Absolute Value Equations and Inequalities

Intuition

This chapter has been one argument the whole way through, and every lesson pinned down a different reason you get forced into a case analysis. In the first lesson the culprit was an expression that might vanish: multiplying by it can add a root, dividing by it can lose one. In the second it was a coefficient you could not inspect, a parameter hiding whether you were even allowed to divide. In the third it was a multiplier whose sign was unknown, since an inequality carries an order and a negative multiplier flips it. This lesson closes the argument with the cleanest reason of all. The absolute value is defined by cases: the rule $\lvert x \rvert = x$ when $x \ge 0$ and $\lvert x \rvert = -x$ when $x < 0$ is a fork in the road built into the symbol. So every equation and inequality containing the bars inherits that fork. You will not be surprised, then, that solving $\lvert X \rvert = Y$ honestly needs a split, or that the split can hand you candidates that do not actually solve the equation. What may surprise you is that the same lesson holds the chapter’s first split that is a genuine equivalence. That split is the one place where two cases collapse back into a single true statement with nothing to check, and it is worth the wait.

Learning goals

Read the definition as a case split, and $\sqrt{x^2}$ as $\lvert x \rvert$
Require $Y \ge 0$ when solving $\lvert X \rvert = Y$
Split $\lvert X \rvert = \lvert Y \rvert$ with no check needed
Turn a band or two rays out of an inequality
Use breakpoints to solve a sum of two absolute values

The definition is a case split

You already met the absolute value as distance from zero: $\lvert x \rvert$ is the length you get by dropping a number’s sign, so $\lvert 5 \rvert = 5$ and $\lvert -5 \rvert = 5$ . That picture is the right intuition, and this lesson leans on the exact rule behind it, the piecewise definition

\lvert x \rvert = \begin{cases} x & \text{if } x \ge 0, \\ -x & \text{if } x < 0. \end{cases}

Read it as an instruction that branches. To evaluate $\lvert x \rvert$ you first ask a yes-or-no question about $x$ , is it nonnegative, and only then do you know which formula applies. The bars are not a single operation the way adding or squaring is; they are a hidden if. Every difficulty in this lesson traces back to that one fact. When the quantity inside the bars contains an unknown, you cannot answer the branching question in advance. So you are forced to carry both branches until something later settles which one you are in.

There is a second way to write the same rule, with no visible cases at all, and it will earn its keep when we set two absolute values equal. For every real number $x$ ,

\lvert x \rvert = \sqrt{x^2}.

Squaring throws away the sign of $x$ , and the square root symbol, which always returns the nonnegative root, hands back the size without it.

Why $\lvert x \rvert = \sqrt{x^2}$ #

The square root symbol is defined to return the nonnegative number whose square is its input. So $\sqrt{x^2}$ is, by definition, the nonnegative number whose square is $x^2$ . The claim is that $\lvert x \rvert$ is exactly that number, and two facts settle it. First, $\lvert x \rvert$ squares to $x^2$ , since $\lvert x \rvert^2 = x^2$ whether $x$ is positive or negative. Second, $\lvert x \rvert \ge 0$ by the meaning of distance. There is only one nonnegative number whose square is $x^2$ , so $\lvert x \rvert$ must be it, that is $\lvert x \rvert = \sqrt{x^2}$ .

The two cases of the definition confirm the same thing, and the second case is where intuition slips. When $x \ge 0$ , we have $\sqrt{x^2} = x$ , which agrees with $\lvert x \rvert = x$ . When $x < 0$ , the number $-x$ is positive and squares to $x^2$ , so $\sqrt{x^2} = -x$ , which agrees with $\lvert x \rvert = -x$ . The two branches of the piecewise rule are precisely the two ways $\sqrt{x^2}$ can resolve.

This is the source of a stubborn error. Writing $\sqrt{x^2} = x$ is correct only when $x \ge 0$ ; for a negative $x$ it reports the wrong sign, claiming a negative value for a root that is always nonnegative. With $x = -5$ it would give $\sqrt{25} = -5$ , which is false, since $\sqrt{25} = 5$ . The honest identity carries the bars, $\sqrt{x^2} = \lvert x \rvert$ , and those bars are the case split of this section wearing a different costume.

