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2.3 Inequalities, Absolute Value, and Systems of Equations

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Adamson University College Entrance Test AdUCET

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1 Introduction & Exam Overview

1.1 AdUCET Overview, Format, and Admission Process 1.2 Test-Taking and Time-Management Strategies

2 Chapter 2: Mathematics I — Numbers, Algebra, and Functions

2.1 Number Sense, Operations, and Fraction–Decimal–Percent Conversions 2.2 Algebra, Polynomials, and Linear/Quadratic Equations 2.3 Inequalities, Absolute Value, and Systems of Equations 2.4 Radicals, Rational Expressions, and Scientific Notation 2.5 Coordinate Geometry, Lines, and Functions

3 Chapter 3: Mathematics II — Geometry, Measurement, and Trigonometry

3.1 Plane Geometry: Angles, Triangles, Polygons, and Circles 3.2 Solid Mensuration, Measurement, and Unit Conversion 3.3 Right-Triangle Trigonometry and Applications

4 Chapter 4: Mathematics III — Sequences, Proportion, and Data

4.1 Arithmetic and Geometric Sequences and Series 4.2 Ratio, Proportion, Percentage, Work, and Mixture Problems 4.3 Descriptive Statistics and Data Interpretation 4.4 Probability: Rules, Compound Events, and Conditional Probability 4.5 Counting Principles: Permutations and Combinations

5 Chapter 5: English I — Reading Comprehension and Vocabulary

5.1 Reading Comprehension: Main Idea, Purpose, and Passage Structure 5.2 Inference, Supporting Detail, and Author's Tone 5.3 Word Roots, Prefixes, and Suffixes 5.4 Vocabulary in Context, Synonyms, Antonyms, and Confusables

6 Chapter 6: English II — Grammar, Usage, and Written Expression

6.1 Subject-Verb Agreement and Verb Tenses 6.2 Pronouns, Modifiers, and Parallel Structure 6.3 Sentence Boundaries, Clauses, and Punctuation 6.4 Sentence Correction, Redundancy, and Idiomatic Usage 6.5 Verbal Analogies and Relationship Bridges 6.6 Paragraph Organization, Coherence, and Transitions

7 Chapter 7: General Science I — Biology and Chemistry

7.1 Cell Biology, Transport, and Cellular Energetics 7.2 Genetics, Heredity, and Molecular Biology 7.3 Human Body Systems, Ecology, and Evolution 7.4 Matter, Phase Changes, and Atomic Structure 7.5 Periodic Trends, Chemical Bonding, Reactions, Acids and Bases

8 Chapter 8: General Science II — Physics and Earth Science

8.1 Mechanics: Motion, Newton's Laws, Work, and Energy 8.2 Waves, Sound, Light, Heat, Electricity, and Magnetism 8.3 Earth's Structure, Plate Tectonics, and the Rock Cycle 8.4 The Atmosphere, Weather, the Water Cycle, and Astronomy

9 Chapter 9: Aptitude and Abstract Reasoning

9.1 Figure Series and Pattern Progression 9.2 Matrix Reasoning and Figural Analogies 9.3 Spatial Reasoning: Rotation, Reflection, and Chirality 9.4 Paper Folding, Cube Nets, and Orthographic Views 9.5 Deductive Reasoning, Syllogisms, and Venn Diagrams 9.6 Number, Letter, and Alphanumeric Series 9.7 Odd-One-Out, Classification, and Logic Puzzles

2.3 Inequalities, Absolute Value, and Systems of Equations

Key Takeaways Explain This Topic AI

Multiplying or dividing an inequality by a negative number reverses the symbol, so − 5 x ≥ − 3 -5x \ge -3 − 5 x ≥ − 3 becomes x ≤ 3 5 x \le \frac{3}{5} x ≤ 5 3 ​ ; adding or subtracting a negative changes nothing.

∣ u ∣ < c \lvert u \rvert < c ∣ u ∣ < c is a conjunction giving the single interval − c < u < c -c < u < c − c < u < c , while ∣ u ∣ > c \lvert u \rvert > c ∣ u ∣ > c is a disjunction giving u < − c u < -c u < − c or u > c u > c u > c — never one chain.

∣ u ∣ = c \lvert u \rvert = c ∣ u ∣ = c has no solution whenever c c c is negative, because an absolute value can never be less than zero.

An elimination that collapses to a false statement such as 0 = − 3 0 = -3 0 = − 3 means the system is inconsistent with no solution; collapsing to a true statement such as 6 = 6 6 = 6 6 = 6 means it is dependent with infinitely many.

