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Absolute Value Inequalities

Module by: Kenny M. Felder. E-mail the author

Summary: This module introduces absolute value inequalities.

Here’s one of my favorite problems:

x < 10 x < 10 size 12{ \lline x \rline <"10"} {}

Having seen that the solution to x=10x=10 size 12{ \lline x \rline ="10"} {} is x=±10x=±10 size 12{ \lline x \rline = +- "10"} {}, many students answer this question x<±10x<±10 size 12{ \lline x \rline < +- "10"} {}. However, this is not only wrong: it is, as discussed above, relatively meaningless. In order to approach this question you have to—you guessed it!—step back and think.

Here are two different, perfectly correct ways to look at this problem.

  1. What numbers work? 4 works. –4 does too. 0 works. 13 doesn’t work. How about –13? No: if x=13x=13 size 12{x= - "13"} {} then x=13x=13 size 12{ \lline x \rline ="13"} {}, which is not less than 10. By playing with numbers in this way, you should be able to convince yourself that the numbers that work must be somewhere between –10 and 10. This is one way to approach the answer.
  2. The other way is to think of absolute value as representing distance from 0. 55 size 12{ \lline 5 \rline } {} and 55 size 12{ \lline - 5 \rline } {} are both 5 because both numbers are 5 away from 0. In this case, x<10x<10 size 12{ \lline x \rline <"10"} {} means “the distance between x and 0 is less than 10”—in other words, you are within 10 units of zero in either direction. Once again, we conclude that the answer must be between –10 and 10.
Figure 1: All numbers whose absolute value is less than 10; - 10 < x < 10 - 10 < x < 10
Number line showing the interval from (-10,10)

It is not necessary to use both of these methods; use whichever method is easier for you to understand.

More complicated absolute value problems should be approached in the same three steps as the equations discussed above: algebraically isolate the absolute value, then think, then algebraically solve for x x size 12{x} {} . However, as illustrated above, the think step is a bit more complicated with inequalities than with equations.

Example 1: Absolute Value Inequality

3(2x+38)<153(2x+38)<15 size 12{ - 3 \( \lline 2x+3 \rline - 8 \) < - "15"} {}

  • Step 1:: Algebraically isolate the absolute value
    • 2x+38>52x+38>5 size 12{ lline 2x+3 rline - 8>5} {} (don’t forget to switch the inequality when dividing by –3!)
    • 2x+3>132x+3>13 size 12{ lline 2x+3 rline >"13"} {}
  • Step 2:: Think!
    • As always, forget the 2x+32x+3 size 12{2x+3} {} in this step. The absolute value of something is greater than 13. What could the something be?
    • We can approach this in two ways, just as the previous absolute value inequality. The first method is trying numbers. We discover that all numbers greater than 13 work (such as 14, 15, 16)—their absolute values are greater than 13. Numbers less than –13 (such as –14,–15,–16) also have absolute values greater than 13. But in-between numbers, such as –12, 0, or 12, do not work.
    • The other approach is to think of absolute value as representing distance to 0. The distance between something and 0 is greater than 13. So the something is more than 13 away from 0—in either direction.
    • Either way, we conclude that the something must be anything greater than 13, OR less than –13!
      Figure 2: The absolute value of something is greater than 13; something < - 13 something < - 13 OR something > - 13 something > - 13
      Number line showing the interval from (-infinity,-13) and (13, infinity)
  • Step 3:: Algebraically solve (both inequalities) for xx size 12{x} {}
    2 x + 3 < -13 OR 2 x + 3 > 13 2 x < -16 OR 2 x > 10 x < -8 OR x>5 2 x + 3 < -13 OR 2 x + 3 > 13 2 x < -16 OR 2 x > 10 x < -8 OR x>5 size 12{x>5} {}
    (1)
    Figure 3: Any x x -value which is less than –8 or greater than 5 will make the original inequality true; x < - 8 x < - 8 OR x > 5 x > 5
    Number line showing the interval from (-infinity,-8) and (5, infinity)

Many students will still resist the think step, attempting to figure out “the rules” that will always lead from the question to the answer. At first, it seems that memorizing a few rules won’t be too hard: “greater-than problems always lead to OR answers” and that kind of thing. But those rules will fail you when you hit a problem like the next one.

Example 2: Absolute Value Inequality

x 3 + 10 > 7 x 3 + 10 > 7 size 12{ lline x - 3 rline +"10">7} {}

  • Step 1:: Algebraically isolate the absolute value
    • x 3 > 3 x 3 > 3 size 12{ lline x - 3 rline > - 3} {}
  • Step 2:: Think!
    • The absolute value of something is greater than –3. What could the something be? 2 works. –2 also works. And 0. And 7. And –10. And...hey! Absolute values are always positive, so the absolute value of anything is greater than –3!

All numbers work

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