Why does "less than" give AND but "greater than" give OR?

Picture distance on a number line. |x| 5 collects points farther than 5 units away, and there are two separate territories, one to the left of -5 and one to the right of 5. A number lands in one or the other, never both, so the connector is OR.

How do I know when an absolute-value problem has no solution?

Can I split the cases before isolating the absolute value?

No, and doing so is one of the most common sources of wrong answers. The two-case rule applies only when the bars stand alone on one side. In 2|x - 1| + 5 = 13 you must first subtract 5 and divide by 2 to reach |x - 1| = 4. Splitting early also hides sign flips: dividing by a negative coefficient reverses an inequality and can turn an OR problem into an AND problem.

Do I always have to check my answers?

Check whenever a variable appears outside the bars, as in |3x - 1| = x + 5, because one case can produce a value that makes the non-absolute side negative, which is impossible. When the other side is a plain positive constant, both cases always work, but substituting back still catches arithmetic slips. For inequalities, testing one value inside your region and one outside is the fastest way to confirm you chose the right connector.

ALG1-3.3

Absolute-Value Equations & Inequalities

Q: Can I split the cases before isolating the absolute value?

No, and doing so is one of the most common sources of wrong answers. The two-case rule applies only when the bars stand alone on one side. In 2|x - 1| + 5 = 13 you must first subtract 5 and divide by 2 to reach |x - 1| = 4. Splitting early also hides sign flips: dividing by a negative coefficient reverses an inequality and can turn an OR problem into an AND problem.

Q: Do I always have to check my answers?

Check whenever a variable appears outside the bars, as in |3x - 1| = x + 5, because one case can produce a value that makes the non-absolute side negative, which is impossible. When the other side is a plain positive constant, both cases always work, but substituting back still catches arithmetic slips. For inequalities, testing one value inside your region and one outside is the fastest way to confirm you chose the right connector.

Learn to solve absolute-value equations and inequalities by reading |x| as distance: split into two cases, spot no-solution cases, and use AND vs OR.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on Absolute-Value Equations & Inequalities, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

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What this lesson covers

Every absolute-value problem in this lesson comes from one idea:

|x|

measures distance from zero on the number line, and distance is never negative. Once you read the bars that way,

|x| = 7

stops being mysterious — it asks "which numbers sit 7 units from 0?" and there are two of them,

7

and

-7

.

That single question also explains why

|x| = -7

has no solution at all, why

|x| < 7

describes a stretch of numbers squeezed between

-7

and

7

, and why

|x| > 7

describes two separate pieces heading off in opposite directions. In this lesson you will isolate the absolute-value bars, split equations into two cases, translate "less than" into an AND compound inequality and "greater than" into an OR, and check your answers so a false case never sneaks through.

Absolute Value Means Distance from Zero

The formal definition says

|x| = x

when

x \ge 0

and

|x| = -x

when

x < 0

. The picture is easier:

|x|

is how far

x

is from

0

, measured without direction. So

|6| = 6

and

|-6| = 6

, because both numbers sit six units from the origin.

That distance reading extends to expressions inside the bars.

|2x - 5|

is the distance from

2x - 5

to zero.

|x - 3|

can also be read as the distance between

x

and

3

, which is why

|x - 3| = 4

means "

x

is 4 units away from 3" and gives

x = 7

x = -1

.

Two consequences follow immediately and both show up all unit long. First, an absolute value can never output a negative number, so an equation like

|4x + 1| = -9

has no solution — no distance is negative. Second, an absolute value can equal zero in exactly one way:

|4x + 1| = 0

forces

4x + 1 = 0

, a single answer.

Where students go wrong early: treating the bars like parentheses and "distributing" a coefficient through them, writing

3|x - 2|

|3x - 6|

— that one happens to be true, but

-3|x-2|

is not

|-3x+6|

, since the left side is negative or zero and the right side is never negative. Bars are a function, not a grouping symbol you can slide numbers into.

Always isolate the absolute-value expression on one side before you split into cases.

Splitting an Equation into Two Cases

To solve

|ax + b| = c

with

c > 0

, write the two cases that produce that distance:

ax + b = c \quad \text{or} \quad ax + b = -c

Solve each linear equation separately. You normally get two answers.

The order of operations matters. In

2|x + 4| - 3 = 9

, you cannot split yet, because the bars are not alone. Add 3 to get

2|x+4| = 12

, divide by 2 to get

|x + 4| = 6

, and only then split into

x + 4 = 6

x + 4 = -6

, giving

x = 2

x = -10

Right side after isolating	What happens	Number of solutions
$c > 0$	Split into two cases	2
$c = 0$	One case only	1
$c < 0$	Impossible — distance is never negative	0 (empty set)

A harder version has variables on both sides, like

|3x - 1| = x + 5

. The same split works —

3x - 1 = x + 5

3x - 1 = -(x + 5)

— giving

x = 3

and

x = -1

. But now you must check both in the original equation, because the right side could turn out negative for some candidate. Here both check out, since

x + 5

is positive at

x = 3

and at

x = -1

.

