Worked examples — Absolute value equations and inequalities

This page is the "throw everything at it" workout for Absolute value equations and inequalities. Before we solve, we map out every kind of case the topic can throw at you, then hit each one with a full worked example. If you have never seen the bars $||$ before, read the parent note first — but every symbol used here is re-explained the moment it appears.

The scenario matrix

Here is the complete list of situations. Each worked example is tagged with the cell(s) it covers, so you can see the whole surface is painted.

Cell	Scenario	Where the trap hides
A	$|A|=b$, $b>0$ (plain equation)	forgetting the negative branch
B	$|A|=b$ with $b<0$ or $b=0$ (degenerate)	$b<0$ → no solution; $b=0$ → one solution
C	$|A|=(\text{expression in }x)$	extraneous roots, RHS-sign check
D	$|A|<b$ (less-than, "between", AND)	writing OR instead of AND
E	$|A|>b$ (greater-than, "outside", OR)	writing AND instead of OR
F	Degenerate inequality: $|A|<0$, $|A|\ge 0$, $|A|>0$	empty set / all reals
G	Sum of two absolute values $=$ constant (critical points)	missing an empty region
H	Word problem (tolerance / distance)	translating words → $||$
I	Exam twist: $|A|=|B|$	two clean cases, not four

Now the examples.

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Forecast: guess how many answers before reading. (Two — distance $10$ has a left and a right hit.)

1. Read it as distance. $|4x-6|=10$ says "the number $4x-6$ is exactly $10$ away from $0$." Why this step? Anchoring to distance tells us there are two positions: $+10$ and $-10$.
2. Branch 1 (inside is positive): $4x-6=10\Rightarrow 4x=16\Rightarrow x=4$. Why this step? If $4x-6\ge 0$, the bars change nothing, so the inside literally equals $10$.
3. Branch 2 (inside is negative): $4x-6=-10\Rightarrow 4x=-4\Rightarrow x=-1$. Why this step? If $4x-6<0$, the bars flip its sign; a flipped number equal to $10$ means the original was $-10$.

Verify: $|4(4)-6|=|10|=10$ ✓ and $|4(-1)-6|=|-10|=10$ ✓. Answer: $x\in \{-1,4\}$.

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Forecast: one of these has zero answers and one has exactly one. Which is which?

1. (a) Look at the right side. It is $-3$, negative. Why this step? A distance can never be negative, so no $x$ can make the left side negative.
2. (a) Conclusion: no solution. The solution set is $\varnothing$ (the empty set). Why this step? We stop early — there is nothing to solve once the RHS is impossible.
3. (b) Right side is $0$. Distance $0$ from zero means the inside is $0$: $5x-2=0$. Why this step? Only one point is $0$ units from $0$ — itself. So the two branches collapse into one.
4. (b) Solve: $5x=2\Rightarrow x=\frac{2}{5}$.

Verify: (a) impossible confirmed. (b) $|5(\frac{2}{5})-2|=|0|=0$ ✓. Answer: (a) no solution; (b) $x=\frac{2}{5}$ (a single, "doubled" root).

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Forecast: the right side contains $x$, so it can be negative. Predict: at least one candidate will be thrown out.

1. Gatekeeper condition. Since the left side is a distance ($\ge 0$), we need $2x-5\ge 0$, i.e. $x\ge \frac{5}{2}$. Why this step? Any answer with $2x-5<0$ is instantly impossible — a distance can't equal a negative.
2. Branch 1 (assume $x-4\ge 0$): $x-4=2x-5\Rightarrow -4+5=2x-x\Rightarrow x=1$. Why this step? If the inside is $\ge 0$, the bars vanish and the inside equals the RHS directly.
3. Test Branch 1 twice. Global gatekeeper: $x=1\ge \frac{5}{2}$? No — reject. Branch assumption: $x-4=1-4=-3\ge 0$? No — also fails. Reject $x=1$ (extraneous on both counts). Why this step? A valid root must pass both the RHS-sign gate and the branch's own sign assumption; $x=1$ fails both.
4. Branch 2 (assume $x-4<0$): $x-4=-(2x-5)=-2x+5\Rightarrow 3x=9\Rightarrow x=3$. Why this step? If the inside is negative, the bars flip its sign, so we set $x-4$ equal to the negative of the RHS.
5. Test Branch 2 twice. Global gatekeeper: $x=3\ge \frac{5}{2}$? Yes. Branch assumption: $x-4=3-4=-1<0$? Yes. Both conditions hold — keep $x=3$. Why this step? Only a root that respects both the gatekeeper and the sign assumption it was derived under is genuine.

