Study guide
5.2 Solving & Graphing Linear Inequalities
ACCUPLACER Quantitative Reasoning, Algebra, and Statistics Exam
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1 Chapter 1: Rational Numbers & Operations
1.1 Rational Number Arithmetic & Signed Numbers 1.2 Fractions, Decimals, & Mixed Numbers Operations 1.3 Absolute Value, Number Line, & Estimation
2 Chapter 2: Ratios, Proportions & Rates
2.1 Ratios, Unit Rates, & Dimensional Analysis 2.2 Proportions & Direct/Inverse Variation 2.3 Percent Applications, Markup, Discount, & Simple Interest
3 Chapter 3: Exponents & Scientific Notation
3.1 Integer Exponents & Laws of Exponents 3.2 Scientific Notation & Operations
4 Chapter 4: Algebraic Expressions
4.1 Evaluating & Translating Algebraic Expressions 4.2 Simplifying, Combining Like Terms, & Factoring
5 Chapter 5: Linear Equations & Inequalities
5.1 Solving Single-Variable Linear Equations 5.2 Solving & Graphing Linear Inequalities 5.3 Systems of Linear Equations
6 Chapter 6: Linear Applications & Graphing
6.1 Slope & Rate of Change 6.2 Forms of Linear Equations & Graphing 6.3 Linear Modeling & Inequality Graphs
7 Chapter 7: Probability & Set Theory
7.1 Fundamental Probability Concepts 7.2 Set Theory, Operations, & Venn Diagrams
8 Chapter 8: Descriptive Statistics & Data Displays
8.1 Measures of Center: Mean, Median, Mode, & Weighted Average 8.2 Measures of Spread: Range, IQR, Variance, & Standard Deviation 8.3 Data Displays: Histograms, Box Plots, & Scatter Plots
9 Chapter 9: Geometry Concepts & Measurement
9.1 2D Geometry: Perimeter, Area, & Circles 9.2 3D Geometry: Surface Area & Volume of Solids 9.3 Triangles, Angles, & the Pythagorean Theorem
5.2 Solving & Graphing Linear Inequalities
Key Takeaways Explain This Topic AI
A linear inequality establishes an order relationship (<, >, ≤, ≥) between expressions, producing an infinite continuum of solutions on the real number line rather than discrete isolated numbers.
The Negative Inversion Rule requires reversing the direction of the inequality symbol whenever both sides are multiplied or divided by a negative number.
On a real number line graph, strict inequalities (<, >) use open circles (○) indicating exclusion, while inclusive inequalities (≤, ≥) use solid closed circles (●) indicating inclusion.
Interval notation represents sets using round parentheses ( ) for non-included boundaries and infinities (±∞), and square brackets [ ] for included boundary values.
Compound inequalities consist of conjunctions ('AND', requiring the intersection of conditions) or disjunctions ('OR', requiring the union of conditions).
Last updated: August 2026
Inequality Symbols & Order Axioms
An inequality is a mathematical statement comparing two expressions using relational order symbols. Unlike linear equations, which typically yield discrete point solutions, linear inequalities in one variable produce continuous intervals of real numbers.
