Absolute value equations and inequalities: Visual walkthrough

This page rebuilds the central result of Absolute value equations and inequalities from nothing but a number line. By the end you will see — not just memorise — why an absolute-value equation has two answers, why a "less-than" inequality traps a number between two walls, and why a "greater-than" one flings it outside them.

We assume you know only one thing: a number line — a straight road with $0$ in the middle, positive numbers to the right, negative to the left. Everything else we build.

Step 1 — What the bars actually mean

WHAT. The symbol $∣ x ∣$ is read "the absolute value of $x$ ". We are going to forget every rule you may have heard and define it as one single idea:

WHY this idea and not "just drop the minus sign"? "Drop the minus sign" is a trick; "distance from zero" is a reason. Distance is always a positive length (you cannot walk $- 4$ steps), which instantly explains why $∣ x ∣$ can never be negative. Every rule below falls out of this one picture.

PICTURE. Look at the figure. The number $- 4$ sits four steps left of zero; the number $3$ sits three steps right. Both distances are drawn as coloured spans. The span for $- 4$ has length $4$ , so $∣ - 4∣ = 4$ . The span for $3$ has length $3$ , so $∣3∣ = 3$ .

Figure — Absolute value equations and inequalities

$∣ - 4∣ = 4$ — the $4$ is the length of the yellow span, not the position.
$∣3∣ = 3$ — the $3$ is the length of the blue span.

The parent note wrote this same idea as a formula: $∣ x ∣ = two rules, one distance {x - x if x \geq 0 if x < 0$ Read it as: if $x$ is already on the right ( $x \geq 0$ ), its distance is $x$ . If $x$ is on the left ( $x < 0$ ), we flip it with $- x$ to report a positive length. (See Piecewise functions for this two-rule shape.)

Step 2 — Turn an equation into a distance question

WHAT. We now attack $∣ A ∣ = b$ , where $A$ is some expression and $b$ is any real number on the right-hand side. To start simple, take the cleanest case $A = x$ : $∣ x ∣ = b .$ We will first reason as if $b$ is a genuine walking distance (so $b \geq 0$ ), then in Step 3 we deliberately test what happens when $b$ is zero or when someone erroneously treats a negative $b$ as a length.

WHY read it as a question? Because the definition is a question. $∣ x ∣ = b$ literally says: "which points $x$ sit exactly $b$ steps from zero?" An equation you can ask out loud is an equation you can see.

PICTURE. Stand at $0$ . Walk out a distance $b$ (taking $b \geq 0$ for now). You can go right — landing on $+ b$ — or left — landing on $- b$ . Two landing spots, both exactly $b$ away.

The green dot at $+ b$ is the right answer: $x = b$ .
The red dot at $- b$ is the left answer: $x = - b$ .
The two arrows have equal length $b$ — that equality is the whole point.

This is the birth of the "two cases" (valid while $b > 0$ ): $∣ x ∣ = b ⟺ walked right x = b OR walked left x = - b .$

Step 3 — The degenerate cases: $b = 0$ and $b < 0$

WHAT. Before trusting the rule, we test its extremes — the moments where it might break. In Step 2 we assumed $b$ was a real walking distance; now we drop that assumption and let $b$ be any real number.

WHY. A rule you only checked on friendly numbers is a rule you don't yet own. Distances of $0$ and negative target lengths are exactly where students get surprised.

PICTURE (left half). If $b = 0$ , the two arrows shrink to nothing. The right landing and the left landing collapse onto the same spot — zero itself. So $∣ x ∣ = 0$ has one solution, $x = 0$ , not two.

PICTURE (right half). If someone erroneously treats a negative $b$ as a length, they are asking us to walk a negative number of steps. That is impossible — you cannot be $- 3$ steps from anywhere. So $∣ x ∣ = b$ with $b < 0$ has no solution. The figure shows the empty target as a crossed-out line.

Step 4 — Shift the centre: $∣ x - c ∣ = b$

WHAT. Real problems rarely centre on zero. Let $c$ be a fixed real number — it will play the role of the new "middle of the road." Consider $∣ x - c ∣ = b$ . The quantity $∣ x - c ∣$ is the distance between $x$ and $c$ ; written out in one dimension, that distance is $dist (x, c) = ∣ x - c ∣ = (x - c)^{2},$ which is the one-dimensional shadow of the Distance formula $(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$ with the $y$ -parts set to zero.

WHY does $x - c$ measure distance to $c$ ? Because subtracting $c$ slides the whole road so that $c$ lands where $0$ used to be (this is a horizontal shift). Distance-from-zero of $x - c$ is the same as distance-from- $c$ of $x$ . (The centre $c$ is the value that makes the inside zero: solve $x - c = 0$ .)

PICTURE. The centre marker moves from $0$ to $c$ . Now we walk $b$ steps left and right from $c$ , landing on $c + b$ and $c - b$ .

$x - c = b ⟹ x = c + b$ — the right landing, $b$ past the centre.
$x - c = - b ⟹ x = c - b$ — the left landing, $b$ before the centre.

Step 5 — Less-than: trapped between two walls

WHAT. Now $∣ x - c ∣ < b$ with $b > 0$ . Same centre, same radius, but the sign is $<$ , not $=$ .

WHY does this become a between? "Distance less than $b$ " means "closer than $b$ steps." Every such point lies inside the circle of radius $b$ — on the line, that circle is the segment from $c - b$ to $c + b$ . You must satisfy the left wall AND the right wall at once, so it is an AND.

PICTURE. The two walls at $c - b$ and $c + b$ are drawn as vertical bars. The interior — the allowed zone — is shaded. Points on the walls are excluded (open circles) because $<$ is strict.

