5.2 Absolute Value Equations, Inequalities & the Absolute Value Function

5.2 Absolute Value Equations, Inequalities & the Absolute Value Function

Skill 4 of Competency 2 names absolute value twice: once for solving equations and inequalities, and once for interpreting the vertex and the line of symmetry of an absolute value function in context. Both halves are tested.

What absolute value means

|x| is the distance from x to zero on the number line, and distance is never negative. So |−7| = 7 and |7| = 7. More usefully, |a − b| is the distance between a and b, which is the interpretation that unlocks every modeling item.

Formally, |x| = x when x ≥ 0 and |x| = −x when x < 0. The second line is not "negative x"; it is the opposite of a negative number, which is positive.

Solving absolute value equations

Isolate the absolute value expression first, then split into two cases.

Solve 3|2x − 5| + 4 = 19 Isolate: 3|2x − 5| = 15, so |2x − 5| = 5 Case 1: 2x − 5 = 5 → 2x = 10 → x = 5 Case 2: 2x − 5 = −5 → 2x = 0 → x = 0

Check both: 3|2(5) − 5| + 4 = 3(5) + 4 = 19 ✓, and 3|2(0) − 5| + 4 = 3(5) + 4 = 19 ✓.

Splitting before isolating is the classic failure. Writing "2x − 5 = 19 or 2x − 5 = −19" from the original equation ignores the multiplication by 3 and the added 4.

Solution counts depend on the isolated right-hand side:

The no-solution case is heavily tested. If isolating yields |x + 3| = −4, stop and answer "no solution" rather than producing two invalid cases.

Absolute value inequalities

Two patterns, and the mnemonic is worth memorizing.

+---------------------------------------------------------------------------+
|   "LESS THAN"  ->  AND    |x| < a   becomes   -a < x < a                  |
|                            (a single bounded interval, "between")         |
|                                                                           |
|   "GREATER"    ->  OR     |x| > a   becomes   x < -a  or  x > a           |
|                            (two rays pointing outward, "beyond")          |
+---------------------------------------------------------------------------+

Think in distance terms. |x| < 5 says "x is less than 5 units from zero," which is the interval between −5 and 5. |x| > 5 says "x is more than 5 units from zero," which is everything outside that interval.

Solve |3x + 2| ≤ 11. Conjunction: −11 ≤ 3x + 2 ≤ 11 → −13 ≤ 3x ≤ 9 → −13/3 ≤ x ≤ 3

Solve |x − 4| > 6. Disjunction: x − 4 < −6 or x − 4 > 6 → x < −2 or x > 10

Two edge cases: |x| > −3 is true for all real numbers, since absolute values are always at least 0 and therefore always exceed a negative. And |x| < −3 has no solution, for the same reason.

The graph of an absolute value function

The parent f(x) = |x| is a V with its vertex at the origin, made of the ray y = x for x ≥ 0 and the ray y = −x for x < 0.

The general form is

f(x) = a|x − h| + k

with these key features:

• Vertex: (h, k). Note the sign flip on h — f(x) = |x − 3| + 2 has vertex (3, 2), while f(x) = |x + 3| + 2 has vertex (−3, 2).
• Line of symmetry: x = h, a vertical line through the vertex.
• Direction: a > 0 opens upward and the vertex is the minimum; a < 0 opens downward and the vertex is the maximum.
• Width: |a| > 1 narrows the V; 0 < |a| < 1 widens it. The slopes of the two branches are a and −a.
• Range: y ≥ k when a > 0; y ≤ k when a < 0.

For f(x) = −2|x + 1| + 8: the vertex is (−1, 8), the line of symmetry is x = −1, the graph opens downward, the maximum value is 8, the range is y ≤ 8, and the branches have slopes −2 and 2.

To find the x-intercepts, set the function to zero: −2|x + 1| + 8 = 0 gives |x + 1| = 4, so x = 3 or x = −5. Notice these are symmetric about x = −1, exactly 4 units on each side — the symmetry is a free check on your arithmetic.

Modeling with absolute value

Any "within a tolerance of a target" statement is an absolute value inequality.

A machine fills bottles to 500 mL with an allowed error of 8 mL. Acceptable volumes satisfy |v − 500| ≤ 8, which resolves to 492 ≤ v ≤ 508.

The structure is always |measured − target| ≤ tolerance. Items reverse this too: given |t − 68| ≤ 3 for a thermostat, interpret it as "the temperature stays within 3 degrees of 68°, so between 65° and 71°."

Distance-from-a-point contexts follow the same pattern. If a delivery hub is at mile marker 240 and a truck must stay within 45 miles, its position satisfies |p − 240| ≤ 45, giving 195 ≤ p ≤ 285. And a minimum separation reverses the symbol: staying more than 45 miles away is |p − 240| > 45.