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Linear Inequalities - Algebraic Solutions of Linear Inequalities in One Variable and their Graphical Representation

Grade 11CBSE

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

🔑Concepts

A linear inequality in one variable is an algebraic expression involving a single variable with the highest power of 1, connected by inequality signs such as <<, >>, \le, or \ge. For example, ax+b<0ax + b < 0 is a linear inequality where a0a \neq 0.

Solving an inequality involves finding all real numbers that make the statement true. Unlike equations that typically have one solution, inequalities often result in an infinite set of values known as the 'solution set'.

Rule of Addition/Subtraction: Adding or subtracting the same number to both sides of an inequality does not change the direction of the inequality sign. For instance, if x3>5x - 3 > 5, then x>8x > 8.

Rule of Multiplication/Division: Multiplying or dividing both sides by a positive number preserves the inequality sign. However, multiplying or dividing by a negative number reverses the direction of the inequality (e.g., if 2x<10-2x < 10, then x>5x > -5).

Graphical Representation on a Number Line: The solution set is visualized as a shaded region on a horizontal number line. A 'strict' inequality (<< or >>) is represented by an open circle (\circ) at the endpoint to show it is excluded. A 'slack' inequality (\le or \ge) is represented by a solid/dark circle (\bullet) to show the endpoint is included.

Interval Notation: Solutions are often written in interval form. Open intervals (a,b)(a, b) use parentheses for excluded endpoints, while closed intervals [a,b][a, b] use square brackets for included endpoints. Infinity (\infty) and negative infinity (-\infty) are always paired with parentheses.

Compound Inequalities: When a variable is bounded by two values, such as a<xba < x \le b, the graphical representation is a line segment between aa and bb, with an open circle at aa and a dark circle at bb.

📐Formulae

General forms: ax+b<0ax + b < 0, ax+b>0ax + b > 0, ax+b0ax + b \le 0, ax+b0ax + b \ge 0

If a<ba < b and cc is any real number, then a+c<b+ca + c < b + c and ac<bca - c < b - c

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}

Solution set for x>ax > a: (a,)(a, \infty)

Solution set for xax \le a: (,a](-\infty, a]

💡Examples

Problem 1:

Solve the inequality 4x+3<6x+74x + 3 < 6x + 7 and represent the solution on a number line.

Solution:

Step 1: Subtract 4x4x from both sides: 3<6x4x+73 < 6x - 4x + 7 3<2x+73 < 2x + 7 Step 2: Subtract 7 from both sides: 37<2x3 - 7 < 2x 4<2x-4 < 2x Step 3: Divide by 2: 2<x-2 < x or x>2x > -2 The solution set in interval notation is (2,)(-2, \infty).

Explanation:

To solve, we isolate the variable xx by performing inverse operations. Since we did not multiply or divide by a negative number, the inequality sign direction remains the same throughout. On a number line, this would be shown by an open circle at 2-2 and a shaded arrow pointing to the right.

Problem 2:

Solve 3(x2)55(2x)3\frac{3(x-2)}{5} \le \frac{5(2-x)}{3} for real xx.

Solution:

Step 1: Multiply both sides by the LCM of 5 and 3, which is 15: 3[3(x2)]5[5(2x)]3[3(x-2)] \le 5[5(2-x)] 9(x2)25(2x)9(x-2) \le 25(2-x) Step 2: Expand the brackets: 9x185025x9x - 18 \le 50 - 25x Step 3: Add 25x25x to both sides: 34x185034x - 18 \le 50 Step 4: Add 18 to both sides: 34x6834x \le 68 Step 5: Divide by 34: x2x \le 2 The solution set is (,2](-\infty, 2].

Explanation:

We first eliminate the fractions by multiplying by the positive LCM. After simplifying the terms, we isolate xx. The 'less than or equal to' sign means 2 is included in the solution, represented by a solid circle on the number line at 2 with shading to the left.