Absolute value - modulus — definition, number line interpretation: Worked examples

This page is a drill through all the cases the modulus can throw at you. On Absolute value - modulus — definition, number line interpretation we learned the rule $∣ x ∣ = x$ if $x \geq 0$ and $∣ x ∣ = - x$ if $x < 0$ , plus the picture "distance from zero". Here we hunt down every corner where that rule bites: negatives, zero, nested bars, inequalities, word problems, and an exam trap.

Before any example, one honest reminder of the whole machine — and two facts we will lean on repeatedly, which we now name and number so later references are unambiguous:

A quick word on one more callout you'll see below:

The scenario matrix

Every problem about $∣ x ∣$ falls into one of these case classes. The goal of this page is to leave no cell untouched.

Cell	What makes it tricky	Covered by
A. Positive input	Bars do nothing	Ex 1
B. Negative input	Bars flip the sign	Ex 1
C. Zero / degenerate	$	0
D. Distance $∣ b - a ∣$ , both orders	Sign of $b - a$	Ex 3 (figure)
E. Equation $∣ stuff ∣ = k$ , $k > 0$	Two branches	Ex 4 (figure)
F. Equation $= k$ with $k < 0$ or $k = 0$	No solution / one solution	Ex 5
G. Nested / algebraic $∣ x ∣ = ∣ \dots ∣$	Both sides carry bars	Ex 6
H. Inequality $∣ u ∣ \geq k$ ("outside")	Two unbounded intervals	Ex 7 (figure)
I. Triangle inequality — all four sign combos	When is it equality?	Ex 8
**J. Word problem + " $\leq k$ " band	Translate English → bars	Ex 9 (figure)
K. Exam twist ( $∣ x ∣ = x$ etc.)	Hidden sign condition	Ex 10

We work them in order. Ten examples, eleven cells (Ex 1 covers A and B).

Cells A & B — the plain input

Worked example Example 1 (Cells A, B): evaluate four numbers

Compute $∣7∣$ , $∣ - 3∣$ , $∣ - 2.5∣$ , and $∣ 4 - 9 ∣$ .

Forecast: which of these does the bar actually change? Guess before reading.

$∣7∣ = 7$ . Why this step? $7 \geq 0$ , so we are in the top case of the rule — the bars leave it alone. (Cell A.)
$∣ - 3∣ = - (- 3) = 3$ . Why this step? $- 3 < 0$ , so we use the bottom case $∣ x ∣ = - x$ . Negating a negative gives a positive. (Cell B.)
$∣ - 2.5∣ = 2.5$ . Same reasoning — the rule doesn't care about whole numbers vs decimals, only the sign.
$∣ 4 - 9 ∣ = ∣ - 5 ∣ = 5$ . Why this step? First simplify the inside ( $4 - 9 = - 5$ ), then apply the rule. The bars act on the finished inside, not on the raw subtraction.

Verify: all four outputs are $\geq 0$ (fact P1). And $∣ - 3∣ = 3$ matches "3 steps from home whichever way you walked."

Cell C — zero and degeneracy

Cell D — distance on the line, both orders

Worked example Example 3 (Cell D): distance is order-blind

Find the distance between $- 4$ and $7$ , computing it both ways.

Forecast: will $∣7 - (- 4) ∣$ and $∣ - 4 - 7∣$ agree?

Figure (s01): a number line from $- 6$ to $8$ ; magenta dot at $- 4$ , violet dot at $7$ , a wide orange double-headed arrow spanning the two labelled $∣7 - (- 4) ∣ = 11$ , and two smaller arrows below splitting the trip into a jump of $4$ (from $- 4$ to $0$ ) and a jump of $7$ (from $0$ to $7$ ).