When the right side carries the unknown

Solving $\lvert X \rvert = c$ for a number $c$ was a one-glance affair: you check the sign of $c$ once. A positive $c$ splits into $X = c$ or $X = -c$ , the value $c = 0$ gives the single solution $X = 0$ , and a negative $c$ gives none. Now let the right side be an expression $Y$ that contains the unknown, as in $\lvert x - 1 \rvert = 2x - 4$ . That single up-front check is gone, because the sign of $Y$ depends on the very $x$ you are solving for. Splitting anyway is where solutions get manufactured.

Passing from $\lvert X \rvert = Y$ to ” $X = Y$ or $X = -Y$ ” looks like the same safe move, but it is not an equivalence. The left side $\lvert X \rvert$ can never be negative, so the original equation secretly demands $Y \ge 0$ as well, and nothing in the two split equations remembers that demand. Dropping it is exactly a non-reversible step of the kind the first lesson warned about. You can travel forward from $\lvert X \rvert = Y$ to the pair of equations, but you cannot travel back. The return trip fails because a solution of $X = -Y$ at a place where $Y$ is negative satisfies the split without satisfying the original. The step can add solutions the equation never had. The honest statement keeps the dropped condition in plain sight:

\lvert X \rvert = Y \quad\Longleftrightarrow\quad Y \ge 0 \ \text{ and } \ \bigl(X = Y \ \text{ or } \ X = -Y\bigr).

Why the condition $Y \ge 0$ cannot be dropped#

Read the equivalence in both directions at a value of $x$ , with $X$ and $Y$ standing for the two sides evaluated there.

Forward. Suppose $\lvert X \rvert = Y$ . The left side is an absolute value, so $Y = \lvert X \rvert \ge 0$ , which gives the condition $Y \ge 0$ for free. And $\lvert X \rvert = Y$ with $Y \ge 0$ says $X$ is a number whose distance from zero is $Y$ , and there are exactly two such numbers, $Y$ and $-Y$ , so $X = Y$ or $X = -Y$ . The right side holds.

Backward. Suppose $Y \ge 0$ and either $X = Y$ or $X = -Y$ . If $X = Y$ then $\lvert X \rvert = \lvert Y \rvert = Y$ , the last step because $Y \ge 0$ . If $X = -Y$ then $\lvert X \rvert = \lvert -Y \rvert = \lvert Y \rvert = Y$ again. Either way $\lvert X \rvert = Y$ , the original.

The backward direction is where $Y \ge 0$ does its work, and dropping it breaks the argument. Suppose $Y$ were negative and $X = -Y$ . Then $\lvert X \rvert = \lvert Y \rvert = -Y$ , a positive number, while $Y$ itself is negative, so $\lvert X \rvert = Y$ is false. That is precisely a candidate the bare split accepts and the original rejects. So the split without its side condition is a one-way street. The way to make it safe is to solve as usual, and then keep only the candidates for which $Y \ge 0$ .

In practice this means the same discipline as clearing denominators in the first lesson. Run the convenient non-reversible step, then check each candidate in the original and discard the extraneous ones. Substituting into the original is the surest check, but testing $Y \ge 0$ is faster and catches exactly the same failures.

Worked example 1 Solve $\lvert x - 1 \rvert = 2x - 4$

Here $X = x - 1$ and $Y = 2x - 4$ . Split into the two cases, treating each result as a candidate until the condition $Y \ge 0$ is checked.

Case $X = Y$ :

x - 1 = 2x - 4 \quad\Longrightarrow\quad 3 = x, \qquad x = 3.

Case $X = -Y$ :

x - 1 = -(2x - 4) = -2x + 4 \quad\Longrightarrow\quad 3x = 5, \qquad x = \tfrac{5}{3}.

