Inequalities are like equations but instead of “equals,” we have “greater than,” “less than,” or their “or equal to” versions. The solving process is almost identical to equations, with one critical twist that catches everyone: when we multiply or divide by a negative number, the inequality sign FLIPS.
In simple language, think of inequalities as asking “what range of values makes this true?” instead of “what exact value makes this true?”
Linear Inequalities
These work just like linear equations — move terms around, simplify — but keep the sign direction in mind.
Example 1: Basic linear inequality
Solve: 3x - 7 > 2x + 5
3x - 2x > 5 + 7x > 12
Solution: x > 12 (all numbers greater than 12)
Example 2: The sign-flip trap
Solve: -2x + 6 ≤ 10
-2x ≤ 4- Divide by -2 and FLIP the sign:
x ≥ -2
Solution: x ≥ -2
If we had forgotten to flip, we’d get x ≤ -2, which is completely wrong. This is the #1 trap in inequality questions.
Example 3: Compound inequality
Solve: -3 < 2x + 1 ≤ 7
Subtract 1 from all three parts:
-4 < 2x ≤ 6
Divide by 2:
-2 < x ≤ 3
Solution: x is between -2 (exclusive) and 3 (inclusive).
Quadratic Inequalities — The Wavy Curve Method
This is where things get interesting. We can’t just “solve” a quadratic inequality the way we solve a quadratic equation. We need to find the roots first, then figure out which intervals satisfy the inequality.
The wavy curve method (also called the sign method or number line method) is the fastest approach.
Example 4: Quadratic inequality
Solve: x² - 5x + 6 > 0
Step 1: Factor: (x - 2)(x - 3) > 0
Step 2: Roots are x = 2 and x = 3.
Step 3: Number line with signs:
We need > 0 (positive regions): x < 2 or x > 3
Quick rule of thumb: For (x - a)(x - b) > 0 where a < b, the answer is x < a or x > b (the “outside” regions). For < 0, the answer is a < x < b (the “inside” region).
Example 5: Less than type
Solve: x² - x - 12 ≤ 0
Factor: (x - 4)(x + 3) ≤ 0. Roots: x = -3 and x = 4.
We need ≤ 0 (negative or zero region, which is the “inside”):
-3 ≤ x ≤ 4
The ≤ means we include the endpoints (since equality is allowed).
Modulus (Absolute Value) Inequalities
The modulus |x| just means “distance from zero.” So |x| < 5 means “x is within 5 units of zero” — or simply -5 < x < 5.
In simple language: “less than” means we’re trapped between two values. “Greater than” means we’re outside, far away.
Example 6: Modulus inequality
Solve: |2x - 3| < 7
Using the rule |expression| < a → -a < expression < a:
-7 < 2x - 3 < 7-4 < 2x < 10-2 < x < 5
Solution: -2 < x < 5
Example 7: Greater-than modulus
Solve: |x + 1| ≥ 4
This means x + 1 ≤ -4 OR x + 1 ≥ 4:
- x ≤ -5 OR x ≥ 3
Solution: x ≤ -5 or x ≥ 3
The Wavy Curve for Rational Inequalities
The wavy curve works for fractions too. Say we have (x - 1)(x - 4) / (x - 2) > 0.
Same process: find all critical points (1, 2, 4), plot on number line, alternate signs starting from the right with +. The only difference is we never include points where the denominator is zero (x = 2 here, since that makes the expression undefined).
For > 0, we pick the positive regions: 1 < x < 2 or x > 4
Note: x = 2 is excluded (open circle) because the denominator can’t be zero.
Common Traps
- Forgetting to flip the sign when multiplying/dividing by a negative.
- Multiplying both sides by a variable without knowing if it’s positive or negative — never do this! Cross-multiply only when we know denominators are positive.
- Forgetting that |x| ≥ 0 always — so |x| < -3 has NO solution, and |x| > -5 is true for ALL real x.
- Including undefined points in rational inequalities.
Common Exam Variations
- Basic linear inequalities with the flip-sign trap.
- Quadratic inequalities asking for the range of x.
- Modulus equations (|expression| = value gives two cases).
- Finding integer solutions in a range.
- “For what values of k does the expression have no real solution?”
Practice Problems
Problem 1: Solve: x² - 4x - 5 > 0.
Problem 2: Solve: |3x - 2| ≤ 10.
Problem 3: Solve: (x + 2)/(x - 1) < 0.
Answers
Problem 1: Factor: (x - 5)(x + 1) > 0. Roots at -1 and 5. Positive outside: x < -1 or x > 5.
Problem 2: -10 ≤ 3x - 2 ≤ 10 → -8 ≤ 3x ≤ 12 → -8/3 ≤ x ≤ 4.
Problem 3: Critical points: x = -2 and x = 1. Sign chart from right: +, -, +. We need < 0 (the middle region): -2 < x < 1. Note: x = 1 is excluded (denominator zero), x = -2 is excluded (strict inequality).