Solve a quadratic inequality by moving all terms to one side and testing regions: x 2 ≤ 5 x − 6 x^2 \le 5x - 6 x 2 ≤ 5 x − 6 becomes ( x − 2 ) ( x − 3 ) ≤ 0 (x-2)(x-3) \le 0 ( x − 2 ) ( x − 3 ) ≤ 0 , whose solution is 2 ≤ x ≤ 3 2 \le x \le 3 2 ≤ x ≤ 3 .

Last updated: August 2026

2.3 Inequalities, Absolute Value, and Systems of Equations

Inequalities and absolute value ask a different question from the equations handled in the previous section: not "which single number works?" but "which set of numbers works?" Solutions are intervals and regions, and entrance-test items in this family tend to hide one procedural rule — a reversed symbol, a disjunction, an excluded case — inside otherwise routine algebra. Solving plain linear and quadratic equations, including the quadratic formula and discriminant, is covered separately; this section assumes you can already do that and focuses on what changes when the equals sign becomes an inequality or a system.

1. Linear Inequalities and the Sign-Flip Rule

Every operation that is legal on an equation is legal on an inequality, with one exception : multiplying or dividing both sides by a negative number reverses the direction of the symbol. Adding or subtracting a negative number does not reverse anything.

Worked Example 1

Solve − 3 ( x − 4 ) ≥ 2 x + 9 -3(x - 4) \ge 2x + 9 − 3 ( x − 4 ) ≥ 2 x + 9 .

Distribute: − 3 x + 12 ≥ 2 x + 9 -3x + 12 \ge 2x + 9 − 3 x + 12 ≥ 2 x + 9 .

Collect the variables on the left and the constants on the right: subtract 2 x 2x 2 x and 12 12 12 from both sides, giving − 5 x ≥ − 3 -5x \ge -3 − 5 x ≥ − 3 . Here that deliberately leaves a negative coefficient, which sets up the sign flip in the next step.

Divide by − 5 -5 − 5 and flip : x ≤ 3 5 x \le \frac{3}{5} x ≤ 5 3 ​ .

Check with x = 0 x = 0 x = 0 : the left side is − 3 ( 0 − 4 ) = 12 -3(0-4) = 12 − 3 ( 0 − 4 ) = 12 and the right side is 9 9 9 ; since 12 ≥ 9 12 \ge 9 12 ≥ 9 is true and 0 ≤ 3 5 0 \le \frac{3}{5} 0 ≤ 5 3 ​ , the direction is right.

In interval notation the answer is ( − ∞ , 3 5 ] \left(-\infty, \frac{3}{5}\right] ( − ∞ , 5 3 ​ ] .

  • Inequality Interval notation Endpoint on the number line
  • x < 3 x < 3 x < 3 ( − ∞ , 3 ) (-\infty, 3) ( − ∞ , 3 ) open circle at 3
  • x ≤ 3 x \le 3 x ≤ 3 ( − ∞ , 3 ] (-\infty, 3] ( − ∞ , 3 ] closed circle at 3
  • x > − 2 x > -2 x > − 2 ( − 2 , ∞ ) (-2, \infty) ( − 2 , ∞ ) open circle at − 2 -2 − 2
  • − 1 < x ≤ 5 -1 < x \le 5 − 1 < x ≤ 5 ( − 1 , 5 ] (-1, 5] ( − 1 , 5 ] open at − 1 -1 − 1 , closed at 5

x ≤ − 5 x \le -5 x ≤ − 5 or x ≥ 4 x \ge 4 x ≥ 4 ( − ∞ , − 5 ] ∪ [ 4 , ∞ ) (-\infty, -5] \cup [4, \infty) ( − ∞ , − 5 ] ∪ [ 4 , ∞ ) two closed circles, gap between

Infinity always takes a parenthesis — you can never include it.

2. Compound Inequalities

An "and" (conjunction) statement keeps only the overlap; an "or" (disjunction) statement keeps everything in either piece. A three-part chain such as − 5 < 2 x − 3 ≤ 7 -5 < 2x - 3 \le 7 − 5 < 2 x − 3 ≤ 7 is an "and" written compactly: add 3 across all three parts to get − 2 < 2 x ≤ 10 -2 < 2x \le 10 − 2 < 2 x ≤ 10 , then halve all three to get − 1 < x ≤ 5 -1 < x \le 5 − 1 < x ≤ 5 , or ( − 1 , 5 ] (-1, 5] ( − 1 , 5 ] .

3. Absolute Value Equations and Inequalities

Absolute value measures distance from zero, so it is never negative. That single fact generates the whole table.