Now test each against $Y = 2x - 4 \ge 0$ . At $x = 3$ the right side is $2(3) - 4 = 2$ , which is nonnegative, and $\lvert 3 - 1 \rvert = 2$ checks, so $x = 3$ is a genuine solution. At $x = \frac{5}{3}$ the right side is $2 \cdot \frac{5}{3} - 4 = \frac{10}{3} - \frac{12}{3} = -\frac{2}{3}$ , which is negative, so this candidate is extraneous. The original would read $\lvert \frac{5}{3} - 1 \rvert = -\frac{2}{3}$ , an absolute value equal to a negative number, which is impossible.

\text{candidates } x = 3, \ \tfrac{5}{3}; \qquad \text{solution set } \{3\}.

The algebra offered two candidates and the check kept one. Skipping the check would have reported the false solution $\frac{5}{3}$ .

Worked example 2 Solve $\lvert 2x + 1 \rvert = x - 3$

Split with $X = 2x + 1$ and $Y = x - 3$ .

Case $X = Y$ :

2x + 1 = x - 3 \quad\Longrightarrow\quad x = -4.

Case $X = -Y$ :

2x + 1 = -(x - 3) = -x + 3 \quad\Longrightarrow\quad 3x = 2, \qquad x = \tfrac{2}{3}.

Check each against $Y = x - 3 \ge 0$ . At $x = -4$ the right side is $-4 - 3 = -7 < 0$ , extraneous. At $x = \frac{2}{3}$ the right side is $\frac{2}{3} - 3 = -\frac{7}{3} < 0$ , also extraneous. Both candidates fail.

\text{solution set } \varnothing.

This one unsettles people. Both cases were solved correctly, each produced a clean number, and yet the equation has no solution at all. The reason is visible without any of the algebra: the right side $x - 3$ must be nonnegative for the equation to stand. So only $x \ge 3$ is even eligible, and neither candidate reached that far. When a required condition like $Y \ge 0$ looks hard to satisfy, expect an empty answer.

Check your understanding

Solving $\lvert x - 5 \rvert = 3x$ produces the two candidates $x = \frac{5}{4}$ and $x = -\frac{5}{2}$ . Which is extraneous, and why?

x = -\frac{5}{2}

, because the right side

3x

is negative there B

x = \frac{5}{4}

, because the right side

3x

is negative there C neither, both are genuine solutions D

x = -\frac{5}{2}

, because an absolute value can be negative

Setting two absolute values equal

Now put an absolute value on the right as well, $\lvert X \rvert = \lvert Y \rvert$ . After a chapter spent exposing splits that quietly fail, this is the one that does not. Here the case split is a true equivalence with no side condition attached, and so is the shortcut that replaces it:

\lvert X \rvert = \lvert Y \rvert \quad\Longleftrightarrow\quad X = Y \ \text{ or } \ X = -Y \quad\Longleftrightarrow\quad X^2 = Y^2.

The side condition disappears for one plain reason: both sides are absolute values, hence both nonnegative from the start, so there is no hidden demand like $Y \ge 0$ left to forget. The very step that was dangerous a moment ago, squaring, becomes exact.

Why $\lvert X \rvert = \lvert Y \rvert$ is equivalent to $X^2 = Y^2$ #

Read the chain in two hops and confirm each hop is reversible.

First hop: $\lvert X \rvert = \lvert Y \rvert$ is equivalent to $X^2 = Y^2$ . Both $\lvert X \rvert$ and $\lvert Y \rvert$ are nonnegative, and on nonnegative numbers squaring is one-to-one, undone by the nonnegative square root, so for such numbers $a = b$ holds exactly when $a^2 = b^2$ . Squaring both sides of $\lvert X \rvert = \lvert Y \rvert$ gives $\lvert X \rvert^2 = \lvert Y \rvert^2$ , that is $X^2 = Y^2$ . Conversely, taking the nonnegative square root of $X^2 = Y^2$ gives $\sqrt{X^2} = \sqrt{Y^2}$ , that is $\lvert X \rvert = \lvert Y \rvert$ by the identity of the first section. Neither direction leaks.

Second hop: $X^2 = Y^2$ is equivalent to $X = Y$ or $X = -Y$ . Move everything to one side and factor the difference of squares:

X^2 - Y^2 = 0 \quad\Longleftrightarrow\quad (X - Y)(X + Y) = 0.

