12.3 Absolute Value Equations & Inequalities

12.3 Absolute Value Equations & Inequalities

Quick Answer: Absolute value measures distance from zero, so $|x|$ is never negative. Solve $|expression| = a$ (with $a > 0$) by writing two equations: $expression = a$ or $expression = -a$. For inequalities, $|x| < a$ means $x$ is between $-a$ and $a$; $|x| > a$ means $x$ is outside that interval. Isolate the absolute value before you split cases.

Absolute value items appear alongside linear equations and inequalities on college-placement math. PERT expects the two-case method for equations and the “inside/outside” reading for inequalities. Students who remember only “absolute value is always positive” but forget the second case lose easy points.

Absolute Value as Distance

The absolute value of a real number $x$, written $|x|$, is the distance between $x$ and 0 on the number line.

Formal piecewise definition:

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

You rarely need the piecewise form on PERT if you stay fluent with the distance idea and the two-case method.

Key Properties

• $|x| \ge 0$ for every real $x$
• $|x| = 0$ if and only if $x = 0$
• $|-x| = |x|$
• $|xy| = |x||y|$
• Distance between $a$ and $b$ is $|a - b|$

Absolute Value Equations: Two Cases

Core Rule

If $a > 0$, then:

$$|expression| = a \quad \Longleftrightarrow \quad expression = a \quad \text{or} \quad expression = -a$$

If $a = 0$:

$$|expression| = 0 \quad \Longleftrightarrow \quad expression = 0$$

If $a < 0$:

$$|expression| = a \quad \text{has no solution}$$

because absolute value cannot equal a negative number.

Process Checklist

1. Isolate the absolute-value expression on one side.
2. Look at the other side: positive → two cases; zero → one case; negative → no solution.
3. Solve each linear equation from the cases.
4. Check solutions in the original equation (especially if you multiplied or divided).

Worked Example 1: Simple Two Cases

Problem: Solve $|x| = 6$.

Cases: $x = 6$ or $x = -6$.

Solutions: $x = 6,\, x = -6$.

Graph: Two points on the number line at −6 and 6 (distances of 6 from 0).

Worked Example 2: Linear Inside

Problem: Solve $|2x - 3| = 7$.

Case 1: $2x - 3 = 7 \Rightarrow 2x = 10 \Rightarrow x = 5$.

Case 2: $2x - 3 = -7 \Rightarrow 2x = -4 \Rightarrow x = -2$.

Check: $|2(5) - 3| = |7| = 7$. $|2(-2) - 3| = |-4 - 3| = 7$. Both work.

Trap: Solving only $2x - 3 = 7$ and reporting $x = 5$ alone. Distance interpretation: $2x - 3$ is 7 units from 0, so it can land at +7 or −7.

Worked Example 3: Isolate First

Problem: Solve $3|x + 1| - 5 = 10$.

Step 1 — Isolate the absolute value:

$$3|x + 1| = 15 \Rightarrow |x + 1| = 5$$

Step 2 — Split cases:

• $x + 1 = 5 \Rightarrow x = 4$
• $x + 1 = -5 \Rightarrow x = -6$

Check: $3|4+1| - 5 = 3(5) - 5 = 10$. $3|-6+1| - 5 = 3(5) - 5 = 10$. Both valid.

If you split before isolating—for example, writing $3(x+1) - 5 = 10$ only—you destroy the absolute-value structure and miss solutions.

Worked Example 4: No Solution and Single Solution

Problem A: Solve $|x - 4| = -2$.

Absolute value is never negative. No solution.

Problem B: Solve $|3x + 6| = 0$.

Only $3x + 6 = 0 \Rightarrow x = -2$. One solution.

Problem C: After isolation you get $|x| = -1$ or $|2x - 1| + 4 = 2$ which simplifies to $|2x - 1| = -2$. Again, no solution.

Absolute Value Inequalities: Inside vs Outside

Think distance again. For $a > 0$:

Memory hooks:

• Less than → and (between the opposites)
• Greater than → or (outside both ends)

Some students use the phrase “less thAND, greatOR.”