Verify: $|3-4|=|-1|=1$ and $2(3)-5=1$. Equal ✓. Answer: $x=3$ only.

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Forecast: "less than" packs the answer into ONE stretch of the line. Predict a single interval.

1. Unfold with the between-rule. $|A|\le 7⟺-7\le A\le 7$. Here $A=2x+1$: $-7\le 2x+1\le 7.$ Why this step? Being at most $7$ from zero means sitting inside $[-7,7]$ — both a lower and an upper wall must hold, so it's an AND.
2. Subtract $1$ from all three parts: $-8\le 2x\le 6$. Why this step? We isolate the $x$-term; whatever we do to the middle we do to both outer pieces.
3. Divide all three parts by $2$ (positive, so no flip): $-4\le x\le 3$. Why this step? Dividing by a positive number keeps the inequality directions the same.

The figure below draws this answer on the number line. Read it like this: the thick red bar is the solution set; the two solid red dots at $x=-4$ and $x=3$ are filled because the endpoints are included ($\le$). Anything off the red bar fails the inequality.

Figure 1 — $|2x+1|\le 7$ is one closed interval $[-4,3]$; filled dots mean the ends belong to the answer.

Verify: endpoints: $|2(-4)+1|=|-7|=7\le 7$ ✓; $|2(3)+1|=|7|=7\le 7$ ✓; a point outside, $x=4$: $|9|=9>7$, correctly excluded ✓. Answer: $x\in [-4,3]$.

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Forecast: "greater than" spills the answer into TWO far pieces. Predict a union of two rays.

1. Unfold with the outside-rule. $|A|>4⟺A>4\text{ or }A<-4$. Here $A=3x-2$. Why this step? Being more than $4$ from zero means fleeing the interval $[-4,4]$ — you can escape to the right or to the left, so it's an OR.
2. Right piece: $3x-2>4\Rightarrow 3x>6\Rightarrow x>2$. Why this step? This is the escape to the positive side.
3. Left piece: $3x-2<-4\Rightarrow 3x<-2\Rightarrow x<-\frac{2}{3}$. Why this step? This is the escape to the negative side.

The figure below shows the two escape rays. Read it like this: the two red arrows shoot outward forever; the hollow (unfilled) red circles at $x=-\frac{2}{3}$ and $x=2$ mean those exact points are excluded (strict $>$). The clear middle band is the forbidden gap.

Figure 2 — $|3x-2|>4$ is two open rays; hollow dots mean the boundary points are NOT in the answer.

Verify: test $x=3$ (right region): $|7|=7>4$ ✓; test $x=-1$ (left region): $|-5|=5>4$ ✓; test $x=0$ (the gap): $|-2|=2\ngtr 4$, correctly excluded ✓. Answer: $x\in \left(-\infty ,-\frac{2}{3}\right)\cup (2,\infty )$.

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Forecast: one is empty, one is everything, and one is everything except a single point. Match them up first.

1. (a) Can a distance be strictly below $0$? Never. Why this step? $|\cdot |$ is $\ge 0$ always, so nothing beats $0$. Solution: $\varnothing$.
2. (b) Is a distance always $\ge 0$? Yes, for every real number. Why this step? The statement is automatically true, so all reals work: $(-\infty ,\infty )$.
3. (c) Distance is $>0$ except at the one spot where it equals $0$, namely $x-5=0\Rightarrow x=5$. Why this step? Only $x=5$ gives distance $0$; every other point gives a strictly positive distance.

Verify: (a) $x=5$ gives $|0|=0\nless 0$; no other point smaller — empty ✓. (b) $x=100$: $|95|\ge 0$ ✓. (c) $x=5$ excluded, $x=6$: $|1|=1>0$ ✓. Answer: (a) $\varnothing$; (b) all real $x$; (c) all $x\ne 5$, i.e. $(-\infty ,5)\cup (5,\infty )$.