$∣ x - c ∣ < b ⟺ right of left wall - b < x - c AND left of right wall x - c < b ⟺ c - b < x < c + b .$

For the non-strict version, the walls themselves are allowed, so the open circles fill in and the strict $<$ become $\leq$ : $∣ x - c ∣ \leq b ⟺ c - b \leq x \leq c + b (closed interval [c - b, c + b]) .$

Step 6 — Greater-than: flung outside the walls

WHAT. Flip to $∣ x - c ∣ > b$ , $b > 0$ .

WHY does this become an OR of two rays? "Distance greater than $b$ " means "farther than $b$ steps." Those points live outside the segment — and outside splits into two disconnected pieces: everything past the right wall, and everything before the left wall. A point cannot be in both at once, so it is an OR (a union, Linear inequalities).

PICTURE. The middle segment is now the forbidden zone. The two outer rays are shaded. The walls are open circles again (strict $>$ ).

$∣ x - c ∣ > b ⟺ past right wall x - c > b OR before left wall x - c < - b ⟺ x > c + b OR x < c - b .$

The non-strict $\geq$ case. Just as $\leq$ filled in the walls for less-than, $\geq$ fills in the walls here — the two boundary points now belong to the answer. The open circles become closed circles, and the strict $>$ become $\geq$ : $∣ x - c ∣ \geq b ⟺ x \geq c + b OR x \leq c - b .$ In interval form this is $(- \infty, c - b] \cup [c + b, \infty)$ — note the square brackets on the inner ends, the closed-circle rays.

Step 7 — When the target itself is a variable

WHAT. In $∣2 x + 5∣ = 3 x - 1$ the right side is not a fixed length — it moves with $x$ . So we must also demand it be a legal length: $3 x - 1 \geq 0$ .

WHY the extra guard? A distance can never equal a negative number (Step 3). If $3 x - 1$ turns out negative, any solution we "find" is a mirage — an extraneous root.

PICTURE. We simply plot the right side $y = 3 x - 1$ against $x$ and mark the horizontal axis $y = 0$ . Wherever the yellow line sits below the axis, the target is negative — an illegal length — so any root landing there is thrown out. This is an ordinary one-line graph (a function of $x$ ), not a 2-D region: the axes are labelled $x$ and "value of right side."

To match the piecewise definition of Step 1 exactly, we split the inside $2 x + 5$ at $2 x + 5 \geq 0$ versus $2 x + 5 < 0$ .

Case 1 — inside $\geq 0$ , i.e. $2 x + 5 \geq 0$ , so $∣2 x + 5∣ = 2 x + 5$ : $2 x + 5 = 3 x - 1 ⟹ x = 6.$

Sign-of-inside guard: $2 (6) + 5 = 17 \geq 0$ ✓ — the assumption holds.
RHS guard: $3 (6) - 1 = 17 \geq 0$ ✓ — legal length. Keep $x = 6$ .

Case 2 — inside $< 0$ , i.e. $2 x + 5 < 0$ , so $∣2 x + 5∣ = - (2 x + 5)$ : $- (2 x + 5) = 3 x - 1 ⟹ x = - \frac{4}{5} .$

Sign-of-inside guard: $2 (- \frac{4}{5}) + 5 = \frac{17}{5} > 0$ ✗ — we assumed $2 x + 5 < 0$ , contradiction.
RHS guard: $3 (- \frac{4}{5}) - 1 = - \frac{17}{5} < 0$ ✗ — not a legal length either. Reject $x = - \frac{4}{5}$ .

Answer: $x = 6$ only.

The one-picture summary

Every result above is one diagram read three ways. Fix a centre $c$ and a radius $b$ . Then:

Equation $= b$ : the two dots on the walls.
Less-than $< b$ : the inside shaded (AND). With $\leq$ , the walls fill in — closed interval.
Greater-than $> b$ : the outside shaded (OR). With $\geq$ , the walls fill in — closed-ended rays.

Recall Feynman retelling — say it like a story

Absolute value is just "how far from a spot." Every absolute-value problem picks a centre $c$ (solve inside $= 0$ ) and a radius $b$ (the number on the other side). If it's an equation, you walk $b$ steps each way and land on two dots: $c - b$ and $c + b$ . If the radius is zero, both dots merge into one; if the radius is negative, there's nowhere to walk, so no answer. If it's a less-than inequality, you want to stay close — so you're stuck between the two walls, an AND, an interval. Strict $<$ leaves the walls out (open); $\leq$ pulls them in (closed). If it's a greater-than inequality, you want to be far — so you flee to either the far-left or far-right, an OR, two rays. Strict $>$ leaves the walls out; $\geq$ pulls them in, giving closed-ended rays with square brackets. And if the number on the right is secretly a moving expression, you must double-check it never goes below zero, or the distance you promised is a lie.

Recall Quick self-test

$∣ x ∣ = b$ has how many solutions when $b = 0$ ? ::: Exactly one, $x = 0$ . $∣ A ∣ < b$ (b>0) becomes which connective? ::: AND — between the walls, one interval. $∣ A ∣ \leq b$ — how does the interval change vs $<$ ? ::: Same interval but closed: $[c - b, c + b]$ , walls included. $∣ A ∣ > b$ (b>0) becomes which connective? ::: OR — outside the walls, two rays with $\cup$ . $∣ A ∣ \geq b$ — how do the rays change vs $>$ ? ::: Same rays but closed-ended: $(- \infty, c - b] \cup [c + b, \infty)$ , walls included. Why check the RHS sign in $∣ A ∣ = 3 x - 1$ ? ::: A distance can't equal a negative number; negative RHS gives extraneous roots. Enforcing RHS $\geq 0$ also makes the sign-of-inside check redundant.