A very common slip is forgetting the parentheses on the negative case: writing

3x - 1 = -x + 5

instead of

3x - 1 = -x - 5

. Negate the entire right side, every term.

Less Than Means AND: A Band Around the Center

Read

|x| < 5

as "the distance from

x

to zero is under 5 units." Those numbers form one connected stretch: everything strictly between

-5

and

5

. In symbols,

|x| < 5 \iff -5 < x < 5 \iff x > -5 \text{ and } x < 5.

The general rule for

c > 0

|ax + b| < c \iff -c < ax + b < c

, and the same pattern with

\le

throughout. Solve the three-part inequality by doing the same operation to all three parts. For

|2x - 6| \le 4

: write

-4 \le 2x - 6 \le 4

, add 6 to get

2 \le 2x \le 10

, divide by 2 to get

1 \le x \le 5

. The graph is a single segment with closed circles at 1 and 5.

A memory hook many classes use is "less thAND": less-than gives AND. The solution is one interval, and it always contains the center point where the inside expression equals zero.

Special cases follow from the distance idea again.

|x + 2| < -1

has no solution, since no distance is less than a negative number.

|x + 2| \le 0

has exactly one solution,

x = -2

, because only zero distance qualifies. And something like

|x + 2| > -1

is true for every real number.

Where students go wrong: flipping only one inequality sign when dividing by a negative, or writing the answer as

-5 > x > 5

, which reads "

x

is below

-5

and above 5" — impossible. In a correct three-part inequality the smaller number is always on the left.

Greater Than Means OR: Two Pieces Heading Outward

Now read

|x| > 5

as "the distance from

x

to zero is more than 5 units." Those points are not between anything — they lie far out on both ends. The solution splits:

|x| > 5 \iff x < -5 \text{ or } x > 5.

For

c > 0

, the rule is

|ax + b| > c \iff ax + b > c

ax + b < -c

, with

\ge

behaving the same way. Solve

|3x + 1| \ge 7

by writing

3x + 1 \ge 7

3x + 1 \le -7

. The first gives

x \ge 2

; the second gives

3x \le -8

, so

x \le -\frac{8}{3}

. The graph is two rays pointing away from each other, with a gap in the middle.

Form	Compound type	Solution shape	Interval notation
$\\|x\\| < c$	AND	one segment between	$(-c,\ c)$
$\\|x\\| > c$	OR	two outward rays	$(-\infty,\ -c) \cup (c,\ \infty)$

The negative-constant cases reverse from the previous section.

|x - 4| > -2

is true for all real numbers, because every distance beats a negative number.

|x - 4| < -2

has no solution.

Two frequent errors: writing an OR answer as a single three-part inequality such as

5 < x < -5

, which describes nothing, and joining OR pieces with the word "and." Say it out loud — a number cannot be both below

-5

and above 5, so "or" is the only honest connector. Test one number from each region against the original inequality to confirm your shading before you finish.

Building and Checking Absolute-Value Models

Absolute value is the natural tool whenever a problem talks about a tolerance: how far a measured value may drift from a target. If a machine fills bottles to a target of 500 milliliters with a tolerance of 8 milliliters, the acceptable amounts

v

satisfy

|v - 500| \le 8

, which unpacks to

492 \le v \le 508

. The number subtracted inside the bars is always the target, and the number on the right is always the allowed error.

The reverse translation is just as useful. Given

460 \le t \le 480

, find the center by averaging the endpoints,

\frac{460 + 480}{2} = 470

, and the radius by halving the width,

\frac{480 - 460}{2} = 10

. The compact form is

|t - 470| \le 10

.

Checking is not optional in this topic. For an equation, substitute each answer back into the original, bars and all. For an inequality, pick a test value inside your solution region and one outside, and verify that the first satisfies the original statement while the second does not.

One more caution: when the isolating step involves dividing by a negative, the inequality direction flips before you split. From

-2|x - 1| > -10

, divide by

-2

and flip to get

|x - 1| < 5

— now it is an AND problem, giving

-4 < x < 6

. Students who split first, while a negative coefficient is still attached to the bars, almost always land on the wrong connector.

Key terms

Absolute value.: The distance of a number from zero on the number line, written $|x|$ ; it is never negative.
Isolating the absolute value.: Using inverse operations to get the expression in bars alone on one side before splitting into cases.
Case split.: Rewriting $|ax+b| = c$ as the two equations $ax + b = c$ and $ax + b = -c$ .
AND compound inequality.: A statement true only where both parts hold; $|x| < c$ becomes $-c < x < c$ , one connected interval.
OR compound inequality.: A statement true where at least one part holds; $|x| > c$ becomes $x < -c$ or $x > c$ , two separate rays.
Empty set.: The solution set with no members, written $\varnothing$ ; it occurs for $|expr| = c$ or $|expr| < c$ when $c < 0$ .
Extraneous solution.: A value produced by the case split that fails in the original equation, usually because it would force an absolute value to equal a negative amount.
Tolerance.: The maximum allowed distance between a measurement and its target value, modeled by $|measurement - target| \le tolerance$ .

Worked example

Solve and graph:

-3|2x - 4| + 15 \ge 3

Isolate the absolute value first. Subtract 15 from both sides:

-3|2x - 4| \ge -12

.