A product is zero exactly when one of its factors is zero, so this holds precisely when $X - Y = 0$ or $X + Y = 0$ , that is $X = Y$ or $X = -Y$ . Every step is an equivalence, so the full chain is one.

Set this beside the first lesson. Squaring an ordinary equation is a textbook non-reversible step, the kind that breeds extraneous roots, because two numbers with equal squares need not be equal. What rescues it here is that the quantities being squared, $\lvert X \rvert$ and $\lvert Y \rvert$ , are already known to be nonnegative. And on the nonnegative numbers, equal squares really do force equal values. The equivalence is not luck; it is bought by the nonnegativity the bars guarantee.

Worked example 3 Solve $\lvert 2x - 1 \rvert = \lvert x + 4 \rvert$

Because both sides are absolute values, split with no check needed. Set the insides equal, then set one equal to the other’s opposite.

Case $X = Y$ :

2x - 1 = x + 4 \quad\Longrightarrow\quad x = 5.

Case $X = -Y$ :

2x - 1 = -(x + 4) = -x - 4 \quad\Longrightarrow\quad 3x = -3, \qquad x = -1.

Both are genuine, so the solution set is $\{-1, 5\}$ . Nothing is discarded, because setting two absolute values equal carries no side condition.

The squaring route lands in the same place. From $\lvert 2x - 1 \rvert = \lvert x + 4 \rvert$ , square both sides:

(2x - 1)^2 = (x + 4)^2 \quad\Longrightarrow\quad 4x^2 - 4x + 1 = x^2 + 8x + 16.

Collect on one side and simplify:

3x^2 - 12x - 15 = 0 \quad\Longrightarrow\quad x^2 - 4x - 5 = 0 \quad\Longrightarrow\quad (x - 5)(x + 1) = 0,

so $x = 5$ or $x = -1$ , exactly the two solutions from the case split. Squaring was safe here for the reason just proved, not by accident.

Check your understanding

How many real solutions does $\lvert x - 6 \rvert = \lvert x + 2 \rvert$ have?

A exactly two,

x = 6

and

x = -2

B exactly one,

x = 2

C none D every real number

Inequalities, read as distance

The two absolute value inequalities are already familiar, and the distance meaning is what makes them memorable instead of memorized. The statement $\lvert X \rvert < c$ says $X$ lies within $c$ of zero, so it is trapped in a band. The statement $\lvert X \rvert > c$ , by contrast, says $X$ lies farther than $c$ from zero, so it splits into two rays:

\lvert X \rvert < c \iff -c < X < c,

\lvert X \rvert > c \iff X < -c \ \text{ or } \ X > c.

Both equivalences hold for every real $c$ , not only for the positive values where the band is a genuine interval and the rays point outward. That is exactly why reciting them fails while reading them does not: you have to see what each one says once $c$ reaches zero and turns negative. Push each to its boundary and past it.

Take $\lvert X \rvert < c$ when $c \le 0$ . A distance cannot be less than a nonpositive number, because distances start at $0$ : if $c = 0$ the demand $\lvert X \rvert < 0$ is impossible, and for $c < 0$ it is more impossible still. So $\lvert X \rvert < c$ has solution set $\varnothing$ whenever $c \le 0$ , with no band to draw. The template $-c < X < c$ needs no special case: at $c = 0$ it degenerates to $0 < X < 0$ , already empty, and for $c < 0$ it reads with $-c > c$ , ends crossed, empty again. The formula still tells the truth, but only if you read a crossed band as empty rather than as a slip of the pen.

Take $\lvert X \rvert > c$ when $c < 0$ . Every distance is at least $0$ , which already exceeds any negative $c$ , so $\lvert X \rvert > c$ holds for every real number and the solution set is all of $\mathbb{R}$ . Two boundary readings deserve their own names. The inequality $\lvert X \rvert > 0$ holds for every $X$ except the single place where the distance is $0$ , so it describes all $x$ with $X \neq 0$ . And $\lvert X \rvert \ge 0$ excludes nothing at all, holding for every real number, since a distance is always at least zero. These are not trick questions; they are the honest values of the formula at and beyond its boundary. A solver who only remembers “band or two rays” will misreport every one of these honest values.

