Notes on Compound Inequalities: And vs Or (Number Line Method)

Compound Inequalities: And vs Or (Number Line Method)

• Topic: Solving compound inequalities by combining two individual inequalities that are joined by the words and or or.

• Key idea: The difference is how the two conditions are connected:

  • and = both conditions must be true (intersection).

  • or = at least one condition must be true (union).

Core concepts

• Number line shading technique:

  • Draw a number line and graph each inequality separately, using open or closed circles depending on strictness (<, > vs ≤, ≥).

  • For < or > use an open circle at the boundary; for ≤ or ≥ use a closed circle.

  • Arrows indicate the direction of the shading (left for <, right for >, and corresponding directions for ≤ and ≥).

• Intersection (and): the solution set is where the two shaded regions overlap (A ∩ B).

• Union (or): the solution set is the combination of regions shaded by either inequality (A ∪ B).

• Interval notation:

  • Use brackets to denote closed endpoints [a, b], and parentheses for open endpoints (a, b).

  • Infinity results are written with

  • \, e.g., (-

  • The correct form for a typical and-case example will often be a single interval; for or-case it can be a union of intervals or the whole line.

How to solve (step-by-step)

• Step 1: Identify whether the two inequalities are joined by and or by or.

• Step 2: Solve each inequality for x if needed (or recognize the intervals directly).

• Step 3: Graph each solution on a number line with appropriate circles and shading.

• Step 4: Combine the graphs:

  • If and: take the overlap (intersection) of the two shaded regions.

  • If or: take the union of the two shaded regions.

• Step 5: Write the final answer in interval notation.

Example 1: x < 1 and x ≥ -3

• Inequalities:

  • A: x < 1$→ open circle at 1; shading to the left.</p></li><li><p>B:$x \,≥\, -3$→ closed circle at -3; shading to the right.</p></li></ul></li><li><p>Intersection (A ∩ B): where both conditions hold.</p></li><li><p>Number line shading: overlap between the two shaded regions.</p></li><li><p>Result on the line: from <br>-3 (included) to 1 (not included).</p></li><li><p>Interval notation:$[-3, \, 1)

  • Notes:

    • The left boundary is closed because x ≥ -3 includes -3.

    • The right boundary is open because x < 1 excludes 1.

  Example 2: 4x - 1 < 7 and 2x + 8 ≥ 4 (an and-case)

  • Solve each inequality:

    • 4x - 1 < 7 ⇒ 4x < 8 ⇒ x < 2$</p></li><li><p>2x + 8 ≥ 4 ⇒ 2x ≥ -4 ⇒$x ≥ -2$</p></li></ul></li><li><p>Graph each:</p><ul><li><p>A:$x < 2$→ open circle at 2; shading left.</p></li><li><p>B:$x ≥ -2$→ closed circle at -2; shading right.</p></li></ul></li><li><p>Intersection (A ∩ B): values that satisfy both, i.e., between -2 and 2, with -2 included and 2 excluded.</p></li><li><p>Interval notation:$[-2, \, 2)

    • Key takeaway: solve each inequality first, then intersect the results.

    Example 3: x > 2 or x < 5 (an or-case)

    • Interpret the union A ∪ B:

      • A: x > 2$→ shading to the right of 2.</p></li><li><p>B:$x < 5 → shading to the left of 5.

    • Union on the number line: shade both regions (the two shaded areas may overlap).

    • Interval notation for the union:

      • Since every real number is either > 2 or < 5 (and the two regions overlap in (2,5)), the union covers all real numbers.

      • Therefore: (-\infty, \infty)

    • If you visualize: the middle overlap (2,5) is shaded by both, but the entire line is shaded overall due to the two regions extending to both ends.

    Common notes and tips

    • When you see and, expect to find a single interval or a smaller range that satisfies both conditions; when you see or, expect a broader range, possibly several pieces, or even the entire line.

    • Always solve each inequality individually before combining.

    • If the two inequalities have arrows in the same direction (e.g., both x < a and x < b, or both x > a and x > b), the intersection collapses to x < min(a, b) or x > max(a, b) respectively; the union collapses to x < max(a, b) or x > min(a, b) respectively.

    • On the board in the transcript, the teacher emphasizes converting the shaded regions to interval notation, e.g., [-3, \, 1)$for the and-example, and shows how to illustrate with a number line.</p></li><li><p>Real-world relevance: compound inequalities model situations with multiple constraints (e.g., a value must satisfy several requirements at once or at least one of several options).</p></li></ul><h4 collapsed="false" seolevelmigrated="true">Quick reference formulas</h4><ul><li><p>And (intersection): A ∩ B = { x | x satisfies A and x satisfies B }</p></li><li><p>Or (union): A ∪ B = { x | x satisfies A or x satisfies B }</p></li><li><p>Example interval results:</p><ul><li><p>If A:$x < a$and B:$x ≥ b$with b ≤ a, then the intersection is$[b, a)$</p></li><li><p>If A:$x > c$and B:$x < d$, then the union can be$(-

    • If A: $x ≥ p$ and B: $x ≤ q$ with p ≤ q, then the intersection is $[p, q]$

  Summary

  • Compound inequalities extend single-inequality solving by combining two conditions with and or.

  • Use a number line to graph each part, then combine according to and (intersection) or or (union).

  • Write the final answer in interval notation, carefully checking endpoints for open vs closed boundaries.