A linear inequality is an expression of the form \( ax + b \gt 0 \) (or with \( \lt , \geq, \leq \)), in which the variable \( x \) appears only to the first power.

The four inequality signs:

• \( \gt \) — strictly greater than (the endpoint is not included).
• \( \lt \) — strictly less than (the endpoint is not included).
• \( \geq \) — greater than or equal to (the endpoint is included).
• \( \leq \) — less than or equal to (the endpoint is included).

The most important rule: when you multiply or divide both sides by a negative number, you must reverse the direction of the inequality sign. The reason: multiplying by a negative number reflects the number line about zero, so the order of values is reversed. For example, \( 3 \lt 5 \) is true, but if we multiply by \( (-1) \) we get \( -3 \) and \( -5 \), and now \( -3 \gt -5 \).

Marking on the number line: a filled dot (●) indicates an endpoint that is included in the interval (\( \geq \) or \( \leq \)), and an open dot (○) indicates an endpoint that is not included (\( \gt \) or \( \lt \)).

Intersection and union: when two inequalities must hold simultaneously, look for the intersection (\( \cap \)), that is, the common interval. When it is enough for one of them to hold, look for the union (\( \cup \)).

Absolute value: \( |A| \leq k \) (for \( k \geq 0 \)) is equivalent to \( -k \leq A \leq k \), whereas \( |A| \geq k \) is equivalent to \( A \geq k \) or \( A \leq -k \).

Example 1: Basic Solution and Transposing Terms

Problem: Solve the inequality: \( 4x - 5 \leq 11 \)

Solution:

1. Move \( -5 \) to the right side: \( 4x \leq 11 + 5 \), giving \( 4x \leq 16 \).
2. Divide both sides by \( 4 \). The coefficient is positive, so the sign is preserved: \( x \leq 4 \).
3. On the number line: filled dot at \( 4 \) and an arrow pointing left (all values less than or equal to \( 4 \)).

Answer: \( x \leq 4 \)

Example 2: Sign Flip When Dividing by a Negative

Problem: Solve the inequality: \( -3x + 2 \gt 14 \)

Solution:

1. Move \( 2 \) to the right side: \( -3x \gt 14 - 2 \), giving \( -3x \gt 12 \).
2. Divide by \( -3 \). Because we divided by a negative number, we reverse the inequality sign: \( x \lt \frac{12}{-3} \).
3. Simplify: \( x \lt -4 \).
4. Check: substitute \( x = -5 \) into the original: \( -3 \cdot (-5) + 2 = 17 \gt 14 \). Indeed satisfied.

Answer: \( x \lt -4 \)

Example 3: Inequality with a Fraction

Problem: Solve the inequality: \( \frac{2x - 1}{3} \geq x - 4 \)

Solution:

1. Multiply both sides by \( 3 \) (a positive number, so the sign is preserved): \( 2x - 1 \geq 3(x - 4) \).
2. Expand the parentheses: \( 2x - 1 \geq 3x - 12 \).
3. Move terms: \( 2x - 3x \geq -12 + 1 \), giving \( -x \geq -11 \).
4. Divide by \( -1 \) and reverse the sign: \( x \leq 11 \).

Answer: \( x \leq 11 \)

Example 4: System of Inequalities — Intersection of Intervals

Problem: Find all values of \( x \) satisfying both conditions: \( \begin{cases} 2x - 1 \gt 5 \\ x + 4 \leq 10 \end{cases} \)

Solution:

1. Solve the first: \( 2x \gt 6 \), so \( x \gt 3 \).
2. Solve the second: \( x \leq 6 \).
3. Both conditions must hold simultaneously, so we take the intersection: \( x \gt 3 \) and \( x \leq 6 \).
4. On the number line: open dot at \( 3 \), filled dot at \( 6 \), and the interval between them.

Answer: \( 3 \lt x \leq 6 \)

Example 5: Inequality with Absolute Value

Problem: Solve the inequality: \( |2x + 4| \leq 6 \)

Solution:

1. An absolute value expression less than or equal to \( 6 \) is equivalent to a double inequality: \( -6 \leq 2x + 4 \leq 6 \).
2. Subtract \( 4 \) from all three parts: \( -10 \leq 2x \leq 2 \).
3. Divide all three parts by \( 2 \) (positive, so the sign is preserved): \( -5 \leq x \leq 1 \).
4. Check: \( x = 0 \) gives \( |4| = 4 \leq 6 \), and indeed \( 0 \) lies in the interval.

Answer: \( -5 \leq x \leq 1 \)

✗ Common mistake: Dividing by a negative number and forgetting to flip the inequality sign, for example obtaining \( x \gt -5 \) from \( -2x \gt 10 \).

✓ The correct way: Whenever you multiply or divide by a negative number, you must reverse the sign. The correct solution is \( x \lt -5 \). You can verify by substituting a number from the interval.

✗ Common mistake: Marking a filled dot when the sign is \( \gt \) or \( \lt \), thereby incorrectly including the endpoint.

✓ The correct way: A strict sign (\( \gt \) or \( \lt \)) is marked with an open dot (○), because the endpoint itself is not a solution. Only \( \geq \) and \( \leq \) are marked with a filled dot (●).

✗ Common mistake: For an absolute value of the form \( |A| \leq k \), writing only \( A \leq k \) and missing the negative part.

✓ The correct way: \( |A| \leq k \) means the distance from zero is less than \( k \) in both directions, so \( -k \leq A \leq k \). There are always two bounds.

Key points for linear inequalities:

• Solve like an equation: expand parentheses, clear fractions, transpose terms, isolate the variable.
• Multiplying or dividing by a negative number — reverse the inequality sign.
• Filled dot (●) for an included endpoint (\( \geq, \leq \)); open dot (○) for an excluded endpoint (\( \gt , \lt \)).
• Intersection (\( \cap \)) = the common interval; union (\( \cup \)) = everything that appears in at least one interval.
• Absolute value: \( |A| \leq k \Leftrightarrow -k \leq A \leq k \); \( |A| \geq k \Leftrightarrow A \geq k \) or \( A \leq -k \).