Contents / Algebra / Inequalities & Absolute Value

Chapter 3

Inequalities & Absolute Value

Absolute value, linear and quadratic inequalities, and compound inequality techniques.

Introduction

Absolute value, linear and quadratic inequalities, and compound inequality techniques.

The absolute value of a real number x, written |x|, is its distance from zero on the number line. Because distance is never negative, |x| ≥ 0 for every real x, and |x| = 0 only when x = 0.

Absolute value is one of the first functions students meet that is not a straight line or simple power; its graph is a V and it appears in distance formulas, error analysis, and defining piecewise functions.

3.1Definition by Cases

Definition 3.1 (Absolute value). For any real number $x$ ,

∣ x ∣ = {x, - x, x \geq 0, x < 0.

This piecewise definition captures the idea that absolute value "folds" the negative half of the number line onto the positive side.

Intuition. Absolute value is the "no negatives allowed" rule for distance.

Negative numbers are on the left of zero; absolute value measures how far left or right without caring about direction.

Example 3.2. |−4| = 4 because −4 is four steps left of zero; distance is 4.

Solution.

|3.5| = 3.5 because it is already on the right side.

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Exercises for §3.1 →

3.2Important Properties

Theorem 3.3 (Properties of absolute value). For all real numbers $x$ and $y$ :

$∣ x ∣ \geq 0$ , with equality only at $x = 0$ .
$∣ - x ∣ = ∣ x ∣$ .
$∣ x y ∣ = ∣ x ∣ ∣ y ∣$ .
$\frac{x}{y} = \frac{∣ x ∣}{∣ y ∣}$ when $y \neq = 0$ .
$∣ x + y ∣ \leq ∣ x ∣ + ∣ y ∣$ (the triangle inequality).

These make absolute value a norm and a useful tool in analysis.

Intuition. Flipping the sign of x never changes its distance from zero.

Distances multiply and divide nicely; the triangle inequality says the direct path is never longer than going via an extra point.

Example 3.4. | (−3) × 4 | = |−12| = 12, and |−3| × |4| = 3 × 4 = 12, matching the product rule.

Exercises for §3.2 →

3.3Distance Interpretation

Definition 3.5 (Distance on the number line). The distance between the points $x$ and $a$ on the real line is

∣ x - a ∣.

This interpretation turns many absolute-value equations and inequalities into statements about "points that are a certain distance from a center."

Intuition. |x − 5| = 2 means "x is exactly two units away from 5," so x = 7 or x = 3.

It is the number-line version of "how far is the house from the school?"

Example 3.6. | x − 10 | = 4 means the points exactly 4 units from 10: x = 14 and x = 6.

Exercises for §3.3 →

3.4Graphs of Absolute Value Functions

The graph $y = ∣ x ∣$ is the basic V with vertex $(0, 0)$ , opening upward. Shifting it moves that vertex around the plane.

Proposition 3.7 (The shifted V). The graph of

y = ∣ x - h ∣ + k

is the basic V shifted $h$ units horizontally and $k$ units vertically. Its vertex is at $(h, k)$ , and the minimum value of $y$ is $k$ .

Intuition. The V is like a valley; the bottom is the closest you can get to the x-axis.

Shifting the V moves the "bottom point" around the plane.

Example 3.8. Graph $y = ∣ x - 3∣ + 2$ .

Solution. Compare with $y = ∣ x - h ∣ + k$ : here $h = 3$ and $k = 2$ , so the vertex is at $(3, 2)$ . The inside $∣ x - 3∣$ is $0$ exactly at $x = 3$ , giving the minimum value $y = 0 + 2 = 2$ ; the graph touches $y = 2$ there and never goes lower. To the right of the vertex the bars open to $y = (x - 3) + 2 = x - 1$ , a ray of slope $+ 1$ . For example at $x = 5$ , $y = ∣5 - 3∣ + 2 = 4$ . To the left the bars open to $y = - (x - 3) + 2 = - x + 5$ , a ray of slope $- 1$ . For example at $x = 1$ , $y = ∣1 - 3∣ + 2 = 4$ . Sanity check the symmetry: $x = 1$ and $x = 5$ are each $2$ units from the vertex line $x = 3$ and both give $y = 4$ , confirming the V is symmetric about $x = 3$ .□