Order one: $∣7 - (- 4) ∣ = ∣7 + 4∣ = ∣11∣ = 11$ . Why this step? Subtracting coordinates gives displacement; the bars strip its sign to give magnitude. Look at the figure: the orange arrow spans from $- 4$ up to $7$ .
Order two: $∣ - 4 - 7∣ = ∣ - 11∣ = 11$ . Why this step? Swapping the order flips the sign of the inside ( $- 11$ instead of $+ 11$ ), but the bars erase that flip — that's why $∣ b - a ∣ = ∣ a - b ∣$ (see Distance and metric spaces).
Sanity split at zero: from $- 4$ to $0$ is $4$ ; from $0$ to $7$ is $7$ ; total $4 + 7 = 11$ . Why this step? Because $- 4$ and $7$ lie on opposite sides of zero, the two jumps add. (If both were on the same side you'd subtract.)

Verify: all three routes give $11$ ✓. Symmetry of distance is confirmed: $∣7 - (- 4) ∣ = ∣ - 4 - 7∣ = 11$ . Linked idea: this is the metric $d (a, b) = ∣ a - b ∣$ from Distance and metric spaces.

Cell E — equation $∣ inside ∣ = k$ with $k > 0$

Worked example Example 4 (Cell E): two branches from one bar

Solve $∣2 x - 6∣ = 8$ .

Forecast: how many solutions, and are they symmetric about anything?

Figure (s02): the V-shaped graph $y = ∣2 x - 6∣$ in violet, a dashed orange horizontal line $y = 8$ , and two magenta dots where they meet, at $x = - 1$ and $x = 7$ , each with a dotted drop-line to the axis.

Read it as distance: "the inside $2 x - 6$ is $8$ units from zero." Since $8 > 0$ , that inside could be $+ 8$ or $- 8$ . Why this step? A positive right-hand side always admits both a positive and a negative inside — hence two branches (see Solving absolute value equations).
Branch one: $2 x - 6 = 8 \Rightarrow 2 x = 14 \Rightarrow x = 7$ . Why this step? This is the case where the inside is already non-negative, so the bars did nothing.
Branch two: $2 x - 6 = - 8 \Rightarrow 2 x = - 2 \Rightarrow x = - 1$ . Why this step? This is the case where the inside was negative and the bars flipped it; we undo that by setting it equal to $- 8$ .

Verify: $∣2 (7) - 6∣ = ∣8∣ = 8$ ✓ and $∣2 (- 1) - 6∣ = ∣ - 8∣ = 8$ ✓. In the figure the V-shaped graph $y = ∣2 x - 6∣$ crosses the height $y = 8$ at exactly $x = - 1$ and $x = 7$ .

Cell F — right-hand side $\leq 0$

Worked example Example 5 (Cell F): when the bar

forbids a solution Solve (a) $∣ x + 1∣ = - 3$ and (b) $∣ x + 1∣ = 0$ .

Forecast: one of these has no solutions and one has exactly one. Which is which?

(a) $∣ x + 1∣ = - 3$ : no solution. Why this step? By fact P1, $∣ \dots ∣ \geq 0$ always. It can never equal a negative number, so we reject immediately — no algebra needed.
(b) $∣ x + 1∣ = 0$ : by fact P2 the inside must itself be zero, so $x + 1 = 0 \Rightarrow x = - 1$ . Exactly one solution. Why this step? Distance zero pins you to a single point, so the usual "two branches" collapses into one.

Verify: (a) plug any $x$ : $∣ x + 1∣$ is never $- 3$ ✓ (no solution is correct). (b) $∣ - 1 + 1∣ = ∣0∣ = 0$ ✓.

Cell G — bars on both sides

Worked example Example 6 (Cell G):

$∣ x - 1∣ = ∣2 x + 4∣$ Solve $∣ x - 1∣ = ∣2 x + 4∣$ .

Forecast: two equal distances-from-zero — what two relationships can the two insides have?