The Four Inequality Relations
- Symbol Verbal Meaning Type Boundary Status
- < < < "Strictly less than" Strict Inequality Endpoint is excluded (open circle ∘ \circ ∘ , parenthesis ( )
- > > > "Strictly greater than" Strict Inequality Endpoint is excluded (open circle ∘ \circ ∘ , parenthesis ) )
≤ \le ≤ "Less than or equal to", "at most", "no more than" Inclusive Inequality Endpoint is included (solid circle ∙ \bullet ∙ , bracket ] )
≥ \ge ≥ "Greater than or equal to", "at least", "no less than" Inclusive Inequality Endpoint is included (solid circle ∙ \bullet ∙ , bracket [ )
Transformation Properties of Linear Inequalities
Solving a linear inequality mirrors the process of solving a linear equation, with one critical, non-negotiable exception regarding negative numbers:
Addition & Subtraction Properties : Adding or subtracting any real number to both sides leaves the inequality symbol unchanged : If a < b , then a + c < b + c and a − c < b − c \text{If } a < b, \text{ then } a + c < b + c \quad \text{and} \quad a - c < b - c If a < b , then a + c < b + c and a − c < b − c
Multiplication & Division by a Positive Number ( c > 0 c > 0 c > 0 ) : Multiplying or dividing both sides by a positive number leaves the inequality symbol unchanged : If a < b and c > 0 , then a c < b c and a c < b c \text{If } a < b \text{ and } c > 0, \text{ then } ac < bc \quad \text{and} \quad \frac{a}{c} < \frac{b}{c} If a < b and c > 0 , then a c < b c and c a < c b
Multiplication & Division by a Negative Number ( c < 0 c < 0 c < 0 ) — The Sign Reversal Rule : Multiplying or dividing both sides by a negative number must reverse the direction of the inequality symbol : If a < b and c < 0 , then a c > b c and a c > b c \text{If } a < b \text{ and } c < 0, \text{ then } ac > bc \quad \text{and} \quad \frac{a}{c} > \frac{b}{c} If a < b and c < 0 , then a c > b c and c a > c b
Why Does the Sign Reverse?
Consider the true numerical statement 2 < 5 2 < 5 2 < 5 . If we multiply both sides by − 1 -1 − 1 , we obtain − 2 -2 − 2 and − 5 -5 − 5 . On the real number line, − 2 -2 − 2 is located to the right of − 5 -5 − 5 , which means − 2 > − 5 -2 > -5 − 2 > − 5 . Multiplication by a negative reflects the quantities across the origin ( 0 0 0 ), reversing their relative order.
Graphing on the Real Number Line & Interval Notation
Solutions to linear inequalities are communicated using three complementary formats: inequality notation, number line graphs, and interval notation.
Master Notation Comparison Table
Inequality Statement Number Line Graph Description Interval Notation Set-Builder Notation
x > a x > a x > a Open circle at a a a , shaded ray extending to the right ( a , ∞ ) (a, \infty) ( a , ∞ ) { x ∈ R ∣ x > a } \{x \in \mathbb{R} \mid x > a\} { x ∈ R ∣ x > a }
x ≥ a x \ge a x ≥ a Solid circle at a a a , shaded ray extending to the right [ a , ∞ ) [a, \infty) [ a , ∞ ) { x ∈ R ∣ x ≥ a } \{x \in \mathbb{R} \mid x \ge a\} { x ∈ R ∣ x ≥ a }
x < a x < a x < a Open circle at a a a , shaded ray extending to the left ( − ∞ , a ) (-\infty, a) ( − ∞ , a ) { x ∈ R ∣ x < a } \{x \in \mathbb{R} \mid x < a\} { x ∈ R ∣ x < a }
x ≤ a x \le a x ≤ a Solid circle at a a a , shaded ray extending to the left ( − ∞ , a ] (-\infty, a] ( − ∞ , a ] { x ∈ R ∣ x ≤ a } \{x \in \mathbb{R} \mid x \le a\} { x ∈ R ∣ x ≤ a }
a < x < b a < x < b a < x < b Open circles at a a a and b b b , shaded line segment between ( a , b ) (a, b) ( a , b ) { x ∈ R ∣ a < x < b } \{x \in \mathbb{R} \mid a < x < b\} { x ∈ R ∣ a < x < b }
a ≤ x ≤ b a \le x \le b a ≤ x ≤ b Solid circles at a a a and b b b , shaded line segment between [ a , b ] [a, b] [ a , b ] { x ∈ R ∣ a ≤ x ≤ b } \{x \in \mathbb{R} \mid a \le x \le b\} { x ∈ R ∣ a ≤ x ≤ b }
a ≤ x < b a \le x < b a ≤ x < b Solid circle at a a a , open circle at b b b , segment between [ a , b ) [a, b) [ a , b ) { x ∈ R ∣ a ≤ x < b } \{x \in \mathbb{R} \mid a \le x < b\} { x ∈ R ∣ a ≤ x < b }
x < a or x ≥ b x < a \text{ or } x \ge b x < a or x ≥ b Open circle at a a a (left ray) and solid circle at b b b (right ray) ( − ∞ , a ) ∪ [ b , ∞ ) (-\infty, a) \cup [b, \infty) ( − ∞ , a ) ∪ [ b , ∞ ) { x ∈ R ∣ x < a or x ≥ b } \{x \in \mathbb{R} \mid x < a \text{ or } x \ge b\} { x ∈ R ∣ x < a or x ≥ b }
All Real Numbers Entire number line shaded ( − ∞ , ∞ ) (-\infty, \infty) ( − ∞ , ∞ ) R \mathbb{R} R
Important
The infinity symbols ∞ \infty ∞ and − ∞ -\infty − ∞ represent unbounded directions, not real numbers. Therefore, infinity is always enclosed by a round parenthesis ) or ( , never a square bracket ] or [ .