Exercises for §3.4 →

Common Mistakes

Thinking |−5| = −5: Absolute value is never negative.
Writing |x| = x as an identity: True only for x ≥ 0.
Forgetting the triangle inequality direction: |x + y| can be smaller than |x| + |y| when x and y have opposite signs.
Confusing |x| with x²: |x| gives distance; x² gives area of square or squared distance.
Dropping the bars incorrectly when solving: |x| = 3 becomes x = 3 or x = −3; you cannot simply remove bars without considering cases.

An inequality compares two expressions and tells which is larger, smaller, or that they are not equal. The four symbols <, >, ≤, ≥ (and ≠) turn expressions into statements that are true for ranges of values rather than single points.

Solving inequalities follows the same steps as equations with the single extra rule about reversing direction when multiplying or dividing by negatives.

3.5Solving Linear Inequalities

Apply inverse operations to isolate the variable, exactly as with equations.

Proposition 3.9 (Operations that preserve the direction). The direction of the inequality symbol stays the same when you

add or subtract any number, or
multiply or divide by a positive number.

Intuition. Adding the same amount to two piles never changes which pile is bigger.

Multiplying by a positive number keeps the order; the piles just get scaled the same way.

Example 3.10. Solve 5x − 3 > 12.

Solution.

Add 3: 5x > 15.
Divide by 5: x > 3.

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Exercises for §3.5 →

3.6The Flipping Rule

Theorem 3.11 (The flipping rule). When you multiply or divide both sides of an inequality by a negative number, the inequality symbol must be reversed:

a < b and c < 0 ⟹ a c > b c .

This is the only operation that changes the direction. Always check the sign of the number you are multiplying or dividing by.

Intuition. Multiplying by a negative is like turning the number line around; left becomes right.

The "greater than" relationship flips when everything is mirrored.

Example 3.12. −4x < 12.

Solution.

Divide by −4 and flip: x > −3.
Verification: pick x = −2 (> −3). −4(−2) = 8 which is indeed < 12. Pick x = −4 (< −3). −4(−4) = 16 which is not < 12. Correct flip.

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Exercises for §3.6 →

3.7Interval Notation

Interval notation gives a compact symbolic description of solution sets.

Notation (Intervals).

$(a, b)$ open interval: $a < x < b$ .
$[a, b]$ closed interval: $a \leq x \leq b$ .
$(a, b]$ half-open, and so on.
$(- \infty, b)$ or $[a, \infty)$ for unbounded rays.
$\cup$ for the union of disjoint intervals.

Intuition. Parentheses say "not including the fence post"; square brackets say "including the fence post."

Infinity symbols mean the solution goes on forever in that direction.

Example 3.13. The solution x ≤ 4 or x > 7 is written (−∞, 4] ∪ (7, ∞).

Exercises for §3.7 →

3.8Graphing on the Number Line

Every solution set of a one-variable inequality can be pictured on the real number line.

Use open circles for strict endpoints, closed circles for included endpoints, and arrows or heavy shading for the included region.

Intuition. The number line is a road; circles are gates that are open or locked, and the shading tells which stretch of road is allowed.

Example 3.14. Graph the solution of $3 x + 1 < 7$ on the number line.

Solution. Solve first: subtract $1$ to get $3 x < 6$ , then divide by the positive $3$ to get $x < 2$ . The symbol is strict $<$ , so the endpoint $2$ is not included — draw an open circle at $2$ . The solution is all numbers less than $2$ , so shade the road to the left of $2$ with an arrow toward $- \infty$ . Sanity check: $x = 0$ is left of $2$ and $3 (0) + 1 = 1 < 7$ ✓; $x = 2$ gives $3 (2) + 1 = 7 \neq < 7$ , confirming the endpoint is correctly open.

