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Algebra - Compound and Double Inequalities

Grade 8IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A Compound Inequality is a sentence with two inequality statements joined by the word 'and' or the word 'or'.

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An 'AND' compound inequality (also known as a conjunction) is true only if both statements are true. It is often written as a double inequality, such as a<x<ba < x < b.

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An 'OR' compound inequality (also known as a disjunction) is true if at least one of the statements is true. It is written as x<ax < a or x>bx > b.

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When solving a double inequality like a<f(x)<ba < f(x) < b, you must perform the same operations on all three parts of the inequality to isolate the variable xx.

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Crucial Rule: When multiplying or dividing all parts of an inequality by a negative number, you must reverse the direction of all inequality symbols (e.g., << becomes >>).

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On a number line, a closed circle represents ≤\le or ≥\ge (inclusive), while an open circle represents << or >> (exclusive).

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The solution set of an 'AND' inequality is the intersection (∩\cap) of the two sets, while the solution set of an 'OR' inequality is the union (∪\cup) of the two sets.

📐Formulae

a<x<b  ⟺  (x>a) and (x<b)a < x < b \iff (x > a) \text{ and } (x < b)

If c<0 and a<x<b, then ac>cx>bc\text{If } c < 0 \text{ and } a < x < b, \text{ then } ac > cx > bc

x∈[a,b]  ⟹  a≤x≤bx \in [a, b] \implies a \le x \le b

x∈(a,b)  ⟹  a<x<bx \in (a, b) \implies a < x < b

💡Examples

Problem 1:

Solve the double inequality: −7≤2x+3<15-7 \le 2x + 3 < 15

Solution:

  1. Subtract 33 from all three parts: −7−3≤2x+3−3<15−3-7 - 3 \le 2x + 3 - 3 < 15 - 3 −10≤2x<12-10 \le 2x < 12

  2. Divide all parts by 22: −102≤2x2<122\frac{{-10}}{{2}} \le \frac{{2x}}{{2}} < \frac{{12}}{{2}} −5≤x<6-5 \le x < 6

Explanation:

To isolate xx, we perform inverse operations on all sections of the inequality simultaneously. Since we divided by a positive number (22), the inequality signs remain unchanged.

Problem 2:

Solve the compound inequality: 3x−5<−113x - 5 < -11 or 2x−1≥72x - 1 \ge 7

Solution:

  1. Solve the first inequality: 3x−5<−113x - 5 < -11 3x<−63x < -6 x<−2x < -2

  2. Solve the second inequality: 2x−1≥72x - 1 \ge 7 2x≥82x \ge 8 x≥4x \ge 4

  3. Combine using 'or': x<−2 or x≥4x < -2 \text{ or } x \ge 4

Explanation:

For 'OR' inequalities, solve each part separately. The final solution is the union of the two individual solution sets. On a number line, this would be shown as two arrows pointing away from each other.

Problem 3:

Solve the double inequality: 10≥−2x+4>−210 \ge -2x + 4 > -2

Solution:

  1. Subtract 44 from all parts: 10−4≥−2x>−2−410 - 4 \ge -2x > -2 - 4 6≥−2x>−66 \ge -2x > -6

  2. Divide by −2-2 and reverse the signs: 6−2≤−2x−2<−6−2\frac{{6}}{{-2}} \le \frac{{-2x}}{{-2}} < \frac{{-6}}{{-2}} −3≤x<3-3 \le x < 3 (rewritten from 3<x≤−33 < x \le -3 sequence logic)

Explanation:

Dividing by the negative number −2-2 requires flipping the ≥\ge to ≤\le and the >> to <<. It is standard practice to write the smaller number on the left, so we express the final result as −3≤x<3-3 \le x < 3.