Divide both sides by

-3

. Because the divisor is negative, the inequality sign flips:

|2x - 4| \le 4

.

Now read the form. The absolute value is less than or equal to a positive number, so this is an AND situation — the inside expression must sit between

-4

and

4

-4 \le 2x - 4 \le 4

Add 4 to all three parts:

0 \le 2x \le 8

. Divide all three parts by 2:

0 \le x \le 4

.

Graph a closed circle at 0, a closed circle at 4, and shade the segment between them. In interval notation the answer is

[0,\ 4]

.

Check with test values. Inside, take

x = 2

-3|2(2) - 4| + 15 = -3(0) + 15 = 15

, and

15 \ge 3

is true. Outside, take

x = 5

-3|10 - 4| + 15 = -18 + 15 = -3

, and

-3 \ge 3

is false. At the boundary

x = 4

-3|4| + 15 = 3

, and

3 \ge 3

is true, confirming the closed circle.

The step that decides everything here is flipping the sign when dividing by

-3

. Splitting before that division would have produced an OR answer with two rays, which the test values would have exposed as wrong.

Practice questions

Which statement is equivalent to

|5 - x| > 3

$-3 < 5 - x < 3$
$5 - x > 3$ or $5 - x < -3$
$5 - x > 3$ and $5 - x > -3$
$5 - x > -3$

Answer: $5 - x > 3$ or $5 - x < -3$

A greater-than absolute value asks for points whose distance from zero exceeds 3, and those points lie outside the band in two separate directions, so the correct connector is OR. The inside expression must be above 3 or below

-3

. The first choice is the AND form that belongs to

|5 - x| < 3

. The third choice uses AND, and its two parts overlap into a single ray, which cannot describe two outward pieces. Solving the correct pair gives

x < 2

x > 8

A thermostat is supposed to hold a greenhouse at 68 degrees Fahrenheit, and the actual temperature

T

is allowed to drift at most 2.5 degrees from that target. Write an absolute-value inequality for the acceptable temperatures, then state the range as a compound inequality.

Answer: $|T - 68| \le 2.5$ , which gives $65.5 \le T \le 70.5$ .

The phrase "at most 2.5 degrees from 68" is a distance statement: the distance between

T

and 68 cannot exceed 2.5, so

|T - 68| \le 2.5

. Because the absolute value is less than or equal to a positive number, unpack it as an AND inequality:

-2.5 \le T - 68 \le 2.5

. Adding 68 to all three parts gives

65.5 \le T \le 70.5

. Notice the target sits at the center of the interval and the tolerance is half the interval's width, a relationship you can use to move back and forth between the two forms.

Solve

|4x + 3| = 2x + 9

and identify any extraneous solutions.

Answer: $x = 3$ and $x = -2$ ; neither is extraneous.

Split into two cases. Case one:

4x + 3 = 2x + 9

gives

2x = 6

, so

x = 3

. Case two negates the whole right side:

4x + 3 = -(2x + 9) = -2x - 9

, so

6x = -12

and

x = -2

. Because a variable appears on the right, check both. At

x = 3

: the left side is

|15| = 15

and the right side is

2(3) + 9 = 15

, true. At

x = -2

: the left side is

|-5| = 5

and the right side is

2(-2) + 9 = 5

, true. Both survive. Had a candidate made the right side negative, it would have been extraneous, since an absolute value cannot equal a negative number.

FAQ

Why does "less than" give AND but "greater than" give OR?: Picture distance on a number line. $|x| < 5$ collects every point closer than 5 units to zero, and those points form one unbroken stretch between $-5$ and $5$ , so $x$ must satisfy both bounds at once — that is AND. $|x| > 5$ collects points farther than 5 units away, and there are two separate territories, one to the left of $-5$ and one to the right of 5. A number lands in one or the other, never both, so the connector is OR.
How do I know when an absolute-value problem has no solution?: Isolate the bars first, then look at the number on the other side. If $|expr| = c$ or $|expr| < c$ or $|expr| \le c$ with $c$ negative, there is no solution, because a distance can never be negative. If $|expr| > c$ or $|expr| \ge c$ with $c$ negative, every real number works. The special value zero splits the difference: $|expr| = 0$ has exactly one solution, and $|expr| < 0$ has none while $|expr| \le 0$ has one.
Can I split the cases before isolating the absolute value?: No, and doing so is one of the most common sources of wrong answers. The two-case rule applies only when the bars stand alone on one side. In $2|x - 1| + 5 = 13$ you must first subtract 5 and divide by 2 to reach $|x - 1| = 4$ . Splitting early also hides sign flips: dividing by a negative coefficient reverses an inequality and can turn an OR problem into an AND problem.
Do I always have to check my answers?: Check whenever a variable appears outside the bars, as in $|3x - 1| = x + 5$ , because one case can produce a value that makes the non-absolute side negative, which is impossible. When the other side is a plain positive constant, both cases always work, but substituting back still catches arithmetic slips. For inequalities, testing one value inside your region and one outside is the fastest way to confirm you chose the right connector.

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