For a non-strict example, $x \geq - 1$ is drawn with a filled circle at $- 1$ (the endpoint is included) and shading extending right to $+ \infty$ .□

Exercises for §3.8 →

Common Mistakes

Forgetting to flip when multiplying or dividing by a negative: The most frequent and costly error.
Using a closed circle for a strict inequality in the graph: < and > never include the endpoint.
Writing the answer with the variable on the right side: 5 > x is correct but x < 5 is preferred for readability.
Treating compound "and" inequalities as separate solves without intersecting the results: Both conditions must hold simultaneously.
Dividing by an expression containing the variable without knowing its sign: You cannot decide the flip without cases.

An absolute value equation contains the absolute value of an expression set equal to a number or another expression. Because absolute value "folds" the number line, these equations typically produce two solutions, one on each side of the "fold" point.

The case method (splitting into positive and negative branches) or the property |A| = |B| ⇔ A = B or A = −B are the standard tools.

3.9The Case Method

Method 3.15 (The case method for $∣ A ∣ = k$ ). To solve $∣ A ∣ = k$ with $k \geq 0$ , split into two cases:

Case 1: A = k, Case 2: A = - k .

Solve each equation and collect the solutions.

If the right side is negative there are no solutions, because an absolute value is never negative.

Intuition. The absolute value bars hide whether the inside is positive or negative; we must consider both possibilities explicitly.

It's like the expression could be the "positive twin" or the "negative twin" of the right-hand side.

Example 3.16. Solve $∣ x - 2∣ = 3$ .

Solution.

Case 1: $x - 2 = 3 \to x = 5$ .
Case 2: $x - 2 = - 3 \to x = - 1$ .
Both check: $∣5 - 2∣ = 3$ and $∣ - 1 - 2∣ = 3$ .

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Exercises for §3.9 →

3.10Using Absolute Value Properties

Theorem 3.17 (Equality of two absolute values). When both sides are absolute values,

∣ A ∣ = ∣ B ∣ ⟺ A = B or A = - B .

This often avoids writing full cases and is especially clean when the expressions are linear.

Intuition. Two distances from zero are equal when the numbers themselves are the same or exact opposites.

You still end up with two linear equations, but the setup is quicker.

Example 3.18. Solve $∣ x + 1∣ = ∣ x - 4∣$ .

Solution.

$x + 1 = x - 4$ (no solution) or $x + 1 = - (x - 4) \to 2 x = 3 \to x = 3/2$ .
Check: $∣3/2 + 1∣ = 2.5$ and $∣3/2 - 4∣ = 2.5$ . Good.

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Exercises for §3.10 →

3.11Extraneous Solutions

Although the case method rarely introduces extraneous roots, students frequently square both sides of |A| = B to obtain A² = B² and then must verify. The reason squaring is dangerous is that $A^{2} = B^{2}$ is satisfied by both $A = B$ and $A = - B$ , so it silently adds the second branch even when the original equation only allowed the first.

Proposition 3.19 (The hidden condition). The equation $∣ A ∣ = B$ can only hold when

B \geq 0,

because the left side is a distance.

Any candidate that makes the right side negative, or that fails substitution into the original equation, is extraneous and must be thrown out.

Intuition. Squaring is a "blunt instrument" that can create extra answers that only work after you squared.

Always plug every candidate back into the equation that still has the bars.

Example 3.20. Solve $∣ x ∣ = x - 1$ .

Solution.

The right side $x - 1$ must be $\geq 0$ , so any real solution needs $x \geq 1$ (note this condition before doing algebra).
Square both sides: $x^{2} = (x - 1)^{2} = x^{2} - 2 x + 1$ .
Cancel $x^{2}$ : $0 = - 2 x + 1$ , so $x = 1/2$ .
Check against the required condition: $1/2 \geq 1$ is false, and directly $∣1/2∣ = 1/2$ while $1/2 - 1 = - 1/2$ , so the two sides disagree.
The candidate $x = 1/2$ is extraneous, so the equation has no solution. This matches the fact that $∣ x ∣ = x$ for $x \geq 0$ , which can never equal the strictly smaller value $x - 1$ .

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Exercises for §3.11 →

3.12Examples

Practice with a variety of right-hand sides and expressions inside the bars builds fluency.

Always consider whether the right-hand side can possibly equal an absolute value (must be ≥ 0).

