Chapter 4: Inequalities

4.3 Linear Absolute Value Inequalities

Absolute values are positive magnitudes, which means that they represent the positive value of any number.

For instance, | −5 | and | +5 | are the same, with both having the same value of 5, and | −99 | and | +99 | both share the same value of 99.

When used in inequalities, absolute values become a boundary limit to a number.

Example 4.3.1

Consider | x | < 4.

This means that the unknown x value is less than 4, so | x | < 4 becomes x < 4. However, there is more to this with regards to negative values for x.

| −1 | is a value that is a solution, since 1 <  4.

However, | −5 | < 4 is not a solution, since 5  >  4.

The boundary of | x | < 4 works out to be between −4 and +4.

This means that | x | < 4 ends up being bounded as -4 <  x  < 4.

If the inequality is written as | x | \le 4, then little changes, except that x can then equal −4 and +4, rather than having to be larger or smaller.

This means that | x | \le 4 ends up being bounded as -4 \le  x  \le 4.

Example 4.3.2

Consider |x| > 4.

This means that the unknown x value is greater than 4, so |x| > 4 becomes x > 4. However, the negative values for x must still be considered.

The boundary of |x| > 4 works out to be smaller than −4 and larger than +4.

This means that |x| > 4 ends up being bounded as x < -4  \text{ or }  4 < x.

If the inequality is written as | x | \ge 4, then little changes, except that x can then equal −4 and +4, rather than having to be larger or smaller.

This means that |x| \ge 4 ends up being bounded as  x \le -4  \text{ or }  4 \le x.

When drawing the boundaries for inequalities on a number line graph, use the following conventions:

For ≤ or ≥, use [brackets] as boundary limits.Blank number line with square brackets positioned on it.

For < or >, use (parentheses) as boundary limits. Blank number line with parentheses positioned on it.

Equation Number Line
| x | <4 x < 4. Left parenthesis on −4; right parenthesis on 4.
| x | \le 4 x ≤ 4. Left square bracket on −4; right bracket on 4.
| x | > 4 x is greater than 4. Right parenthesis on −4; left parenthesis on 4. Arrows to both infinities.
| x | \ge 4 x ≥ 4. Right square bracket on −4; left bracket on 4. Arrows to both infinities.

When an inequality has an absolute value, isolate the absolute value first in order to graph a solution and/or write it in interval notation. The following examples will illustrate isolating and solving an inequality with an absolute value.

Example 4.3.3

Solve, graph, and give interval notation for the inequality  -4  - 3 | x | \ge  -16.

First, isolate the inequality:

    \[\begin{array}{rrrrrl} -4&-&3|x|& \ge & -16 &\\ +4&&&&+4& \text{add 4 to both sides}\\ \midrule &&\dfrac{-3|x|}{-3}& \ge & \dfrac{-12}{-3}&\text{divide by }-3 \text{ and flip the sense} \\ \\ &&|x|&\le & 4 && \end{array}\]

At this point, it is known that the inequality is bounded by 4. Specifically, it is between −4 and 4.

This means that -4 \le | x | \le 4.

This solution on a number line looks like:

−4 ≤ | x | ≤ 4. Left square bracket at −4; right bracket at 4.

To write the solution in interval notation, use the symbols and numbers on the number line: [-4, 4].

Other examples of absolute value inequalities result in an algebraic expression that is bounded by an inequality.

Example 4.3.4

Solve, graph, and give interval notation for the inequality | 2x - 4 | \le  6.

This means that the inequality to solve is -6\le 2x - 4\le 6:

\begin{array}{rrrcrrr} -6&\le & 2x&-&4&\le & 6 \\ +4&&&+&4&&+4 \\ \midrule \dfrac{-2}{2}&\le &&\dfrac{2x}{2}&&\le & \dfrac{10}{2} \\ \\ -1 &\le &&x&&\le & 5 \end{array}

−1 ≤ x ≤ 5. Left square bracket on −1; right bracket on 5.

To write the solution in interval notation, use the symbols and numbers on the number line: [-1,5].

Example 4.3.5

Solve, graph, and give interval notation for the inequality 9  -  2 | 4x + 1 |  > 3.

First, isolate the inequality by subtracting 9 from both sides:

Divide both sides by −2 and flip the sense:

negative 2 times the absolute value of 4 x + 1 divided by negative 2 is greater than negative 6 over negative 2
absolute value of 4 x + 1 is less than 3

At this point, it is known that the inequality expression is between −3 and 3, so -3  <  4x + 1  <  3.

All that is left is to isolate x. First, subtract 1 from all three parts:

    \[\begin{array}{rrrrrrr} -3&<&4x&+&1&<&3 \\ -1&&&-&1&&-1 \\ \midrule -4&<&&4x&&<&2 \\ \end{array}\]

Then, divide all three parts by 4:

    \[\begin{array}{rrrrr} \dfrac{-4}{4}&<&\dfrac{4x}{4}&<&\dfrac{2}{4} \\ \\ -1&<&x&<&\dfrac{1}{2} \\ \end{array}\]

−1 < x < ½. Left parenthesis on −1; right parenthesis on ½.

In interval notation, this is written as \left(-1,\dfrac{1}{2}\right).

It is important to remember when solving these equations that the absolute value is always positive. If given an absolute value that is less than a negative number, there will be no solution because absolute value will always be positive, i.e., greater than a negative. Similarly, if absolute value is greater than a negative, the answer will be all real numbers.

This means that:

    \[| 2x - 4| < -6 \text{ has no possible solution } (x \ne \mathbb{R})\]

and the absolute value of 2X minus 4 > negative 6 has every number as a solution and is written negative infinity, infinity

Note: since infinity can never be reached, use parentheses instead of brackets when writing infinity (positive or negative) in interval notation.

Questions

For questions 1 to 33, solve each inequality, graph its solution, and give interval notation.

  1. | x | < 3
  2. | x | \le 8
  3. | 2x | < 6
  4. | x + 3 | < 4
  5. | x - 2 | < 6
  6. | x - 8 | < 12
  7. | x - 7 | < 3
  8. | x + 3 | \le 4
  9. | 3x - 2 | < 9
  10. | 2x + 5 | < 9
  11. 1 + 2 | x - 1 | \le 9
  12. 10 - 3 | x - 2 | \ge 4
  13. 6 -  | 2x - 5 |  > 3
  14. | x | > 5
  15. | 3x |  > 5
  16. | x - 4 | > 5
  17. | x + 3 | > 3
  18. | 2x - 4 | > 6
  19. | x - 5 | > 3
  20. 3 -  | 2 - x | < 1
  21. 4 + 3 | x - 1 |  < 10
  22. 3 - 2 | 3x - 1 | \ge -7
  23. 3 - 2 | x - 5 | \le -15
  24. 4 - 6 | -6 - 3x | \le -5
  25. -2 - 3 | 4 - 2x | \ge -8
  26. -3 - 2 | 4x - 5 | \ge 1
  27. 4 - 5 | -2x - 7 | < -1
  28. -2 + 3 | 5 - x | \le 4
  29. 3 - 2 | 4x - 5 | \ge 1
  30. -2 - 3 | - 3x - 5| \ge  -5
  31. -5 - 2 | 3x - 6 | < -8
  32. 6 - 3 | 1 - 4x | < -3
  33. 4 - 4 | -2x + 6 | > -4

<a class=”internal” href=”/intermediatealgebraberg/back-matter/answer-key-4-3/”>Answer Key 4.3

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