Getting Absolute Value Inequalities Right
I still see students lose points on these every semester. The concept isn't hard, but the mechanics trip people up because they try to memorize rules without understanding what the absolute value is actually doing on the number line. Let me walk through it the way I wish someone had explained it to me. The absolute value |x| represents distance from zero. That's it. When you see |x| < 5, you're asking: which numbers sit less than 5 units away from zero? The answer is -5 < x < 5. When you see |x| > 5, you're asking which numbers are more than 5 units away, and the answer splits into two separate regions: x < -5 or x > 5. The flip side that catches everyone off guard is how the inequality direction changes when you're dealing with "greater than" versus "less than." It's not arbitrary. A < sign means "between," and a > sign means "outside." Think of it that way and you won't need to memorize two separate rule sets.
Here's the standard method for solving these. Take |2x - 3| 7. First, isolate the absolute value expression. It's already isolated here. Then set up the compound inequality. Because it's "less than or equal to," you write -7 2x - 3 7. Solve all three parts simultaneously. Add 3 throughout: -4 2x 10. Divide by 2: -2 x 5. That's your solution set. Now take the other case: |3x + 1| > 4. Because it's "greater than," the compound inequality splits into two separate statements. Either 3x + 1 > 4 or 3x + 1 < -4. Solve each independently. First one: 3x > 3, so x > 1. Second one: 3x < -5, so x < -5/3. Your solution is x < -5/3 or x > 1. I once had a student working a problem where the absolute value expression had a negative coefficient inside, like |-4x + 8|
12. She panicked because of the negative sign in front of the 4x. There's nothing to panic about. |4x + 8| is the same thing as |4x 8| because absolute value makes everything positive regardless. She could have just dropped the negative and moved on. I watched her spend eight minutes rewriting it when it would have taken thirty seconds to recognize that |-a| = |a| for any expression a. That's the kind of small thing that slows people down on tests.
Another edge case I run into constantly is when the absolute value is set against zero or a negative number. |x| < -3 has no solution. Distance can't be negative. |x| > -3 is true for all real numbers, because every number's distance from zero is greater than a negative number. Students routinely miss these and try to solve them algebraically, producing nonsense answers. The fix is simple: before you do any algebra, check whether the number on the other side of the inequality is negative. If it is, stop and think about what that actually means. The trickier problems show up when you have something like |x - 2| + 3 7. You need to isolate the absolute value first by subtracting 3 from both sides, giving |x - 2| 4. Then proceed normally: -4 x - 2 4, so -2 x 6. The mistake most people make here is setting up the compound inequality before isolating the absolute value expression. If you skip that step, your arithmetic falls apart. Graphing these is where the real understanding shows. A graph of y = |x - 2| + 3 forms a V-shape with its vertex at (2, 3). When you want to solve |x - 2| + 3 7 graphically, you're looking for where the V sits on or below the horizontal line y = 7. The intersection points give you the endpoints of your solution interval. This visual approach catches errors that algebraic manipulation sometimes misses, especially when the absolute value appears on both sides of the equation.
Get the Full Details
Speaking of both sides, |x + 1| > |2x - 3| is a type of problem that looks scary but has a straightforward approach. Square both sides to eliminate the absolute values, since both sides are non-negative and squaring preserves the inequality direction. That gives (x + 1)² > (2x - 3)². Expand both sides: x² + 2x + 1 > 4x² - 12x + 9. Rearrange everything to one side and you get 0 > 3x² - 14x + 8, or 3x² - 14x + 8 < 0. Factor or use the quadratic formula to find the critical points, then test intervals. The solution comes out to 2/3 < x
4. This squaring method works whenever both sides of an absolute value inequality are pure absolute value expressions. It doesn't work when you have additional terms mixed in, which is why you need to isolate first. One thing textbooks don't always emphasize: checking your answers. After solving, plug values from your solution set back into the original inequality to verify. Pick a value inside the interval and one outside. If you're solving |2x - 1| 5 and get -2 x 3, test x = 0 (which should work) and x = 4 (which should fail). If both tests behave incorrectly, you've made an algebraic error somewhere. This takes about thirty seconds and prevents a lot of careless mistakes on exams. There's also a computational shortcut for those who work with these frequently. If you're solving |ax + b| c, the solution is always [-b/a - c/|a|, -b/a + c/|a|]. The center point is -b/a, which is where the expression inside the absolute value equals zero, and the radius is c/|a|. For |3x - 6| 9, the center is 6/3 = 2 and the radius is 9/3 = 3, so the solution is [2 - 3, 2 + 3] = [-1, 5]. Check it: -1 gives |3(-1) - 6| = |-9| = 9, which equals the bound, and 5 gives |15 - 6| = 9. It works. This formula only applies to the "less than" form though. The "greater than" form produces a union of two intervals, not a single bounded interval, so the shortcut doesn't help there.
The main bottleneck with these problems is that they combine multiple skill sets—inequality manipulation, absolute value properties, and sometimes piecewise reasoning. Students who are weak on any one of those three tend to struggle disproportionately. If inequality solving is your weak point, spend time there first. Absolute value inequalities are really just a wrapper around linear inequalities, and if you're comfortable moving terms across the equals sign and flipping inequality directions when multiplying or dividing by negatives, you've already done the hardest part.
