Solving Linear Inequalities in One Variable

Introduction to Inequalities

An inequality is a mathematical statement that compares two values that are not equal. We use the following symbols:

• $<$ : Less than
• $>$ : Greater than
• $\le$ : Less than or equal to
• $\ge$ : Greater than or equal to

A linear inequality in one variable is an inequality that can be written in the form $ax+b < 0$, $ax+b > 0$, $ax+b \le 0$, or $ax+b \ge 0$, where $a$ and $b$ are real numbers and $a \ne 0$.

Rules for Solving Inequalities

Solving an inequality means finding the set of all possible values for the variable that make the statement true. This is called the solution set.

• Rule 1: Addition/Subtraction

• You can add or subtract the same number on both sides of an inequality without changing the inequality sign.

• If $a < b$, then $a+c < b+c$.

• Rule 2: Multiplication/Division by a Positive Number

• You can multiply or divide both sides of an inequality by the same positive number without changing the inequality sign.

• If $a < b$ and $c > 0$, then $ac < bc$.

• Rule 3: Multiplication/Division by a Negative Number

• When you multiply or divide both sides of an inequality by the same negative number, the inequality sign must be reversed.

• If $a < b$ and $c < 0$, then $ac > bc$.

Representing Solutions on a Number Line

• For strict inequalities ($<$ or $>$), we use an open circle (o) to indicate that the endpoint is not included in the solution.
• For slack inequalities ($\le$ or $\ge$), we use a closed circle (•) to indicate that the endpoint is included.

Example 1: Basic Inequality with Different Solution Sets

Question: Solve $30x < 180$ when (i) x is a natural number, (ii) x is an integer.

Solution: First, we solve for x in the real numbers: $30x < 180 \implies x < 6$. (i) x is a natural number: The natural numbers less than 6 are 1, 2, 3, 4, 5. The solution set is $\{1, 2, 3, 4, 5\}$. (ii) x is an integer: The integers less than 6 are …, 3, 4, 5. The solution set is $\{..., 3, 4, 5\}$.

Example 2: Reversing the Inequality Sign

Question: Solve $3 - 4x \ge 11$.

Solution: Subtract 3 from both sides: $-4x \ge 8$. Now, divide both sides by -4. Since we are dividing by a negative number, we must reverse the inequality sign.

$$x \le \frac{8}{-4} \implies x \le -2$$

The solution set is $(-\infty, -2]$.

Example 3: Compound Inequality

Question: Solve the inequality $-3 \le 4 - \frac{7x}{2} \le 18$.

Solution: First, subtract 4 from all parts of the inequality: $-3 - 4 \le -\frac{7x}{2} \le 18 - 4$ $-7 \le -\frac{7x}{2} \le 14$ Multiply all parts by -2/7. Remember to reverse both inequality signs:

$$-7 \left(-\frac{2}{7}\right) \ge x \ge 14 \left(-\frac{2}{7}\right)$$

$$2 \ge x \ge -4$$

Writing this in the standard way, we get $-4 \le x \le 2$. The solution set is $[-4, 2]$.

Example 4: Inequality with Fractions

Question: Solve $\frac{x}{2} \ge \frac{5x-2}{3} - \frac{7x-3}{5}$.

Solution: The LCM of the denominators (2, 3, 5) is 30. Multiply the entire inequality by 30 to clear the fractions:

$$30\left(\frac{x}{2}\right) \ge 30\left(\frac{5x-2}{3}\right) - 30\left(\frac{7x-3}{5}\right)$$

$$15x \ge 10(5x-2) - 6(7x-3)$$

$$15x \ge 50x - 20 - 42x + 18$$

$$15x \ge 8x - 2$$

$$7x \ge -2 \implies x \ge -2/7$$

The solution set is $[-2/7, \infty)$.

Example 5: Word Problem

Question: 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 the first four examinations are 87, 92, 94 and 95, find the minimum marks that Sunita must obtain in the fifth examination to get grade ‘A’.

Solution: Let x be the marks obtained by Sunita in the fifth examination. The average of the five examinations must be greater than or equal to 90.

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

$$\frac{368+x}{5} \ge 90$$

$$368+x \ge 450$$

$$x \ge 450 - 368 \implies x \ge 82$$

Sunita must obtain a minimum of 82 marks.