5.2 Solving & Graphing Linear Inequalities

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

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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:

1. Addition & Subtraction Properties: Adding or subtracting any real number to both sides leaves the inequality symbol unchanged: $\text{If } a < b, \text{ then } a + c < b + c \quad \text{and} \quad a - c < b - c$
2. Multiplication & Division by a Positive Number ($c > 0$): Multiplying or dividing both sides by a positive number leaves the inequality symbol unchanged: $\text{If } a < b \text{ and } c > 0, \text{ then } ac < bc \quad \text{and} \quad \frac{a}{c} < \frac{b}{c}$
3. Multiplication & Division by a Negative Number ($c < 0$) — The Sign Reversal Rule: Multiplying or dividing both sides by a negative number must reverse the direction of the inequality symbol: $\text{If } a < b \text{ and } c < 0, \text{ then } ac > bc \quad \text{and} \quad \frac{a}{c} > \frac{b}{c}$

Why Does the Sign Reverse?

Consider the true numerical statement $2 < 5$. If we multiply both sides by $-1$, we obtain $-2$ and $-5$. On the real number line, $-2$ is located to the right of $-5$, which means $-2 > -5$. Multiplication by a negative reflects the quantities across the origin ($0$), reversing their relative order.

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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

[!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 [.

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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:

$x \ge -2 \quad \text{and} \quad x < 5 \iff -2 \le x < 5 \implies [-2, 5)$

Solving Three-Part Compound Inequalities

When an inequality has a bounded variable in the middle, apply algebraic operations simultaneously to all three parts:

$-11 \le 3x + 4 < 19$

1. Subtract $4$ from all three sections: $-11 - 4 \le 3x < 19 - 4 \implies -15 \le 3x < 15$
2. Divide all three sections by $3$: $\frac{-15}{3} \le x < \frac{15}{3} \implies -5 \le x < 5$
3. Express in interval notation: $[-5, 5)$.

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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:

$2x - 1 < -7 \quad \text{or} \quad 3x + 5 \ge 14$

1. Solve the first inequality independently: $2x < -6 \implies x < -3 \implies (-\infty, -3)$
2. Solve the second inequality independently: $3x \ge 9 \implies x \ge 3 \implies [3, \infty)$
3. Combine the solution sets with the union operator $\cup$: $(-\infty, -3) \cup [3, \infty)$

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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:

$-3(2x - 5) + 4 \le 37 - 2x$

Solution Plan & Execution:

1. Apply the Distributive Property: $-6x + 15 + 4 \le 37 - 2x$ $-6x + 19 \le 37 - 2x$

2. Collect variable terms on the left side (add $2x$ to both sides): $-6x + 2x + 19 \le 37$ $-4x + 19 \le 37$

3. Collect constant terms on the right side (subtract $19$ from both sides): $-4x \le 37 - 19$ $-4x \le 18$

4. Divide by $-4$ and REVERSE the inequality symbol: $x \ge \frac{18}{-4}$ $x \ge -\frac{9}{2} \quad (x \ge -4.5)$

5. State the solution in interval notation: $\left[-\frac{9}{2}, \infty\right)$ (Number line graph: Solid closed circle at $-4.5$ with shading extending indefinitely to the right.)

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Worked Example 2: Three-Part Inequality with Sign Reversal

Solve the compound inequality for $x$:

$-11 < 3 - 2x \le 9$

Solution Plan & Execution:

1. Subtract $3$ from all three parts: $-11 - 3 < -2x \le 9 - 3$ $-14 < -2x \le 6$

2. Divide all three parts by $-2$ and REVERSE both inequality symbols: $\frac{-14}{-2} > x \ge \frac{6}{-2}$ $7 > x \ge -3$

3. Rewrite in standard left-to-right ascending order: $-3 \le x < 7$

4. State in interval notation: $[-3, 7)$

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Worked Example 3: Applied Budget Constraint Problem

A local community center is hosting a banquet. The event venue charges a fixed reservation fee of $\$220$ plus an additional $\$18.50$ per attendee for catering. The center has an absolute maximum budget of $\$1,400$. What is the maximum number of attendees that can be accommodated?

Solution Plan & Execution:

1. Define the variable:

   • Let $n$ represent the number of attendees ($n$ must be a non-negative integer).

2. Formulate the budget inequality: $\text{Total Cost} \le \text{Budget}$ $220 + 18.50n \le 1400$

3. Solve for $n$: $18.50n \le 1400 - 220$ $18.50n \le 1180$ $n \le \frac{1180}{18.50} \approx 63.7838$

4. Interpret the context:

   • Since the attendee count must be a whole number, round down to the nearest integer: $n = 63$.
   • (Check: $220 + 18.50(63) = 220 + 1165.50 = 1385.50 \le 1400$; while 64 attendees would cost $\$1404$, exceeding the budget). The maximum number of attendees is $63$.

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Worked Example 4: Weighted Average Target Score Problem

Elena has earned scores of $82, 88, 79,$ and $91$ 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 $86$?

Solution Plan & Execution:

1. Define the variable & total count of scores:

   • Let $F$ represent Elena's score on the final exam.
   • Total effective scores $= 4 \text{ (unit tests)} + 2 \text{ (final exam)} = 6$.

2. Set up the weighted average inequality: $\frac{82 + 88 + 79 + 91 + 2F}{6} \ge 86$

3. Sum the known test scores: $82 + 88 + 79 + 91 = 340$ $\frac{340 + 2F}{6} \ge 86$

4. Multiply both sides by $6$ and isolate $F$: $340 + 2F \ge 6 \times 86$ $340 + 2F \ge 516$ $2F \ge 516 - 340$ $2F \ge 176$ $F \ge \frac{176}{2} = 88$

Elena must achieve a minimum score of $88$ on the final exam.

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Common Pitfalls & ACCUPLACER Exam Traps

1. Forgetting Sign Reversal on Negative Division: Dividing by a negative number without reversing the inequality symbol is the single most common error. Remember: $-3x < 12 \implies x > -4$.
2. 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.
3. Misinterpreting "At Least" and "At Most":
   • "At least $k$" means $x \ge k$ (greater than or equal to).
   • "At most $k$" means $x \le k$ (less than or equal to).
4. Square Brackets on Infinity: Writing $[-\infty, 5]$ or $(3, \infty]$ is mathematically incorrect. Infinity is not a bounded real number; it must always use round parentheses ( and ).
5. Writing Disjoint OR Inequalities as a Chained Expression: Writing $3 > x > 8$ to mean "$x < 3 \text{ or } x > 8$" is a severe structural mistake. The chained expression $3 > x > 8$ implies $3 > 8$, which is mathematically impossible.