  • Form (with c > 0 c > 0 c > 0 ) Meaning Solution
  • ∣ u ∣ = c \lvert u \rvert = c ∣ u ∣ = c distance is exactly c c c u = c u = c u = c or u = − c u = -c u = − c
  • ∣ u ∣ = 0 \lvert u \rvert = 0 ∣ u ∣ = 0 distance is zero u = 0 u = 0 u = 0
  • ∣ u ∣ = − c \lvert u \rvert = -c ∣ u ∣ = − c impossible no solution
  • ∣ u ∣ < c \lvert u \rvert < c ∣ u ∣ < c within c c c of zero − c < u < c -c < u < c − c < u < c ( and )

∣ u ∣ > c \lvert u \rvert > c ∣ u ∣ > c farther than c c c from zero u < − c u < -c u < − c or u > c u > c u > c

∣ u ∣ > − c \lvert u \rvert > -c ∣ u ∣ > − c always true all real numbers

Solve ∣ 3 x − 4 ∣ = 11 \lvert 3x - 4 \rvert = 11 ∣ 3 x − 4 ∣ = 11 by splitting: 3 x − 4 = 11 3x - 4 = 11 3 x − 4 = 11 gives x = 5 x = 5 x = 5 , and 3 x − 4 = − 11 3x - 4 = -11 3 x − 4 = − 11 gives x = − 7 3 x = -\frac{7}{3} x = − 3 7 ​ .

For ∣ x − 3 ∣ < 5 \lvert x - 3 \rvert < 5 ∣ x − 3 ∣ < 5 , rewrite as − 5 < x − 3 < 5 -5 < x - 3 < 5 − 5 < x − 3 < 5 , then add 3 throughout: − 2 < x < 8 -2 < x < 8 − 2 < x < 8 . |x - 3| < 5 -> -2 < x < 8 (one connected interval)

  • <---------o==================o--------->
  • 2 8

For ∣ 2 x + 1 ∣ ≥ 9 \lvert 2x + 1 \rvert \ge 9 ∣ 2 x + 1 ∣ ≥ 9 , split into 2 x + 1 ≥ 9 2x + 1 \ge 9 2 x + 1 ≥ 9 , giving x ≥ 4 x \ge 4 x ≥ 4 , and 2 x + 1 ≤ − 9 2x + 1 \le -9 2 x + 1 ≤ − 9 , giving x ≤ − 5 x \le -5 x ≤ − 5 . The answer is ( − ∞ , − 5 ] ∪ [ 4 , ∞ ) (-\infty, -5] \cup [4, \infty) ( − ∞ , − 5 ] ∪ [ 4 , ∞ ) — two rays, never a single chain. |2x + 1| >= 9 -> x <= -5 or x >= 4 (two separated rays)

  • ==========*----------------------*==========>
  • 5 4

The mnemonic used across Philippine review classes is "less than = and , greater than = or " .

4. Quadratic Inequalities by Sign Chart

Never divide an inequality by a variable expression — its sign is unknown. Move everything to one side, factor, and test signs.

Solve x 2 ≤ 5 x − 6 x^2 \le 5x - 6 x 2 ≤ 5 x − 6 . Rearranged, x 2 − 5 x + 6 ≤ 0 x^2 - 5x + 6 \le 0 x 2 − 5 x + 6 ≤ 0 , which factors as ( x − 2 ) ( x − 3 ) ≤ 0 (x-2)(x-3) \le 0 ( x − 2 ) ( x − 3 ) ≤ 0 . The critical values 2 and 3 cut the number line into three regions: 2 3

  • -----------------|--------------|------------------>
  • (x - 2) neg | pos | pos
  • (x - 3) neg | neg | pos
  • product POS | NEG | POS
  • ^ the only region where the product is <= 0

Because the inequality allows equality, the endpoints are included: 2 ≤ x ≤ 3 2 \le x \le 3 2 ≤ x ≤ 3 , or [ 2 , 3 ] [2, 3] [ 2 , 3 ] .

5. Systems of Linear Equations

Two lines in a plane either cross once, never cross, or coincide.

Substitution — isolate one variable and replace it in the other equation. Best when a coefficient is already 1.

Elimination — scale the equations so one variable cancels when you add or subtract.

Worked Example 2 — elimination

Solve 3 x + 4 y = 10 3x + 4y = 10 3 x + 4 y = 10 and 5 x − 2 y = 8 5x - 2y = 8 5 x − 2 y = 8 . Multiply the second equation by 2 to get 10 x − 4 y = 16 10x - 4y = 16 10 x − 4 y = 16 , then add it to the first: 13 x = 26 13x = 26 13 x = 26 , so x = 2 x = 2 x = 2 . Substituting back, 3 ( 2 ) + 4 y = 10 3(2) + 4y = 10 3 ( 2 ) + 4 y = 10 gives 4 y = 4 4y = 4 4 y = 4 and y = 1 y = 1 y = 1 . Check the second original equation: 5 ( 2 ) − 2 ( 1 ) = 8 5(2) - 2(1) = 8 5 ( 2 ) − 2 ( 1 ) = 8 . The solution is ( 2 , 1 ) (2, 1) ( 2 , 1 ) .