The inside of the left bar is $x - 1$ ; the inside of the right bar is $2 x + 4$ . Equal moduli means these two insides are equal or exact opposites: $x - 1 = 2 x + 4 or x - 1 = - (2 x + 4) .$ Why this step? $∣ A ∣ = ∣ B ∣$ says the insides $A$ and $B$ are the same distance from zero, so either they're the same number or mirror numbers ( $A = \pm B$ ). This is the both-signs cousin of Cell E.
First case: $x - 1 = 2 x + 4 \Rightarrow - 5 = x$ , so $x = - 5$ . Why this step? Straight linear solve of the "insides equal" branch.
Second case: $x - 1 = - 2 x - 4 \Rightarrow 3 x = - 3 \Rightarrow x = - 1$ . Why this step? Straight linear solve of the "insides opposite" branch.

Verify: at $x = - 5$ : $∣ - 5 - 1∣ = 6$ and $∣2 (- 5) + 4∣ = ∣ - 6∣ = 6$ ✓. At $x = - 1$ : $∣ - 1 - 1∣ = 2$ and $∣2 (- 1) + 4∣ = ∣2∣ = 2$ ✓.

Cell H — the "outside" inequality $∣ u ∣ \geq k$

Worked example Example 7 (Cell H):

$∣ x - 3∣ \geq 5$ Solve $∣ x - 3∣ \geq 5$ .

Forecast: an equation gave points; this " $\geq$ " gives regions. Will they be a single band, or two pieces?

Figure (s03): the V-graph $y = ∣ x - 3∣$ in violet, the dashed orange line $y = 5$ , and the two regions where the V is at or above $5$ shaded magenta — the ray $x \leq - 2$ and the ray $x \geq 8$ — with the middle strip left unshaded.

Read it as distance: "the inside $x - 3$ is at least $5$ from zero." Why this step? $∣ u ∣ \geq k$ asks the inside to be far — outside the band $(- k, k)$ — the exact opposite of $∣ u ∣ \leq k$ which asks it to be close. That's why this splits into two unbounded pieces, not a single interval (contrast the word problem, Cell J).
Being at least $5$ from zero means the inside is $\geq 5$ or $\leq - 5$ : $x - 3 \geq 5 or x - 3 \leq - 5.$ Why this step? "Far from zero" happens on either side, so we get two separate branches joined by or, not and.
Solve each: $x - 3 \geq 5 \Rightarrow x \geq 8$ ; and $x - 3 \leq - 5 \Rightarrow x \leq - 2$ . Why this step? Each branch is a plain linear inequality (see Inequalities); note we did not flip any inequality sign because we only added $3$ .
Answer: $x \leq - 2$ or $x \geq 8$ , i.e. $(- \infty, - 2] \cup [8, \infty)$ — two rays shooting off to $\pm \infty$ .

Verify: boundary points sit exactly at height $5$ : $∣ - 2 - 3∣ = ∣ - 5∣ = 5$ and $∣8 - 3∣ = ∣5∣ = 5$ ✓. Test one interior point of each ray: $x = - 10 \Rightarrow ∣ - 13∣ = 13 \geq 5$ ✓; $x = 20 \Rightarrow ∣17∣ = 17 \geq 5$ ✓. Test a middle point that should fail: $x = 3 \Rightarrow ∣0∣ = 0 \neq \geq 5$ ✓ (correctly excluded).

Cell I — triangle inequality, all four sign combos

Here we don't just test numbers — we prove when $∣ x + y ∣ = ∣ x ∣ + ∣ y ∣$ and when it is strict, sign case by sign case. The engine is one fact about any real $u$ : $u \leq ∣ u ∣$ and $- u \leq ∣ u ∣$ (a number never exceeds its own size).

Worked example Example 8 (Cell I): the four sign cases, argued in general

Decide, for every combination of signs of $x$ and $y$ , whether $∣ x + y ∣ \leq ∣ x ∣ + ∣ y ∣$ is equality or strict.

Forecast: in which combos is it a strict " $<$ ", and in which is it exactly " $=$ "? Guess before reading.

