22 HO₂ + HO₂ → O₂ + OH + H₂O 3.7 23 HO₂ + HO₂ → O₂ + O₂ + OH + H₂O 7 x 10⁵ s⁻¹ 24 H⁺ + O₂⁻ → HO₂ 4.5 x 10¹⁰ 25 H⁺ + O₂⁻ → O₂ 2.0 x 10¹⁰ 26 H⁺ + OH⁻ 1.4 x 10¹¹ 27 H⁺ + HO₂⁻ 2 x 10¹⁰ 28 H₂O₂ → HO₂ + H⁺ + OH⁻ 2.5 x 10⁻⁵ s⁻¹ 29 H₂O₂ → H⁺ + OH⁻ 1.4 x 10⁻⁷ s⁻¹ 30 O₂ + O₂ → O₂ + HO₂ + OH⁻ 0.3 31 O₂ + H₂O₂ → O₂ + OH + OH 16 32

(2) O3 + H → O2 + OH k2 = 1.78×10^-11 cm^3 s^-1 (3) O + OH → O2 + H k3 = 4.40×10^-11 cm^3 s^-1 (5) O + HO2 → O2 + OH k5 = 3.50×10^-11 cm^3 s^-1 (6) H2O + O → 2 OH k6 = 5.40×10^-12 cm^3 s^-1 (9) OH + HO2 → O2 + H2O k9 = 4.00×10^-11 cm^3 s^-1 (10) HO2 + HO2 → O2 + H2O2 k10 = 2.50×10^-12 cm s^-1 (11) O + O2 + M → O3 + M k11 = 1.05×10^-34 cm^6 s^-1 (14) H + O2 + M → HO2 + M k14 = 8.08×10^-32 cm^6 s^-1 (15) OH + H + M → H2O + M k15 = 3.31×10^-27 cm^6 s^-1 (16) O2 + hv → 2 O k16 = (1.26×10^-8 s^-1) φ (17) H2O + hv → H + OH k17 = (3.4×10^-6 s^-1) φ (18) O3 + hv → O2 + O k18 = (7.10×10^-8 s^-1) φ