Compound Inequalities: Conjunctions (AND) vs. Disjunctions (OR)
A compound inequality combines two individual inequality statements into a single mathematical condition using logical connectives.
1. Conjunctions ("AND" / Intersection ∩ \cap ∩ )
A conjunction asserts that both inequality conditions must be satisfied simultaneously. The solution set is the mathematical intersection of the two individual solution sets: x ≥ − 2 and x < 5 ⟺ − 2 ≤ x < 5 ⟹ [ − 2 , 5 ) x \ge -2 \quad \text{and} \quad x < 5 \iff -2 \le x < 5 \implies [-2, 5) x ≥ − 2 and x < 5 ⟺ − 2 ≤ x < 5 ⟹ [ − 2 , 5 )
Solving Three-Part Compound Inequalities
When an inequality has a bounded variable in the middle, apply algebraic operations simultaneously to all three parts: − 11 ≤ 3 x + 4 < 19 -11 \le 3x + 4 < 19 − 11 ≤ 3 x + 4 < 19
Subtract 4 4 4 from all three sections: − 11 − 4 ≤ 3 x < 19 − 4 ⟹ − 15 ≤ 3 x < 15 -11 - 4 \le 3x < 19 - 4 \implies -15 \le 3x < 15 − 11 − 4 ≤ 3 x < 19 − 4 ⟹ − 15 ≤ 3 x < 15
Divide all three sections by 3 3 3 : − 15 3 ≤ x < 15 3 ⟹ − 5 ≤ x < 5 \frac{-15}{3} \le x < \frac{15}{3} \implies -5 \le x < 5 3 − 15 ≤ x < 3 15 ⟹ − 5 ≤ x < 5
Express in interval notation: [ − 5 , 5 ) [-5, 5) [ − 5 , 5 ) .
2. Disjunctions ("OR" / Union ∪ \cup ∪ )
A disjunction asserts that at least one of the conditions must be satisfied. The solution set is the mathematical union of the two solution sets: 2 x − 1 < − 7 or 3 x + 5 ≥ 14 2x - 1 < -7 \quad \text{or} \quad 3x + 5 \ge 14 2 x − 1 < − 7 or 3 x + 5 ≥ 14
Solve the first inequality independently: 2 x < − 6 ⟹ x < − 3 ⟹ ( − ∞ , − 3 ) 2x < -6 \implies x < -3 \implies (-\infty, -3) 2 x < − 6 ⟹ x < − 3 ⟹ ( − ∞ , − 3 )
Solve the second inequality independently: 3 x ≥ 9 ⟹ x ≥ 3 ⟹ [ 3 , ∞ ) 3x \ge 9 \implies x \ge 3 \implies [3, \infty) 3 x ≥ 9 ⟹ x ≥ 3 ⟹ [ 3 , ∞ )
Combine the solution sets with the union operator ∪ \cup ∪ : ( − ∞ , − 3 ) ∪ [ 3 , ∞ ) (-\infty, -3) \cup [3, \infty) ( − ∞ , − 3 ) ∪ [ 3 , ∞ )
- Step-by-Step Worked Examples
- Worked Example 1: Multi-Step Inequality with Negative Division
Solve the inequality, state the solution in interval notation, and identify the correct number line graph: − 3 ( 2 x − 5 ) + 4 ≤ 37 − 2 x -3(2x - 5) + 4 \le 37 - 2x − 3 ( 2 x − 5 ) + 4 ≤ 37 − 2 x
Solution Plan & Execution :
Apply the Distributive Property : − 6 x + 15 + 4 ≤ 37 − 2 x -6x + 15 + 4 \le 37 - 2x − 6 x + 15 + 4 ≤ 37 − 2 x − 6 x + 19 ≤ 37 − 2 x -6x + 19 \le 37 - 2x − 6 x + 19 ≤ 37 − 2 x
