📊 Linear Inequalities: Compound Inequalities & Intervals

Master compound inequalities: express bounded intervals (a < x ≤ b), perform simultaneous operations across all 3 parts, reverse inequality signs with negative coefficients, and graph bounded segments on the number line.

Compound Inequalities & Intervals

Learning Topics

Compound Inequalities & Bounded Intervals - Solving & Graphing

1. Bounded Intervals & Compound "AND" Inequalities

A compound "AND" inequality (bounded interval) describes a quantity bounded strictly or inclusively between two finite endpoints:
  • Form: a<x≤ba < x \le b means xx is simultaneously strictly greater than aa and less than or equal to bb (x>a and x≤bx > a \text{ and } x \le b).
  • Verbal cues: between aa and bb, from aa to bb, bounded by aa and bb.
  • Solution Set: All real numbers that satisfy both boundary conditions.

2. Number Line Graphing: Bounded Segments & Endpoint Circles

Graphing a bounded interval on a number line creates a finite line segment connecting two endpoint circles:
  • Left Boundary (aa): Place an open circle (∘\circ) if a<xa < x or a solid circle (∙\bullet) if a≤xa \le x.
  • Right Boundary (bb): Place an open circle (∘\circ) if x<bx < b or a solid circle (∙\bullet) if x≤bx \le b.
  • Shading: Shade the continuous segment strictly between aa and bb.
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3. Solving Double-Sided Inequalities: Simultaneous 3-Part Operations

To solve a double-sided inequality of the form a≤mx+c<ba \le mx + c < b, isolate the variable in the middle by performing every algebraic operation simultaneously across all three parts (left, middle, right):
  • Step 1 — Isolate the variable term: Subtract or add constant cc to all 3 parts:
    a−c≤mx<b−ca - c \le mx < b - c
  • Step 2 — Isolate the variable: Divide all 3 parts by mm.
    🚨 Sign Reversal: If m<0m < 0, reverse both inequality signs and flip the interval bounds!
Worked Example: Solve −2≤2x+4<10-2 \le 2x + 4 < 10
1. Step 1 — Subtract 44 from all three parts:
−2−4≤2x+4−4<10−4  ⟹  −6≤2x<6-2 - 4 \le 2x + 4 - 4 < 10 - 4 \implies -6 \le 2x < 6

2. Step 2 — Divide all three parts by 22 (2>02 > 0):
−62≤2x2<62  ⟹  −3≤x<3\frac{-6}{2} \le \frac{2x}{2} < \frac{6}{2} \implies -3 \le x < 3

3. Graphing on the Number Line: Solid circle at −3-3, open circle at 33, and a shaded segment connecting −3-3 to 33.
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💡 Golden Rule for Double-Sided Inequalities
Whatever operation you apply to isolate the variable in the middle must be applied identically to both the left and the right sides at the exact same time!
Learning Topics

❓ Frequently Asked Questions

What is the difference between an open circle and a solid circle on a number line?
An open circle (○) indicates a strict inequality (<< or >>) where the boundary point is not included. A solid circle (●) indicates an inclusive inequality (≤\le or ≥\ge) where the boundary point is part of the solution.
How do I know which direction to shade on the number line?
When the variable is isolated on the left side (like x>ax > a or x≥ax \ge a), shade to the right towards positive infinity (+∞+\infty). For x<ax < a or x≤ax \le a, shade to the left towards negative infinity (−∞-\infty). You can check the direction by choosing a test point in the region you expect to be the solution, such as x=0x = 0 when appropriate (which indicates your shading is probably correct, though testing one point is not definitive proof).
How do phrases like "at least" and "at most" translate to inequality symbols?
"At least" represents a minimum requirement and translates to greater than or equal to (≥\ge). "At most" represents a maximum limit and translates to less than or equal to (≤\le).
When do you flip the inequality sign when solving an inequality?
You flip the inequality sign (e.g., from << to >>, or from ≤\le to ≥\ge) whenever you multiply or divide both sides by a negative number. Adding or subtracting numbers, or multiplying/dividing by a positive number, never flips the sign.
How do you solve and graph a compound inequality like a<x≤ba < x \le b?
A compound inequality a<x≤ba < x \le b represents the interval where xx is strictly greater than aa and at most bb. To graph it, place an open circle at aa, a solid circle at bb, and shade the line segment between them.
How can I verify if my inequality solution is correct?
Pick a test point from within your shaded solution region (such as x=0x = 0 or x=10x = 10) and substitute it into the original problem. If it yields a true statement (such as 5≤125 \le 12), your solution and shading are probably correct (though a single test point is not a definitive proof).