Worked example 4 Three inequalities at or past the boundary

Read each by distance rather than by a template.

The inequality $\lvert 3x - 5 \rvert < -2$ asks an absolute value to fall below $-2$ . It never can, so no $x$ works and the solution set is $\varnothing$ .

The inequality $\lvert 4 - 7x \rvert > -1$ asks an absolute value to exceed $-1$ . It always does, since it is at least $0$ , so every $x$ works and the solution set is all of $\mathbb{R}$ .

The inequality $\lvert x - 6 \rvert > 0$ asks the distance from $x$ to $6$ to be strictly positive. That holds for every $x$ except $x = 6$ , where the distance is zero, so the solution set is every real number but $6$ :

\lvert x - 6 \rvert > 0 \quad\Longleftrightarrow\quad x \neq 6 \quad\Longleftrightarrow\quad (-\infty, 6) \cup (6, \infty).

None of these needed a case split or a band. Checking the sign of the right side, the same reflex from solving $\lvert X \rvert = c$ , settled all three at a glance.

Worked example 5 Read $\lvert x - 4 \rvert \le 3$ as a distance

The quantity $\lvert x - 4 \rvert$ is the distance from $x$ to $4$ , so $\lvert x - 4 \rvert \le 3$ says $x$ is within $3$ of $4$ . The numbers within $3$ of $4$ run from $4 - 3 = 1$ up to $4 + 3 = 7$ :

-3 \le x - 4 \le 3 \quad\Longrightarrow\quad 1 \le x \le 7,

the interval $[1, 7]$ , centered at $4$ with radius $3$ . Reading a band this way, as a center and a radius, is quicker than solving and it makes the answer easy to picture. In symbols, $\lvert x - a \rvert \le c$ is the closed interval $[a - c, a + c]$ , every point no more than $c$ from $a$ . The strict version $\lvert x - a \rvert < c$ is the open interval $(a - c, a + c)$ , the same band without its two endpoints.

The capstone, a sum of two absolute values

Everything in this chapter has been preparing you for a problem with two absolute values added together, like

\lvert x - 3 \rvert + \lvert x + 1 \rvert = 6.

No single split unlocks it, because two independent bars fork independently. The method is to find the breakpoints, the values where each inside expression changes sign. You then use those breakpoints to cut the number line into intervals on which every bar resolves to one definite formula. Here $x - 3$ changes sign at $x = 3$ and $x + 1$ changes sign at $x = -1$ , so the breakpoints $-1$ and $3$ split the line into three intervals. On each one, both absolute values become ordinary linear expressions, and the equation becomes an ordinary linear equation. The one catch is that a root counts only if it lands inside the interval that produced it.

Worked example 6 Solve $\lvert x - 3 \rvert + \lvert x + 1 \rvert = 6$ by breakpoints

The breakpoints $-1$ and $3$ give three intervals. On each, resolve the two bars with the piecewise definition, then solve the linear equation.

Interval $x < -1$ . Here $x - 3 < 0$ so $\lvert x - 3 \rvert = 3 - x$ , and $x + 1 < 0$ so $\lvert x + 1 \rvert = -x - 1$ . The equation becomes

(3 - x) + (-x - 1) = 6 \quad\Longrightarrow\quad 2 - 2x = 6 \quad\Longrightarrow\quad x = -2.

Since $-2 < -1$ , the root lies in this interval, so keep it.

Interval $-1 \le x \le 3$ . Here $x + 1 \ge 0$ so $\lvert x + 1 \rvert = x + 1$ , while $x - 3 \le 0$ so $\lvert x - 3 \rvert = 3 - x$ . The equation becomes

(3 - x) + (x + 1) = 6 \quad\Longrightarrow\quad 4 = 6,

which is false. The variable cancelled and left a contradiction, so this interval contributes no solution.

