How To Solve An Inequality With Absolute Value

How to Solve an Inequality with Absolute Value

Absolute value inequalities appear frequently in algebra, calculus, and real‑world modeling because they describe distances on a number line. Here's the thing — mastering the technique of solving these inequalities not only sharpens algebraic skills but also builds intuition for more advanced topics such as piecewise functions and optimization. Below is a step‑by‑step guide, complete with explanations, examples, and common pitfalls to avoid No workaround needed..

Easier said than done, but still worth knowing.

1. Understanding the Basics

The absolute value of a number (x), denoted (|x|), measures its distance from zero regardless of direction. Formally:

[ |x| = \begin{cases} x & \text{if } x \ge 0 \ -,x & \text{if } x < 0 \end{cases} ]

When an inequality involves (|x|) (or a more general expression like (|ax+b|)), the goal is to rewrite it as a compound inequality without the absolute value bars. The two primary forms are:

• “Less than” type: (|,\text{expression},| < c)
• “Greater than” type: (|,\text{expression},| > c)

where (c) is a non‑negative constant. If (c < 0), the inequality has either no solution or all real numbers, depending on the direction (see Section 4).

2. General Procedure

Follow these steps for any inequality of the form (|A| , \mathbin{\square} , B), where (A) is an algebraic expression, (B\ge0), and (\square) stands for (<, \le, >,) or (\ge).

1. Isolate the absolute value on one side of the inequality.
   Example: From (2|x-3|+5 \le 11) subtract 5, then divide by 2 to get (|x-3| \le 3).

2. Check the sign of the constant (B).

   • If (B < 0) and the inequality is (<) or (\le), there is no solution because an absolute value can never be negative.
   • If (B < 0) and the inequality is (>) or (\ge), the solution is all real numbers (the absolute value is always ≥ 0 > negative B).

3. Rewrite the inequality without absolute value using the definition:

   For (<) or (\le):
   [ -B \le A \le B ] For (>) or (\ge):
   [ A \le -B \quad \text{or} \quad A \ge B ]

   Notice the “or” for the greater‑than case because the distance must lie outside the interval ([-B, B]) That's the whole idea..

4. Solve each resulting linear (or polynomial) inequality separately.
   This may involve distributing, combining like terms, or factoring.

5. Combine the solution sets according to the logical connector (“and” for the compound inequality, “or” for the two‑part case).
   Express the final answer in interval notation, set‑builder notation, or on a number line.

6. Verify by testing a point from each region (especially the boundaries) in the original inequality to ensure no algebraic mistake was made.

3. Worked Examples

Example 1: (|2x-5| < 7)

1. Absolute value already isolated.
2. (B = 7 \ge 0).
3. Rewrite as (-7 \le 2x-5 \le 7).
4. Add 5 to all parts: (-2 \le 2x \le 12).
5. Divide by 2: (-1 \le x \le 6).
6. Solution: (\boxed{[-1,,6]}).

Example 2: (|x+4| \ge 3)

1. Isolated.
2. (B = 3 \ge 0).
3. Rewrite as (x+4 \le -3) or (x+4 \ge 3).
4. Solve each:
   • (x \le -7)
   • (x \ge -1)
5. Solution: (\boxed{(-\infty,,-7] \cup [-1,,\infty)}).

Example 3: (|5-2x| > -4)

Here (B = -4 < 0) and the inequality is “>”. Since an absolute value is always ≥ 0, it is automatically greater than any negative number.
Solution: all real numbers, (\boxed{(-\infty,\infty)}).

Example 4: (|3x+1| \le -2)

(B = -2 < 0) and the inequality is “≤”. No absolute value can be ≤ a negative number.
Solution: no solution, (\boxed{\varnothing}).

Example 5: Quadratic inside absolute value – (|x^2-4| < 5)

1. Isolated.
2. (B = 5).
3. Rewrite: (-5 < x^2-4 < 5).
4. Add 4: (-1 < x^2 < 9).
5. Solve the two parts:
   • (x^2 > -1) is always true (since squares are ≥ 0).
   • (x^2 < 9) gives (-3 < x < 3).
6. Intersection yields (-3 < x < 3).
   Still, we must also consider where the expression inside the absolute value changes sign because the inequality (-5 < x^2-4 < 5) already accounts for that; no extra case splitting is needed.
   Solution: (\boxed{(-3,,3)}).

4. Special Cases and Graphical Insight

4.1. When the Constant Is Zero

• (|A| < 0) → no solution.
• (|A| \le 0) → solution is where (A = 0).
• (|A| > 0) → all real numbers except where (A = 0).
• (|A| \ge 0) → all real numbers.

4.2. Visualizing on a Number Line

Think of (|A|) as the distance from the point where (A = 0).

• For (|A| < c), you keep points whose distance is less than (c) → an interval centered at the zero of (A).
• For (|A| > c), you keep points whose distance is greater than (c) → two rays extending outward from that center.

Drawing a quick sketch helps avoid sign errors, especially when the expression inside the absolute value shifts sign (e.Also, g. On top of that, , (|x-2|) vs. (|2-x|)).

4.3. Compound Absolute Value Expressions

Sometimes you encounter inequalities like (|x-1| + |x+2| \le 5). The strategy is to identify critical points where each absolute value changes sign (here, (x = 1) and (x = -2)), split the number line into intervals, rewrite the expression without absolute values on each interval,

4.3 Solving Compound Absolute‑Value Inequalities

When more than one absolute value appears in a single inequality, the key is to remove the absolute signs by considering the points where the expressions inside change sign. These points are called critical points (or breakpoints) Took long enough..

Step‑by‑step strategy

1. Identify the critical points.
   For each term (|A_i(x)|) locate the value(s) of (x) that make (A_i(x)=0).
2. Partition the real line.
   The critical points split (\mathbb R) into intervals on which every (A_i(x)) keeps a constant sign (either non‑negative or negative).
3. Rewrite the inequality on each interval.
   Replace each (|A_i(x)|) by (A_i(x)) if (A_i(x)\ge0) on that interval, or by (-A_i(x)) if (A_i(x)<0).
4. Solve the resulting (now linear or polynomial) inequality on that interval.
5. Intersect the solution set with the interval under consideration.
6. Union the pieces obtained from all intervals to form the final solution.

Because the intervals are disjoint, the union automatically respects the original sign pattern.

Example: (\displaystyle |x-1|+|x+2|\le 5)

Critical points: (x=1) and (x=-2).
Intervals: ((-\infty,-2]), ([-2,1]), ([1,\infty)).

| Interval | Sign of (x-1) | Sign of (x+2) | Expression (|x-1|+|x+2|) | Inequality | |----------|----------------|----------------|----------------------------|------------| | (x\le -2) | (x-1<0) → (-,(x-1)=-x+1) | (x+2\le0) → (-,(x+2)=-x-2) | ((-x+1)+(-x-2) = -2x-1) | (-2x-1\le5) | | (-2\le x\le 1) | (x-1\le

Quick note before moving on.