Both $\geq 0$ ( $x \geq 0, y \geq 0$ ): then $x + y \geq 0$ , so $∣ x + y ∣ = x + y$ , while $∣ x ∣ + ∣ y ∣ = x + y$ too. Equality. Why this step? When nothing is negative the bars change nothing on either side, so the two sides are literally the same expression.
Both $\leq 0$ ( $x \leq 0, y \leq 0$ ): then $x + y \leq 0$ , so $∣ x + y ∣ = - (x + y) = (- x) + (- y) = ∣ x ∣ + ∣ y ∣$ . Equality. Why this step? Same direction (both left of zero) means no cancellation; flipping the whole sum equals flipping each part.
Mixed signs (one positive, one negative, both nonzero): say $x > 0 > y$ . Inside the sum, the positive and negative partly cancel, so $∣ x + y ∣$ is smaller than the un-cancelled total $∣ x ∣ + ∣ y ∣$ . Formally, the two " $u \leq ∣ u ∣$ " facts give $x + y \leq ∣ x ∣ + ∣ y ∣$ and $- (x + y) \leq ∣ x ∣ + ∣ y ∣$ , hence $∣ x + y ∣ \leq ∣ x ∣ + ∣ y ∣$ ; the cancellation makes it strict. Strict $<$ . Why this step? Opposite directions mean you backtrack, so net distance is strictly less than total path length (see Triangle inequality).
A zero factor ( $x = 0$ or $y = 0$ ): if $y = 0$ then $∣ x + 0∣ = ∣ x ∣ = ∣ x ∣ + 0 = ∣ x ∣ + ∣ y ∣$ . Equality. Why this step? Adding zero moves nothing, so there is no cancellation to lose.

Conclusion: equality holds exactly when $x$ and $y$ have the same sign or one is zero (cases 1, 2, 4); it is strict exactly when they have opposite signs (case 3).

Verify (numeric spot-checks of each case): case 1 $x = 3, y = 5$ : $∣8∣ = 8 = 3 + 5$ ✓ (equality). Case 2 $x = - 3, y = - 5$ : $∣ - 8∣ = 8 = 3 + 5$ ✓ (equality). Case 3 $x = 3, y = - 5$ : $∣ - 2∣ = 2 < 8$ ✓ (strict). Case 4 $x = 0, y = 7$ : $∣7∣ = 7 = 0 + 7$ ✓ (equality). All obey $\leq$ ; equality/strict pattern matches the argument.

Cell J — word problem meeting the " $\leq k$ " band

Worked example Example 9 (Cell J): drone temperature tolerance

A drone's rotor must stay within $4^{\circ} C$ of the target temperature $6 0^{\circ} C$ . Write this as an absolute-value statement and find the allowed temperature range $T$ .

Forecast: will the answer be a pair of numbers or a whole interval?

Figure (s04): a number line from $54$ to $66$ ; a violet dot at the target $60$ ; magenta dots at $56$ and $64$ ; the band $[56, 64]$ shaded orange; and two magenta arrows of length $4$ each, from $60$ out to $56$ and from $60$ out to $64$ , showing the half-width.

Translate "within $4$ of $60$ " into distance: the inside is $T - 60$ , and we need $∣ T - 60∣ \leq 4$ . Why this step? " $T$ is at most $4$ away from $60$ " is exactly distance from $T$ to $60$ is $\leq 4$ , and distance on the line is $∣ T - 60∣$ (Example 3's idea). This is a " $\leq$ " (a close inequality), so — unlike Cell H's " $\geq$ " — expect a single band, not two rays. See Inequalities.
Unfold the inequality by splitting on the sign of the inside. $∣ T - 60∣ \leq 4$ means the inside is at most $4$ from zero, i.e. it lies between $- 4$ and $4$ : $- 4 \leq T - 60 \leq 4.$ Why this step? $∣ u ∣ \leq 4$ says $u$ is not below $- 4$ and not above $4$ — the two sign branches ( $u \geq 0$ giving $u \leq 4$ , and $u < 0$ giving $- u \leq 4$ i.e. $u \geq - 4$ ) merge into one connected strip joined by and.
Add $60$ to every part of the chain: $- 4 + 60 \leq T \leq 4 + 60$ , i.e. $56 \leq T \leq 64.$ Why this step? Adding the same number to all three parts shifts the interval bodily without changing its width or flipping any inequality.
Final answer: the rotor is safe for $T \in [56, 64]$ — a single band of width $8$ , centred on $60$ .