Intuition. Draw the number line, mark the "center" inside the bars, and measure the given distance on both sides.

The two points symmetric about the center are the solutions.

Example 3.21. Solve $∣2 x + 1∣ = 5$ .

Solution.

$2 x + 1 = 5 \to 2 x = 4 \to x = 2$ .
$2 x + 1 = - 5 \to 2 x = - 6 \to x = - 3$ .
Verify both.

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Exercises for §3.12 →

Common Mistakes

Forgetting that |A| = k has no solution when k < 0: Absolute value cannot equal a negative.
Solving only one case: There are almost always two solutions; both branches must be solved.
Not checking after squaring: x = 1/2 is a common extraneous root for equations like |x| = x − 1.
Dropping the bars without writing the ± case: |x − 3| = 4 is not the same as x − 3 = 4.
Confusing absolute value equations with inequalities: The solution sets look different (points vs intervals).

Absolute value inequalities describe intervals or unions of intervals on the number line. They translate the idea "the distance from a center is less than / greater than a certain amount" into algebraic statements.

The solutions are either a single bounded interval ("and") or two unbounded rays ("or"), depending on whether the inequality is < / ≤ or > / ≥.

3.13Case or Property Approach

Theorem 3.22 (Absolute value inequalities). Let $k > 0$ . Then

∣ A ∣ < k ⟺ - k < A < k

(the "and" form), and

∣ A ∣ > k ⟺ A < - k or A > k

(the "or" form).

The same logic applies to ≤ and ≥, simply including the endpoints.

Intuition. Less than means "inside the fence" around the center: both left fence and right fence conditions must be true at once ("and").

Greater than means "outside the fence": you are either far left or far right ("or").

Example 3.23. Solve $∣ x - 4∣ \leq 1$ .

Solution.

This is a "less than" case, so use the "and" form: $- 1 \leq x - 4 \leq 1$ .
Add $4$ to all three parts: $3 \leq x \leq 5$ .
This is the closed interval $[3, 5]$ , centered at $4$ with radius $1$ — exactly the distance reading of " $x$ is at most $1$ unit from $4$ . "
Sanity check: at $x = 4$ (center), $∣4 - 4∣ = 0 \leq 1$ ✓; at the endpoint $x = 5$ , $∣5 - 4∣ = 1 \leq 1$ ✓; just outside at $x = 6$ , $∣6 - 4∣ = 2 \neq \leq 1$ , correctly excluded.

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Exercises for §3.13 →

3.14"And" versus "Or" Logic

Strictly less/greater than or equal produces the compound connector:

|A| < k or |A| ≤ k → compound "and" (intersection) → one interval.
|A| > k or |A| ≥ k → compound "or" (union) → two rays.

When k = 0 the behavior is special: |A| > 0 is all reals except 0; |A| ≥ 0 is all reals.

Intuition. "Inside the distance limit" requires being on the correct side of both boundaries.

"Outside the distance limit" is satisfied as soon as you cross either boundary.

Example 3.24. Solve $∣ x ∣ > 2$ .

Solution.

"Greater than" gives the "or" form: $x > 2$ or $x < - 2$ . These are two separate rays, and there is no overlap because a single number cannot be both above $2$ and below $- 2$ .
Sanity check: $x = 3$ gives $∣3∣ = 3 > 2$ ✓; $x = 0$ (in the excluded middle) gives $∣0∣ = 0 \neq > 2$ ✓ that it is left out.
Edge case with $k = 0$ : $∣ x ∣ > 0$ is the "or" form $x > 0$ or $x < 0$ , which together is every real number except $0$ . And $∣ x ∣ \geq 0$ is true for every real $x$ with no exceptions, since a distance is always at least $0$ .

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Exercises for §3.14 →

3.15Graphing the Solutions

Translate the algebraic solution set directly onto the number line using the same circle and shading conventions as ordinary inequalities.

< or > : open circles.
≤ or ≥ : closed circles.
"And" interval gets a single shaded segment.
"Or" gets two shaded rays.

Intuition. The graph is the picture of the "allowed zone" described by the distance condition.

The center of the absolute value expression is the center of symmetry of the solution set.