Collect variable terms on the left side (add 2 x 2x 2 x to both sides): − 6 x + 2 x + 19 ≤ 37 -6x + 2x + 19 \le 37 − 6 x + 2 x + 19 ≤ 37 − 4 x + 19 ≤ 37 -4x + 19 \le 37 − 4 x + 19 ≤ 37
Collect constant terms on the right side (subtract 19 19 19 from both sides): − 4 x ≤ 37 − 19 -4x \le 37 - 19 − 4 x ≤ 37 − 19 − 4 x ≤ 18 -4x \le 18 − 4 x ≤ 18
Divide by − 4 -4 − 4 and REVERSE the inequality symbol : x ≥ 18 − 4 x \ge \frac{18}{-4} x ≥ − 4 18 x ≥ − 9 2 ( x ≥ − 4.5 ) x \ge -\frac{9}{2} \quad (x \ge -4.5) x ≥ − 2 9 ( x ≥ − 4.5 )
- State the solution in interval notation : [ − 9 2 , ∞ ) \left[-\frac{9}{2}, \infty\right) [ − 2 9 , ∞ )
- (Number line graph: Solid closed circle at − 4.5 -4.5 − 4.5 with shading extending indefinitely to the right.)
- Worked Example 2: Three-Part Inequality with Sign Reversal
- Solve the compound inequality for x x x : − 11 < 3 − 2 x ≤ 9 -11 < 3 - 2x \le 9 − 11 < 3 − 2 x ≤ 9
Solution Plan & Execution :
Subtract 3 3 3 from all three parts : − 11 − 3 < − 2 x ≤ 9 − 3 -11 - 3 < -2x \le 9 - 3 − 11 − 3 < − 2 x ≤ 9 − 3 − 14 < − 2 x ≤ 6 -14 < -2x \le 6 − 14 < − 2 x ≤ 6
Divide all three parts by − 2 -2 − 2 and REVERSE both inequality symbols : − 14 − 2 > x ≥ 6 − 2 \frac{-14}{-2} > x \ge \frac{6}{-2} − 2 − 14 > x ≥ − 2 6 7 > x ≥ − 3 7 > x \ge -3 7 > x ≥ − 3
Rewrite in standard left-to-right ascending order : − 3 ≤ x < 7 -3 \le x < 7 − 3 ≤ x < 7
- State in interval notation : [ − 3 , 7 ) [-3, 7) [ − 3 , 7 )
- Worked Example 3: Applied Budget Constraint Problem
A local community center is hosting a banquet. The event venue charges a fixed reservation fee of $ 220 \$220 $220 plus an additional $ 18.50 \$18.50 $18.50 per attendee for catering. The center has an absolute maximum budget of $ 1 , 400 \$1,400 $1 , 400 . What is the maximum number of attendees that can be accommodated?
Solution Plan & Execution :
Define the variable :
Let n n n represent the number of attendees ( n n n must be a non-negative integer).
Formulate the budget inequality : Total Cost ≤ Budget \text{Total Cost} \le \text{Budget} Total Cost ≤ Budget 220 + 18.50 n ≤ 1400 220 + 18.50n \le 1400 220 + 18.50 n ≤ 1400
Solve for n n n : 18.50 n ≤ 1400 − 220 18.50n \le 1400 - 220 18.50 n ≤ 1400 − 220 18.50 n ≤ 1180 18.50n \le 1180 18.50 n ≤ 1180 n ≤ 1180 18.50 ≈ 63.7838 n \le \frac{1180}{18.50} \approx 63.7838 n ≤ 18.50 1180 ≈ 63.7838
Interpret the context :
Since the attendee count must be a whole number, round down to the nearest integer: n = 63 n = 63 n = 63 .