Worked Example 5: $|x| < a$

Problem: Solve $|x| < 3$.

Rewrite: $-3 < x < 3$.

Graph: Open circles at −3 and 3, shade between.

Check: $x = 0$ works; $x = 4$ fails; $x = -3$ fails because strict inequality excludes endpoints.

Worked Example 6: Linear Inside a “Less Than”

Problem: Solve $|2x - 1| \le 5$.

Rewrite as compound “and”:

$$-5 \le 2x - 1 \le 5$$

Add 1: $-4 \le 2x \le 6$.

Divide by 2: $-2 \le x \le 3$.

Graph: Closed circles at −2 and 3, shade between.

Worked Example 7: $|x| > a$

Problem: Solve $|x + 2| > 4$.

Split into “or”:

• $x + 2 > 4 \Rightarrow x > 2$
• $x + 2 < -4 \Rightarrow x < -6$

Solution: $x < -6$ or $x > 2$.

Graph: Open circles at −6 and 2; shade left of −6 and right of 2.

Check: $x = 0$: $|0+2| = 2 > 4$? False (0 is in the gap). $x = 5$: $|7| = 7 > 4$ true. $x = -7$: $|-5| = 5 > 4$ true.

Worked Example 8: Isolate Then Interpret

Problem: Solve $2|x - 3| - 1 \ge 7$.

Isolate: $2|x - 3| \ge 8 \Rightarrow |x - 3| \ge 4$.

“Or” cases:

• $x - 3 \ge 4 \Rightarrow x \ge 7$
• $x - 3 \le -4 \Rightarrow x \le -1$

Solution: $x \le -1$ or $x \ge 7$.

Special Inequality Cases

These edge cases appear occasionally in multiple-choice traps.

Connecting Absolute Value to Number-Line Distance

$|x - c| = r$ means $x$ is exactly $r$ units from $c$: solutions $x = c + r$ and $x = c - r$.

$|x - c| < r$ means $x$ lies within $r$ units of $c$: $c - r < x < c + r$.

$|x - c| > r$ means $x$ is more than $r$ units from $c$: $x < c - r$ or $x > c + r$.

This geometric reading prevents mechanical errors when the expression inside is $x - c$ rather than $x$ alone.

Strategy Card for PERT Absolute-Value Items

PERT-Style Traps

1. Forgetting the second case on $|expression| = a$ (only reporting the positive side).
2. Treating $|x| = -3$ as $x = -3$ instead of no solution.
3. Splitting before isolating, which corrupts coefficients outside the absolute value.
4. Using “and” for $>$ (writing $-a < x > a$ nonsense) instead of “or.”
5. Using “or” for $<$ and shading both outer rays when the solution is the middle interval.
6. Wrong endpoints: open vs closed when the inequality is strict vs inclusive.
7. Dropping the absolute value after one step and continuing as if the expression were free of $|\,|\,|$.

Quick Practice Patterns to Memorize

• $|x| = 5$ → $x = 5$ or $x = -5$
• $|x - 2| = 3$ → $x = 5$ or $x = -1$
• $|x| < 4$ → $-4 < x < 4$
• $|x| \ge 4$ → $x \le -4$ or $x \ge 4$
• $|2x + 1| = 0$ → $x = -\frac{1}{2}$
• $|x| = -1$ → no solution

Putting the Chapter Skills Together

Literal equations, linear inequalities, and absolute value share one theme: undo operations carefully and respect structure. Literal equations isolate a letter among other letters. Inequalities isolate a variable and protect the inequality direction—especially under negative multiplications. Absolute value isolates the $|\,\cdot\,|$ chunk first, then branches into geometric cases. On PERT Math, pause long enough to name the structure you are solving; that one-second classification prevents most errors on this topic cluster.

Master the two-case split, the inside/outside maps for inequalities, and the habit of isolation-before-branching. With those tools, absolute-value questions become predictable rather than mysterious.