Two special outcomes carry marks:

Inconsistent (no solution). For 2 x + 6 y = 5 2x + 6y = 5 2 x + 6 y = 5 and x + 3 y = 4 x + 3y = 4 x + 3 y = 4 , doubling the second gives 2 x + 6 y = 8 2x + 6y = 8 2 x + 6 y = 8 ; subtracting produces 0 = − 3 0 = -3 0 = − 3 , a false statement. The lines are parallel with different intercepts.

Dependent (infinitely many solutions). For y = 2 x − 3 y = 2x - 3 y = 2 x − 3 and 4 x − 2 y = 6 4x - 2y = 6 4 x − 2 y = 6 , substitution gives 4 x − 2 ( 2 x − 3 ) = 6 4x - 2(2x - 3) = 6 4 x − 2 ( 2 x − 3 ) = 6 , which collapses to 6 = 6 6 = 6 6 = 6 , a statement true for every x x x . Both equations describe the same line.

6. Systems of Linear Inequalities

A system of inequalities defines a feasible region — the overlap of several half-planes. Graph each boundary line (solid for ≤ \le ≤ or ≥ \ge ≥ , dashed for < < < or > > > ), then shade using a test point ; ( 0 , 0 ) (0,0) ( 0 , 0 ) is easiest whenever it is not on a boundary.

For x ≥ 0 x \ge 0 x ≥ 0 , y ≥ 0 y \ge 0 y ≥ 0 , y ≤ − x + 6 y \le -x + 6 y ≤ − x + 6 and y ≥ 2 x − 3 y \ge 2x - 3 y ≥ 2 x − 3 , the point ( 0 , 0 ) (0,0) ( 0 , 0 ) satisfies all four, so the region contains the origin. The corners are ( 0 , 0 ) (0,0) ( 0 , 0 ) , ( 0 , 6 ) (0,6) ( 0 , 6 ) , the intersection of the two slanted boundaries at ( 3 , 3 ) (3,3) ( 3 , 3 ) — found by solving − x + 6 = 2 x − 3 -x + 6 = 2x - 3 − x + 6 = 2 x − 3 — and ( 1.5 , 0 ) (1.5, 0) ( 1.5 , 0 ) .

7. Translating Words into Inequalities

  • Phrase Symbol
  • at least, minimum of, no less than ≥ \ge ≥
  • at most, maximum of, no more than ≤ \le ≤
  • more than, exceeds, over > > >
  • fewer than, under, below < < <

Two standard AdUCET-style setups:

Minimum score. Ana scored 88, 79 and 92 on three 100-point quizzes and needs an average of at least 87 over four quizzes. Then 88 + 79 + 92 + x 4 ≥ 87 \frac{88+79+92+x}{4} \ge 87 4 88 + 79 + 92 + x ​ ≥ 87 , so 259 + x ≥ 348 259 + x \ge 348 259 + x ≥ 348 and x ≥ 89 x \ge 89 x ≥ 89 .

Budget. A student has ₱600 for a ₱120 bag and notebooks at ₱45 each: 120 + 45 n ≤ 600 120 + 45n \le 600 120 + 45 n ≤ 600 gives 45 n ≤ 480 45n \le 480 45 n ≤ 480 and n ≤ 10.67 n \le 10.67 n ≤ 10.67 . Because notebooks are whole objects, the answer is 10 — always round a counting answer down , whatever the decimal says.

Common Traps

Flipping the symbol when you merely added or subtracted a negative number.

Writing ∣ x ∣ > 5 \lvert x \rvert > 5 ∣ x ∣ > 5 as the impossible chain − 5 > x > 5 -5 > x > 5 − 5 > x > 5 instead of two separate rays.

Forgetting that ∣ u ∣ = c \lvert u \rvert = c ∣ u ∣ = c has no solution when c c c is negative.

Ignoring a check when the right side of an absolute-value equation contains a variable: ∣ x − 2 ∣ = 3 x \lvert x - 2 \rvert = 3x ∣ x − 2 ∣ = 3 x yields x = − 1 x = -1 x = − 1 and x = 1 2 x = \frac{1}{2} x = 2 1 ​ , but x = − 1 x = -1 x = − 1 makes the right side negative and must be rejected.

Reading a collapse to 0 = 0 0 = 0 0 = 0 as " x = 0 x = 0 x = 0 " rather than as infinitely many solutions.

Using only the critical values of a quadratic inequality without testing the sign of each region.