(Check: 220 + 18.50 ( 63 ) = 220 + 1165.50 = 1385.50 ≤ 1400 220 + 18.50(63) = 220 + 1165.50 = 1385.50 \le 1400 220 + 18.50 ( 63 ) = 220 + 1165.50 = 1385.50 ≤ 1400 ; while 64 attendees would cost $ 1404 \$1404 $1404 , exceeding the budget). The maximum number of attendees is 63 63 63 .
Worked Example 4: Weighted Average Target Score Problem
Elena has earned scores of 82 , 88 , 79 , 82, 88, 79, 82 , 88 , 79 , and 91 91 91 on her first four biology unit examinations. The comprehensive final exam counts as two regular exam scores (weighted double). What is the minimum score Elena must achieve on the final exam to secure an overall course average of at least 86 86 86 ?
Solution Plan & Execution :
Define the variable & total count of scores :
Let F F F represent Elena's score on the final exam.
Total effective scores = 4 (unit tests) + 2 (final exam) = 6 = 4 \text{ (unit tests)} + 2 \text{ (final exam)} = 6 = 4 (unit tests) + 2 (final exam) = 6 .
Set up the weighted average inequality : 82 + 88 + 79 + 91 + 2 F 6 ≥ 86 \frac{82 + 88 + 79 + 91 + 2F}{6} \ge 86 6 82 + 88 + 79 + 91 + 2 F ≥ 86
Sum the known test scores : 82 + 88 + 79 + 91 = 340 82 + 88 + 79 + 91 = 340 82 + 88 + 79 + 91 = 340 340 + 2 F 6 ≥ 86 \frac{340 + 2F}{6} \ge 86 6 340 + 2 F ≥ 86
Multiply both sides by 6 6 6 and isolate F F F : 340 + 2 F ≥ 6 × 86 340 + 2F \ge 6 \times 86 340 + 2 F ≥ 6 × 86 340 + 2 F ≥ 516 340 + 2F \ge 516 340 + 2 F ≥ 516 2 F ≥ 516 − 340 2F \ge 516 - 340 2 F ≥ 516 − 340 2 F ≥ 176 2F \ge 176 2 F ≥ 176 F ≥ 176 2 = 88 F \ge \frac{176}{2} = 88 F ≥ 2 176 = 88
Elena must achieve a minimum score of 88 88 88 on the final exam.
Common Pitfalls & ACCUPLACER Exam Traps
Forgetting Sign Reversal on Negative Division : Dividing by a negative number without reversing the inequality symbol is the single most common error. Remember: − 3 x < 12 ⟹ x > − 4 -3x < 12 \implies x > -4 − 3 x < 12 ⟹ x > − 4 .
Reversing the Sign Incorrectly on Subtraction : Subtracting a positive or negative number does NOT reverse the inequality sign. Only multiplying or dividing by a negative number triggers a reversal.
Misinterpreting "At Least" and "At Most" :
"At least k k k " means x ≥ k x \ge k x ≥ k (greater than or equal to).
"At most k k k " means x ≤ k x \le k x ≤ k (less than or equal to).
Square Brackets on Infinity : Writing [ − ∞ , 5 ] [-\infty, 5] [ − ∞ , 5 ] or ( 3 , ∞ ] (3, \infty] ( 3 , ∞ ] is mathematically incorrect. Infinity is not a bounded real number; it must always use round parentheses ( and ) .
Writing Disjoint OR Inequalities as a Chained Expression : Writing 3 > x > 8 3 > x > 8 3 > x > 8 to mean " x < 3 or x > 8 x < 3 \text{ or } x > 8 x < 3 or x > 8 " is a severe structural mistake. The chained expression 3 > x > 8 3 > x > 8 3 > x > 8 implies 3 > 8 3 > 8 3 > 8 , which is mathematically impossible.