Example 3.25. Solve and graph $∣ x + 1∣ \geq 3$ .

Solution.

This is an "or" (greater-than) case: $x + 1 \geq 3$ or $x + 1 \leq - 3$ .
$x + 1 \geq 3 \to x \geq 2$ , and $x + 1 \leq - 3 \to x \leq - 4$ .
Because the symbol is $\geq$ (allows equality), the boundary points $- 4$ and $2$ are included, so use closed circles at $- 4$ and at $2$ .
Shade the ray going left from $- 4$ and the ray going right from $2$ ; the middle strip $(- 4, 2)$ is excluded.
Sanity check: at the center $x = - 1$ , $∣ - 1 + 1∣ = 0$ , which is not $\geq 3$ , confirming the center is correctly left out.

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Exercises for §3.15 →

3.16Examples

A few varied centers and radii build recognition of the pattern.

Always remember that the right-hand side of a < or > absolute value inequality must be non-negative for real solutions to exist in the usual way (negative right side yields empty or all-reals depending on direction).

Intuition. Draw the center point first, mark the radius distance on both sides, then decide whether you want the inside strip or the two outer pieces.

Example 3.26. Solve $∣2 x - 3∣ > 5$ .

Solution.

$2 x - 3 > 5$ or $2 x - 3 < - 5$ .
$2 x > 8 \to x > 4$ or $2 x < - 2 \to x < - 1$ .
Solution: $(- \infty, - 1) \cup (4, \infty)$ .

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Exercises for §3.16 →

Common Mistakes

Using "and" for a > inequality: Greater absolute value means the two outer regions, which is "or".
Forgetting to flip both inequality signs when multiplying or dividing a compound by a negative: In a three-part inequality every part must be handled the same way. Dividing −3 < 2x < 6 by the positive 2 keeps all directions: −3/2 < x < 3. But dividing, say, −6 < −2x < 4 by the negative −2 reverses both symbols: 3 > x > −2, i.e. −2 < x < 3.
Writing a single interval for a > problem: That would be the "inside" solution, the opposite of what > asks.
Ignoring that |A| < 0 has no solution: Absolute value cannot be negative.
Placing the circles at the wrong locations: Solve for the boundary points first by setting |A| = k, then decide open/closed and direction of shading from the compound.

A quadratic inequality is an inequality that can be written with a quadratic expression on one side and zero on the other. The solution is usually an interval or a union of two intervals on the real line.

Because a quadratic changes sign only at its roots (when they exist), we locate the roots and then determine the sign in each interval they create.

3.17Moving to Standard Form and Factoring

Method 3.27 (Putting a quadratic inequality in standard form).

Bring every term to the left side, so the inequality compares a quadratic expression with $0$ .
Factor (or use the quadratic formula) to find the critical numbers, the values where the expression is zero.

Intuition. Getting everything on one side lets you ask a simple question: "where is this expression positive or negative?"

Factoring turns the expression into a product whose sign is easy to read from the factors.

Example 3.28. Put $2 x^{2} - 6 x > 20$ into standard form and factor.

Solution. Bring every term to the left so the right side is $0$ : $2 x^{2} - 6 x - 20 > 0$ . Factor out the common $2$ : $2 (x^{2} - 3 x - 10) > 0$ . Since $2 > 0$ , dividing both sides by it does not change the direction: $x^{2} - 3 x - 10 > 0$ . Factor the quadratic: we need two numbers multiplying to $- 10$ and adding to $- 3$ ; those are $- 5$ and $+ 2$ , giving $(x - 5) (x + 2) > 0$ . The critical numbers (where the expression is zero) are therefore $x = 5$ and $x = - 2$ ; these are the "fence posts" the next section uses to build a sign chart.□

Exercises for §3.17 →

3.18The Sign Chart (or Test-Point) Method

Method 3.29 (The sign chart).

Find the roots and draw them on a number line; they divide the line into intervals.
Pick one convenient test point in each open interval.
Plug the test point into the quadratic (or evaluate the factored form) and record the sign.
Mark the intervals where the sign matches the inequality.
Decide whether to include the roots, based on whether the symbol is $\leq$ or $\geq$ .

