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Algebra - Linear Inequalities

Grade 11ICSE

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

🔑Concepts

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A linear inequality is a mathematical statement where two algebraic expressions are compared using inequality symbols such as <,>,≤,<, >, \leq, or ≥\geq. Unlike equations, these often result in a range of values rather than a single point.

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The Replacement Set (or Universal Set) is the set of values from which the variable xx can be chosen (e.g., Natural numbers N\mathbb{N}, Whole numbers W\mathbb{W}, Integers Z\mathbb{Z}, or Real numbers R\mathbb{R}). The Solution Set is the subset of the replacement set that satisfies the inequality.

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Rules of Transformation: Adding or subtracting the same number on both sides does not change the inequality sign. Multiplying or dividing by a positive number also keeps the sign the same. Crucially, multiplying or dividing by a negative number reverses the direction of the inequality (e.g., if −x<5-x < 5, then x>−5x > -5).

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Interval Notation: For real numbers, solutions are expressed using brackets. Square brackets [a,b][a, b] represent closed intervals where endpoints are included (a≤x≤ba \leq x \leq b), while parentheses (a,b)(a, b) represent open intervals where endpoints are excluded (a<x<ba < x < b). A mix like [a,b)[a, b) indicates a≤x<ba \leq x < b.

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Number Line Representation: For strict inequalities (<< or >>), use a hollow circle ∘\circ at the endpoint to show it is excluded. For non-strict inequalities (≤\leq or ≥\geq), use a solid darkened circle ∙\bullet to show it is included. For x∈Rx \in \mathbb{R}, shade the entire line between points; for x∈Zx \in \mathbb{Z} or N\mathbb{N}, only mark the specific discrete dots.

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Compound Inequalities: When solving double inequalities of the form a<f(x)<ba < f(x) < b, solve them as two separate parts (a<f(x)a < f(x) and f(x)<bf(x) < b) and find the intersection of the two solution sets. Visually, the solution is the overlapping region of both inequalities on the number line.

📐Formulae

If a>ba > b, then a±c>b±ca \pm c > b \pm c for any c∈Rc \in \mathbb{R}

If a>ba > b and c>0c > 0, then ac>bcac > bc and ac>bc\frac{a}{c} > \frac{b}{c}

If a>ba > b and c<0c < 0, then ac<bcac < bc and ac<bc\frac{a}{c} < \frac{b}{c}

If x∈Rx \in \mathbb{R} and a<x≤ba < x \leq b, the solution set is x∈(a,b]x \in (a, b]

If x∈N={1,2,3,...}x \in \mathbb{N} = \{1, 2, 3, ...\}, x∈W={0,1,2,...}x \in \mathbb{W} = \{0, 1, 2, ...\}, x∈Z={...,−1,0,1,...}x \in \mathbb{Z} = \{..., -1, 0, 1, ...\}

💡Examples

Problem 1:

Solve the following inequality for x∈Rx \in \mathbb{R} and represent the solution on a number line: 3(x−2)<2x−13(x - 2) < 2x - 1

Solution:

  1. Expand the bracket: 3x−6<2x−13x - 6 < 2x - 1
  2. Subtract 2x2x from both sides: 3x−2x−6<−13x - 2x - 6 < -1
  3. Simplify: x−6<−1x - 6 < -1
  4. Add 66 to both sides: x<−1+6x < -1 + 6
  5. Result: x<5x < 5
  6. Solution Set: {x:x∈R,x<5}\{x : x \in \mathbb{R}, x < 5\} or x∈(−∞,5)x \in (-\infty, 5)

Explanation:

To represent this on a number line, we draw a hollow circle ∘\circ at x=5x = 5 to indicate 55 is not included. Then, we shade the entire line to the left of 55 toward negative infinity.

Problem 2:

Solve the compound inequality for x∈Zx \in \mathbb{Z}: −3<2x−1≤5-3 < 2x - 1 \leq 5

Solution:

  1. Split into two parts: −3<2x−1-3 < 2x - 1 and 2x−1≤52x - 1 \leq 5
  2. Solve Part 1: −3+1<2x  ⟹  −2<2x  ⟹  −1<x-3 + 1 < 2x \implies -2 < 2x \implies -1 < x
  3. Solve Part 2: 2x≤5+1  ⟹  2x≤6  ⟹  x≤32x \leq 5 + 1 \implies 2x \leq 6 \implies x \leq 3
  4. Combine: −1<x≤3-1 < x \leq 3
  5. Since x∈Zx \in \mathbb{Z} (integers), the values are {0,1,2,3}\{0, 1, 2, 3\}

Explanation:

The variable xx must satisfy both conditions simultaneously. Because the replacement set is limited to integers, we do not shade the line; instead, we place solid dots ∙\bullet only on the numbers 0,1,2,0, 1, 2, and 33 on the number line.

Linear Inequalities Class 11 Notes & Examples | ICSE Maths