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Forecast: two "corners" at $x=-3$ and $x=2$ split the line into three zones. Predict that only the two outer zones can pay the bill of $9$ (the middle keeps the sum fixed at the gap width $5$).

1. Find critical points. Insides vanish at $x=-3$ and $x=2$. Why this step? These are exactly where each piecewise bar switches its sign-rule.

The figure below marks these corners and labels the three regions. Read it like this: the two red dots are the corners where a bar changes its rule; the middle region is annotated "no sol" because the equation collapses to $5=9$ there; the black squares are the two genuine solutions.

Figure 3 — the corners $x=-3,2$ cut the line into three regions; only the outer two yield solutions ($x=-5$ and $x=4$).

2. Region 1, $x<-3$ (both insides negative): $|x+3|=-(x+3)$, $|x-2|=-(x-2)$. Equation: $-(x+3)-(x-2)=9\Rightarrow -2x-1=9\Rightarrow x=-5$. Why this step? Flip both bars because each inside is negative here. Check $-5<-3$ ✓.
3. Region 2, $-3\le x\le 2$ (first $+$, second $-$): $(x+3)-(x-2)=9\Rightarrow 5=9$. False. Why this step? In the middle the two flips cancel; the sum is stuck at the gap width $5$, never $9$. No solution here.
4. Region 3, $x>2$ (both positive): $(x+3)+(x-2)=9\Rightarrow 2x+1=9\Rightarrow x=4$. Check $4>2$ ✓. Why this step? No flips needed; both bars open straight.

Verify: $x=-5$: $|-2|+|-7|=2+7=9$ ✓. $x=4$: $|7|+|2|=7+2=9$ ✓. Answer: $x\in \{-5,4\}$.

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Forecast: "differs by at most" is a between statement — expect a single closed interval centred at $250$.

1. Translate "difference" into bars. The difference between $L$ and target $250$ is $L-250$; its size is $|L-250|$. Why this step? Distance on a line is the absolute value of the gap; the words "differs by" mean exactly this size.
2. Translate "at most $0.4$". $|L-250|\le 0.4$. Why this step? "At most" is $\le$; a bound on distance is a less-than-or-equal absolute-value inequality.
3. Unfold (between-rule): $-0.4\le L-250\le 0.4$. Why this step? Same AND-logic as Example 4.
4. Add $250$ to all parts: $249.6\le L\le 250.4$. Why this step? Recentre on the true length by undoing the "$-250$."

Verify: midpoint $L=250$: $|0|=0\le 0.4$ ✓ (accepted); edge $L=250.4$: $|0.4|=0.4\le 0.4$ ✓; a reject $L=250.5$: $|0.5|=0.5>0.4$, correctly failed ✓. Units: all in mm, consistent ✓. Answer: $L\in [249.6,250.4]$ mm.

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Forecast: two absolute values equal to each other. It looks like four cases, but it collapses to two clean lines. Predict two roots.

1. Key idea: $|A|=|B|$ means the two quantities are the same size, so $A=B$ or $A=-B$. Why this step? Equal distances from zero means the numbers are either identical or exact mirror images. The other two sign combinations are duplicates of these.
2. Case $A=B$: $2x-1=x+4\Rightarrow x=5$. Why this step? Same size, same sign.
3. Case $A=-B$: $2x-1=-(x+4)=-x-4\Rightarrow 3x=-3\Rightarrow x=-1$. Why this step? Same size, opposite sign.

Verify: $x=5$: $|9|=9$ and $|9|=9$ ✓. $x=-1$: $|-3|=3$ and $|3|=3$ ✓. Answer: $x\in \{-1,5\}$.

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Recall check

Recall Why does

$|A|<b$ use AND but $|A|>b$ use OR? Less-than traps you inside one interval (need both walls) ::: AND / between; greater-than lets you flee to either far side ::: OR / two regions.

Recall When is

$|A|=b$ unsolvable, and when does it give exactly one root? Unsolvable when $b<0$ (distance can't be negative) ::: one root when $b=0$ (the two branches merge).

Recall In Example 3, why was

$x=1$ thrown out? It failed both the RHS gatekeeper $2x-5\ge 0$ ($2(1)-5=-3<0$) and its branch assumption $x-4\ge 0$ ($1-4=-3<0$) ::: so it was extraneous.