Interval $x > 3$ . Here both insides are positive, $\lvert x - 3 \rvert = x - 3$ and $\lvert x + 1 \rvert = x + 1$ , so

(x - 3) + (x + 1) = 6 \quad\Longrightarrow\quad 2x - 2 = 6 \quad\Longrightarrow\quad x = 4.

Since $4 > 3$ , the root lies in this interval, so keep it.

Collecting the survivors, the solution set is $\{-2, 4\}$ . The discarded middle interval was not a failure of the algebra; it was the method telling you the sum can never reach $6$ there.

The left side of the equation is the distance from x to 3 plus the distance from x to minus 1. The two anchors minus 1 and 3 are 4 apart, so between them the two distances split that gap and always sum to 4, shown by the shaded middle segment, which is why that interval gives no solution. To reach a sum of 6 you step 1 unit past an anchor, since each unit outside adds 2 to the sum, landing on the two solutions x = minus 2 and x = 4, symmetric about the midpoint 1.

That middle interval rewards a second look, because geometry explains it fully. Read $\lvert x - 3 \rvert$ as the distance from $x$ to $3$ and $\lvert x + 1 \rvert$ as the distance from $x$ to $-1$ . The left side is then the total distance from $x$ to the two fixed anchors $3$ and $-1$ . So the equation asks which points have distances to $3$ and to $-1$ that add up to $6$ . The two anchors are $4$ apart. For any $x$ between them the two distances simply split that gap, so their sum is exactly $4$ wherever $x$ falls in the middle. That constant sum is why the middle interval produced the flat statement $4 = 6$ and no solution. To reach a sum of $6$ you must step outside the anchors. Each unit you travel beyond an endpoint adds $1$ to each of the two distances, and hence $2$ to the sum. Starting from $4$ , you need $2$ more, so you step exactly $1$ unit past an end. One unit left of $-1$ lands on $-2$ , and one unit right of $3$ lands on $4$ . The algebra and the picture agree, and the picture explains why the answer had to be symmetric about $1$ , the midpoint of the anchors.

Worked example 7 When the target equals the gap, $\lvert x - 2 \rvert + \lvert x + 4 \rvert = 6$

The breakpoints are $-4$ and $2$ , and the two anchors are $6$ apart, the same as the number on the right. Walk the three intervals.

Interval $x < -4$ . Both insides are negative, so $(2 - x) + (-x - 4) = 6$ gives $-2x - 2 = 6$ , hence $x = -4$ . But $-4$ is not strictly less than $-4$ , so it sits on the boundary rather than inside this interval; hold it for the next case.

Interval $-4 \le x \le 2$ . Here $\lvert x - 2 \rvert = 2 - x$ and $\lvert x + 4 \rvert = x + 4$ , so

(2 - x) + (x + 4) = 6 \quad\Longrightarrow\quad 6 = 6,

true for every $x$ in the interval. The entire interval $[-4, 2]$ solves the equation.

Interval $x > 2$ . Both insides are positive, so $(x - 2) + (x + 4) = 6$ gives $2x + 2 = 6$ , hence $x = 2$ , again a boundary point already contained in $[-4, 2]$ .

The solution set is the whole interval $[-4, 2]$ . Geometry says the same thing instantly: the anchors are $6$ apart, so between them the two distances always sum to $6$ , matching the target exactly. So every point between the two anchors is a solution of the equation. When the number on the right equals the gap between the anchors, the answer is not two points but the entire segment joining them.

These two examples map out every possibility for $\lvert x - p \rvert + \lvert x - q \rvert = d$ , with $p$ and $q$ the anchors and $g = \lvert p - q \rvert$ the gap between them. Between the anchors the sum is always $g$ , and it grows as you move outside, so three cases exhaust it. If $d < g$ the sum never dips low enough and there is no solution. If $d = g$ the sum equals $d$ across the whole middle segment, giving the interval between the anchors. If $d > g$ there are exactly two solutions, one step outside each anchor and symmetric about the midpoint. You can read off which case you are in before solving, just by comparing the target $d$ with the gap $g$ .

