This is a question bank for the tricky ideas behind Absolute value equations and inequalities — not the arithmetic (that lives in the other decks). Each item below is a concept trap: places where a confident student walks straight into a wall. Read the question, say your answer out loud, then peek.
Before the traps, three anchors you must have in your head — nothing here is a new symbol, just a reminder of what each thing means.
Recall What the bars actually mean
∣x∣ is a distance — how far x sits from 0 on the number line. Distance is never negative, and two different points can be the same distance away (that is where the "two cases" come from).
Picture all five grammars on one line first — the figure below is the map every trap in this page tests against. Read it like this: the black square at 0 is the centre (where the inside expression A equals zero); the labels −b and +b mark the target distance (b=3 in the drawing). A filled dot means the endpoint is included, an open dot means excluded. A teal band between the dots is an "inside/AND" solution; two plum rays running outward are an "outside/OR" solution; two lone orange dots are the two solutions of an equation. Walk one row to fix the reading: on the "∣A∣<b" row the teal band with open ends says "every point strictly between −3 and 3", and on the "∣A∣>b" row the two plum rays with open ends say "everything to the left of −3or to the right of 3, endpoints left out."
The second figure zooms in on the trap that most students trip over: how the outside/OR picture degenerates as the target b shrinks to 0 and then goes negative. Watch the two rays close in, kiss at the centre, and finally swallow the whole line.
Recall The five grammars (assume
b≥0 for the equals/less-than rows)
∣A∣=b → two points at distance b (an or of equations). Needs b≥0; if b<0, no solution.
∣A∣<b → strictly inside the band, endpoints excluded: −b<A<b (an and). Needs b>0; if b≤0, no solution.
∣A∣≤b → inside including the band, endpoints included: −b≤A≤b (an and). Needs b≥0; if b<0, no solution.
∣A∣>b → strictly outside the band, two rays, endpoints excluded: A>b or A<−b (an or). Special cases: if b=0 the two rays meet at the centre and the only excluded point is 0, so the solution is A=0 (all nonzero A); if b<0 every distance already beats b, so the solution is all real A.
∣A∣≥b → outside including the band, endpoints included: A≥b or A≤−b (an or). Special cases: if b=0 the two closed rays cover everything (since ∣A∣≥0 always), so the solution is all real A; if b<0 likewise all real A.
The negative branch is missing. x−2=−5 also works, giving x=−3; geometrically both are 5 units from 2.
"∣x∣>3 means −3>x>3."
That chain is self-contradictory (nothing is both below −3 and above 3). Greater-than is outside, an or: x>3orx<−3.
"∣x∣<3 means x<3 or x>−3."
Wrong connector. It should be and — −3<x<3 — the two conditions squeeze together into a band, they do not spread apart.
"Solving ∣2x+5∣=3x−1, I get x=6 and x=−54, both valid."
x=−54 is extraneous: it makes the right side 3x−1negative, which no absolute value can equal, so only x=6 survives.
"For ∣x−1∣+∣x+2∣=5 I set x−1=5 and x+2=5 separately."
You cannot split a sum of absolute values term by term. You must use critical points (x=1, x=−2) to break the line into regions and rewrite each bar with a definite sign.
"−5≤x+2≤5, subtract 2 from the middle only: −5≤x≤3."
You must subtract 2 from all three parts; the correct band is −7≤x≤3. What you do to the middle you do to both edges.
"∣A∣=b so A=±b — always two answers."
Only when b>0. If b=0 there is one answer, and if b<0 there are none; the "±" quietly assumes b is positive.
"∣x−1∣=2x+3; I found x=−34 and accepted it because the algebra was clean."
Clean algebra is not a validity check — substitute back. If the right side turns out negative there, it is extraneous and must be discarded.
Why does ∣A∣=b split into two equations but ∣A∣<b into one double inequality?
An equation pins A to a set of isolated points (two, at distance b), while the inequality allows a whole continuous band, which one −b<A<b statement captures at once.
Why must we check the sign of the right side in ∣A∣=expression?
The left side is a distance, so it is ≥0; if the right side can go negative for some x, those x can never be solutions no matter what the casework says.
Why does the "less than" case become an AND while "greater than" becomes an OR?
Near zero there is one connected region (inside the band → both bounds must hold → and); far from zero there are two separate regions (left ray or right ray → or).
Why are the critical points in ∣x−1∣+∣x+2∣=5 exactly x=1 and x=−2?
Those are the inputs where each inside expression hits zero and flips sign; only at a sign flip does the piecewise rule for ∣⋅∣ change, so the line is naturally cut there.
Why is ∣x∣=x2 a distance statement and not just an algebra trick?
Squaring erases the sign and the square root returns the non-negative root, so the output is always the magnitude — precisely the distance from 0, which is what absolute value measures.
Why can a case in the equation give a number that then fails its own assumption?
Each case is solved under the hypothesis that the inside has a certain sign; the algebra does not enforce that hypothesis, so the answer must be tested against it and thrown out if it violates the region it came from.
Why does shifting from ∣x∣ to ∣x−c∣ move the "centre" of every solution to x=c?
∣x−c∣ measures distance from c, not 0; it is a horizontal translation, so all "distance b" answers pivot around c instead of the origin.
Exactly one — the two branches A=0 and A=−0 are the same equation, so the double answer degenerates into a single point.
∣A∣≤0: what is the solution set?
Only where A=0. Since ∣A∣ can never be negative, "≤0" forces "=0", collapsing a band to one point.
∣A∣<0: solution set?
Empty. No distance is strictly less than 0, so nothing qualifies — a common trap disguised as an ordinary inequality.
∣A∣≥0: solution set?
All real numbers. It is always true, so the solution is the entire line (−∞,∞).
∣A∣>0 (i.e. b=0 in the greater-than grammar): solution set?
All nonzeroA — that is, A>0 or A<0, written A=0. The two rays close right up to the centre but leave the single point A=0 excluded, since ∣0∣=0 is not strictly greater than 0.
∣A∣≥b when b<0: solution set?
All real numbers. Every distance already exceeds a negative b, so the "outside including" rays swallow the whole line.
∣x−1∣+∣x+2∣=3 (the smallest possible right side): what happens?
3 is the minimum distance-sum, equal to the gap between the two centres, so the whole interval [−2,1] satisfies it — infinitely many solutions, not two isolated ones.
∣x−1∣+∣x+2∣=2 (below the minimum): solution set?
Empty. The sum of distances to −2 and 1 can never be smaller than the 3-unit gap between them, so no x works.
∣A∣>b at the exact boundary, e.g. does x=3 solve ∣3x−1∣>8?
No — at x=3 we get ∣8∣=8, which is not strictly greater than 8; strict inequalities exclude their endpoints, so the boundary point is left out.