Intuition. The roots are the "fence posts." Between posts the expression keeps the same "mood" (positive or negative).

One quick plug per room tells you which rooms are allowed.

Example 3.30. For (x − 5)(x + 2) > 0 the posts are −2 and 5.

Solution.

Test x = −3: (−)(−) = positive.
Test x = 0: (−)(+) = negative.
Test x = 6: (+)(+) = positive.
Solution: x < −2 or x > 5.

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Exercises for §3.18 →

3.19Leading Coefficient and Parabola Direction

Proposition 3.31 (End behaviour from the leading coefficient). The sign of the leading coefficient $a$ tells you the end behaviour of the parabola:

$a > 0$ : opens up, so the expression is positive far to the left and far to the right.
$a < 0$ : opens down, so the expression is negative far to the left and far to the right.

This lets you fill the outer intervals without testing once you know one middle interval.

Intuition. If the bowl opens upward the far ends are high (positive). The middle may dip below zero between the roots.

Downward bowl starts negative at both infinities.

Example 3.32. Solve $- x^{2} + 5 x + 6 \leq 0$ using the leading-coefficient shortcut.

Solution. The leading coefficient is $- 1 < 0$ , so the parabola opens downward: it is negative at both far ends and positive (if anywhere) in the middle between the roots. Find the roots by solving $- x^{2} + 5 x + 6 = 0$ . Multiply by $- 1$ : $x^{2} - 5 x - 6 = 0$ , which factors as $(x - 6) (x + 1) = 0$ , so the roots are $x = - 1$ and $x = 6$ . Since the bowl opens down, the expression is $\geq 0$ between the roots $[- 1, 6]$ and $\leq 0$ outside them. We want $\leq 0$ , so the solution is the two outer pieces, and because the symbol allows equality the roots are included: $x \leq - 1$ or $x \geq 6$ , i.e. $(- \infty, - 1] \cup [6, \infty)$ . Sanity check a point in each region: at $x = - 2$ , $- (- 2)^{2} + 5 (- 2) + 6 = - 4 - 10 + 6 = - 8 \leq 0$ ✓; at $x = 0$ (between the roots), $6 > 0$ so correctly excluded; at $x = 7$ , $- 49 + 35 + 6 = - 8 \leq 0$ ✓.□

Exercises for §3.19 →

3.20Writing the Solution in Interval Notation

Translate the marked intervals on the sign chart into interval notation, using ∪ for separate pieces.

Include square brackets at roots only when the inequality allows equality.

Intuition. The chart picture becomes the compact math sentence that a computer or another person can read without the drawing.

Example 3.33. Write the solution of $x^{2} - 3 x - 10 > 0$ in interval notation.

Solution. Factored form (from the earlier section) is $(x - 5) (x + 2) > 0$ with critical numbers $- 2$ and $5$ ; the sign chart shows the expression is positive on the two outer intervals. The strict symbol $>$ means the roots themselves give a value of $0$ , which is not $> 0$ , so the endpoints are excluded — use round parentheses: $(- \infty, - 2) \cup (5, \infty)$ . If the inequality had instead been $\geq 0$ , the roots satisfy it (they give exactly $0$ ), so they are included with square brackets: $(- \infty, - 2] \cup [5, \infty)$ . Note that $\pm \infty$ always gets a parenthesis, never a bracket, because infinity is not a number that can be "reached."□

Exercises for §3.20 →

Common Mistakes

Testing the roots themselves for sign: At a root the value is exactly zero; you already know the sign change happens there.
Forgetting to consider the leading coefficient when predicting outer signs: Saves time and catches errors.
Including or excluding roots incorrectly: < and > never include; ≤ and ≥ do.
Not bringing all terms to one side first: Mixed signs on right side make test points confusing.
Assuming every quadratic inequality has solutions: When the parabola never crosses the required side there may be no solution or the entire line.

About the practice questions. They check that you can carry out this chapter's computations correctly, and each one is graded on a single answer. They are not proof exercises: working through them confirms the mechanics, not that you could prove the results yourself. For that, re-read the statements above and try to reconstruct their proofs with the page closed.

Practice this chapter