Verify: endpoints $56$ and $64$ are each exactly $4$ from $60$ ( $∣56 - 60∣ = 4$ , $∣64 - 60∣ = 4$ ) ✓. A point inside, $T = 61$ : $∣61 - 60∣ = 1 \leq 4$ ✓ (allowed). A point outside, $T = 70$ : $∣70 - 60∣ = 10 \neq \leq 4$ ✓ (correctly forbidden). The figure shows the shaded band $[56, 64]$ centred on $60$ with half-width $4$ . Units: all in $^{\circ} C$ , consistent throughout.

Cell K — the exam twist

Worked example Example 10 (Cell K): for which

$x$ is $∣ x ∣ = x$ ? Find every real $x$ with $∣ x ∣ = x$ . Then find every $x$ with $∣ x ∣ = - x$ .

Forecast: infinitely many, finitely many, or none?

$∣ x ∣ = x$ holds exactly when we're in the top branch of the rule, i.e. $x \geq 0$ . Why this step? The rule defines $∣ x ∣ = x$ only for $x \geq 0$ ; for $x < 0$ we'd get $∣ x ∣ = - x \neq = x$ (since $- x > 0 > x$ ). So the solution set is all $x \geq 0$ .
Check a negative to be sure: $x = - 2 \Rightarrow ∣ x ∣ = 2 \neq = - 2$ . Rejected. Good — confirms the boundary.
Now $∣ x ∣ = - x$ : this is the bottom branch, so it holds when $x \leq 0$ . Why this step? For $x < 0$ , $∣ x ∣ = - x$ by definition; and at $x = 0$ , $- x = 0 = ∣0∣$ too, so include the endpoint.

Verify: $∣ x ∣ = x$ for $x = 0, 3$ ✓ but fails for $x = - 1$ ; $∣ x ∣ = - x$ for $x = 0, - 4$ (since $∣ - 4∣ = 4 = - (- 4)$ ) ✓ but fails for $x = 2$ . The two solution sets are $[0, \infty)$ and $(- \infty, 0]$ — overlapping only at $x = 0$ , which is why both equations hold at zero. This "hidden sign condition" is the classic trap. See Piecewise functions for why the branch boundary matters.

Recall

Recall Did every cell get hit?

Which cell class has no solution, and why? ::: RHS negative ( $∣ \dots ∣ = k$ with $k < 0$ ), because a modulus is never negative (fact P1). $∣ A ∣ = ∣ B ∣$ splits into which two equations? ::: $A = B$ or $A = - B$ (insides equal or opposite). $∣ u ∣ \geq k$ (with $k > 0$ ) gives what kind of solution set? ::: Two unbounded rays, $u \geq k$ or $u \leq - k$ — joined by "or". $∣ u ∣ \leq k$ (with $k > 0$ ) gives what kind of solution set? ::: One connected band, $- k \leq u \leq k$ . When is $∣ x + y ∣ = ∣ x ∣ + ∣ y ∣$ (not just $\leq$ )? ::: When $x$ and $y$ have the same sign (or one is zero). For which $x$ does $∣ x ∣ = x$ ? ::: All $x \geq 0$ .

Connections

Solving absolute value equations — the branching in Cells E, F, G
Inequalities — the "outside" ray (Cell H) and the "inside" band (Cell J)
Triangle inequality — Cell I's geometry
Distance and metric spaces — Cell D's $∣ a - b ∣$ metric
Piecewise functions — the branch condition behind Cell K
Number line and ordering — every figure lives here