Check your understanding

For $\lvert x - 5 \rvert + \lvert x + 1 \rvert = d$ , the anchors are $5$ and $-1$ . For which value of $d$ does the equation have infinitely many solutions?

d = 12

d = 4

d = 0

d = 6

Common mistakes

Splitting $\lvert X \rvert = Y$ and reporting both cases without checking $Y \ge 0$ . The plain split into $X = Y$ or $X = -Y$ drops the demand that the right side be nonnegative, so it can add roots the original never had. Treat each result as a candidate and confirm it in the original equation.
Reading $\sqrt{x^2}$ as $x$ . The identity is $\sqrt{x^2} = \lvert x \rvert$ , which equals $x$ only when $x \ge 0$ ; for $x = -5$ it gives $5$ , not $-5$ . The square root symbol always returns the nonnegative root.
Attaching a check to $\lvert X \rvert = \lvert Y \rvert$ . There is no side condition here, since both sides are already nonnegative, so every case-solution is genuine and squaring to $X^2 = Y^2$ is exact rather than a source of extraneous roots.
Trusting an inequality template without reading what it says. Both equivalences hold for every real $c$ : the band $\lvert X \rvert < c \iff -c < X < c$ simply reads as empty when $c \le 0$ , since its ends cross. The two rays $\lvert X \rvert > c \iff X < -c$ or $X > c$ together cover the whole line when $c < 0$ . The templates need no side condition; the fragile one is the equation $\lvert X \rvert = c$ , which does need $c \ge 0$ . The reason is that at $c = -3$ the naive split would report $X = 3$ and $X = -3$ for an equation with no solution.
Mishandling the strict-zero cases. The inequality $\lvert X \rvert > 0$ excludes exactly the point where $X = 0$ (not every $x$ , and not none), while $\lvert X \rvert \ge 0$ is true for every real number.
Keeping a root that falls outside its interval. In a sum of absolute values, each interval’s linear equation is valid only on that interval, so a root outside the interval that produced it must be discarded. An interval whose equation reduces to a true statement instead contributes all of its points.

Key takeaways

The definition $\lvert x \rvert = x$ for $x \ge 0$ and $\lvert x \rvert = -x$ for $x < 0$ is a case split, and the same rule is $\lvert x \rvert = \sqrt{x^2}$ ; the identity $\sqrt{x^2} = \lvert x \rvert$ , not $x$ , is that split in another form.
$\lvert X \rvert = Y$ is equivalent to $Y \ge 0$ and ( $X = Y$ or $X = -Y$ ). The bare split is a non-reversible step that drops $Y \ge 0$ , so its candidates must be checked, exactly like clearing denominators in the first lesson.
$\lvert X \rvert = \lvert Y \rvert$ is equivalent to $X = Y$ or $X = -Y$ , and to $X^2 = Y^2$ ; this split is a genuine biconditional needing no check, and squaring is exact here because both sides are already nonnegative.
$\lvert X \rvert < c \iff -c < X < c$ and $\lvert X \rvert > c \iff X < -c$ or $X > c$ hold for every real $c$ : the band just reads as empty when $c \le 0$ and the rays cover all of $\mathbb{R}$ when $c < 0$ . So read the template rather than memorize a case; by contrast the equation $\lvert X \rvert = c$ genuinely needs $c \ge 0$ .
$\lvert X \rvert > 0$ holds exactly when $X \neq 0$ , and $\lvert X \rvert \ge 0$ holds for every real number; read $\lvert x - a \rvert < c$ as ” $x$ within $c$ of $a$ ”, the open interval $(a - c, a + c)$ .
A sum $\lvert x - p \rvert + \lvert x - q \rvert = d$ is solved by the breakpoints $p$ and $q$ : on each interval every bar resolves to a sign. So solve the linear equation and keep only roots inside their interval; comparing $d$ with the gap $g = \lvert p - q \rvert$ predicts none ( $d < g$ ), a whole segment ( $d = g$ ), or two points ( $d > g$ ).

Absolute Value Equations and Inequalities

Learning goals

The definition is a case split

When the right side carries the unknown

Setting two absolute values equal

Inequalities, read as distance

The capstone, a sum of two absolute values

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Free Response Questions (FRQ)