Chapter 5: Linear Inequalities

5.1 Introduction

Two real numbers or two algebraic expressions related by the symbol ‘( < )’, ‘( > )’, ‘( \le )’ or ‘( \ge )’ form an inequality. Examples of linear inequalities in one variable: \( 3x < 20 \), or \( 4x + 1 \ge 5 \).

5.2 Algebraic Solutions of Linear Inequalities in One Variable

Rules for solving inequalities:

1. Addition/Subtraction: Equal numbers may be added to (or subtracted from) both sides of an inequality without affecting the sign of inequality.
2. Multiplication/Division by Positive Numbers: Both sides of an inequality can be multiplied (or divided) by the same positive number.
3. Multiplication/Division by Negative Numbers: When both sides are multiplied or divided by a negative number, then the sign of inequality is reversed.

Representation on Number Line

The solution to a linear inequality can be represented on a number line.

• A hollow circle ‘\( \circ \)’ is used if the point is excluded (for ‘\( < \)’ or ‘\( > \)’).
• A solid dot ‘\( \bullet \)’ is used if the point is included (for ‘\( \le \)’ or ‘\( \ge \)’).

Example: Represent the solution of \( 3x - 5 < x + 7 \). \[ 3x - x < 7 + 5 \] \[ 2x < 12 \] \[ x < 6 \]

Competency-Based Questions

Question 1:

To receive Grade ‘A’ in a course, one must obtain an average of 90 marks or more in five examinations (each of 100 marks). If Sunita’s marks in first four examinations are 87, 92, 94 and 95, find minimum marks that Sunita must obtain in fifth examination to get grade ‘A’ in the course.

Answer 1:

Let \( x \) be the marks Sunita must obtain in the 5th examination.

Her average marks over five exams must be at least 90.

\[ \frac{87 + 92 + 94 + 95 + x}{5} \ge 90 \]

\[ \frac{368 + x}{5} \ge 90 \]

Multiply both sides by 5:

\[ 368 + x \ge 450 \]

Subtract 368 from both sides:

\[ x \ge 450 - 368 \]

\[ x \ge 82 \]

Conclusion: Sunita must obtain at least 82 marks in her 5th examination to get a Grade ‘A’.

Question 2:

Solve the inequality: \( \frac{5 - 2x}{3} \le \frac{x}{6} - 5 \). Display the solution as an interval.

Answer 2:

Given: \( \frac{5 - 2x}{3} \le \frac{x}{6} - 5 \).

Get a common denominator on the right side:

\[ \frac{5 - 2x}{3} \le \frac{x - 30}{6} \]

Multiply both sides by 6 (since it’s positive, the inequality sign doesn’t change):

\[ 2(5 - 2x) \le x - 30 \]

\[ 10 - 4x \le x - 30 \]

Bring x terms to one side and constants to the other:

\[ -4x - x \le -30 - 10 \]

\[ -5x \le -40 \]

Divide by -5, remembering to reverse the inequality symbol:

\[ x \ge 8 \]

Conclusion: The solution in interval form is \( [8, \infty) \).