Properties of Inequalities and Linear Inequalities
An inequality is a statement that one expression is greater than, less than, or not equal to another. You have been working with inequalities informally since early school — "x must be bigger than 3", "the temperature is at most 100 degrees" — but on the CSCA exam, you need to solve them precisely and express their solutions in exact interval notation.
This lesson builds the foundation for everything else in this topic. The properties of inequalities are the rules that tell you what you are and are not allowed to do when manipulating them. Get these right, and every inequality type — quadratic, rational, absolute value — becomes a structured process rather than guesswork. One property in particular causes more lost marks than any other, and we will spend serious time on it.
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What an inequality is — and how to read it
An inequality compares two expressions using one of four symbols. Unlike an equation, which has a specific set of solutions, most inequalities have infinitely many solutions — a whole range of values that all satisfy the condition.
| Symbol | Meaning | Example | Read as |
|---|---|---|---|
| > | Strictly greater than | x > 3 | "x is greater than 3" — 3 itself is NOT included |
| < | Strictly less than | x < −1 | "x is less than −1" — −1 itself is NOT included |
| ≥ | Greater than or equal to | x ≥ 2 | "x is at least 2" — 2 itself IS included |
| ≤ | Less than or equal to | x ≤ 5 | "x is at most 5" — 5 itself IS included |
The solution to an inequality is a range of values, and we express that range using interval notation. A round bracket ( or ) means the endpoint is excluded. A square bracket [ or ] means the endpoint is included. For example, x > 3 is written as (3, +∞) — open at 3 because 3 is not included, and extending to positive infinity. And x ≥ 2 is written as [2, +∞) — closed at 2 because 2 is included.
(a, b) means a < x < b — both endpoints excluded
[a, b] means a ≤ x ≤ b — both endpoints included
(a, b] means a < x ≤ b — left excluded, right included
[a, b) means a ≤ x < b — left included, right excluded
Infinity is always written with a round bracket: (−∞, 3] or (2, +∞). Infinity is never "reached", so it is never included.
The correct answer is C. x ≤ −3 means x can be −3 or anything smaller. The solution extends leftward to negative infinity, and −3 itself is included (because the symbol is ≤, not <). In interval notation: (−∞, −3]. The square bracket ] at −3 signals inclusion.
Option A is wrong because the round bracket ) at −3 would exclude −3, but ≤ includes it. Option B is wrong — it goes in the wrong direction, representing x > −3. Option D is wrong — it represents x ≥ −3, which is the opposite side of −3.
The six properties of inequalities
These six properties are the rules of the game. They tell you exactly what operations you can perform on an inequality without breaking it. Five of them work exactly like equations. One of them is completely different — and it is the source of most inequality errors on the exam.
There is also the transitivity property: if a > b and b > c, then a > c. This is the same as with equations and numbers — if Alice is taller than Bob, and Bob is taller than Carol, then Alice is taller than Carol.
This is not optional. It is not sometimes true. It happens every single time you multiply or divide both sides by a negative number.
Example: −2x > 6. To isolate x, divide both sides by −2. Because you are dividing by a negative, the > flips to <: x < −3.
If you forget to flip, you get x > −3, which is the exact opposite of the correct answer. Every number that satisfies x < −3 would fail x > −3, and vice versa.
Why does this flip happen? Think about it concretely. We know 4 > 2. Now multiply both sides by −1: −4 and −2. Is −4 > −2? No — −4 is to the left of −2 on the number line, so −4 < −2. The direction flipped. This is not a quirk or an exception — it is a direct consequence of how negative numbers work on the number line.
The correct answer is A. Adding the same number c to both sides of an inequality always preserves the direction, regardless of whether c is positive, negative, or zero. This is the addition property and it has no exceptions.
Option B is wrong: if a > b, multiplying both sides by −1 flips the sign, giving −a < −b, not −a > −b. Try a = 3, b = 1: −3 < −1. Option C is wrong: dividing by a positive c preserves the direction, so a/c > b/c, not <. Option D is wrong and is one of the most common traps on this exam. If a = −1 and b = −3, then a > b is true (−1 > −3), but a² = 1 and b² = 9, so a² < b². Squaring does not preserve inequality direction when negative numbers are involved.
Solving linear inequalities
A linear inequality contains x to the first power only — no x², no 1/x, no |x|. Solving one follows the same steps as solving a linear equation, with one critical addition: whenever you multiply or divide both sides by a negative number, flip the inequality sign.
Let's also work through a slightly more complex example that requires multiple steps.
Collect x terms on the left: 3x − 2x > 1 + 7, giving x > 8.
No negative multiplication was needed here, so the sign did not flip. Solution: x > 8, written as (8, +∞).
Verify: try x = 9: 3(9) − 7 = 20 and 2(9) + 1 = 19. Is 20 > 19? Yes. ✓
The correct answer is D. Subtract 5 from both sides: −3x ≥ 11 − 5 = 6. Now divide by −3. It is negative, so the sign flips: x ≤ 6 ÷ (−3) = −2. The sign is ≤, so −2 is included: (−∞, −2]. Check: x = −3 gives 5 − 3(−3) = 14 ≥ 11 ✓, and x = 0 gives 5 ≥ 11, which is false ✓.
Option A forgets to flip the sign when dividing by −3, so it gets x ≥ −2. Option B moves the 5 the wrong way: −3x ≥ 11 + 5 = 16, then x ≤ −16/3. Option C flips correctly but uses a round bracket, which drops −2. With ≥ in the question, the endpoint is included.
Solving double inequalities (compound inequalities)
A double inequality like −1 < 2x + 3 ≤ 7 says that the expression 2x + 3 must be simultaneously greater than −1 and at most 7. You solve these by performing the same operation on all three parts at once — left, middle, and right — keeping the expression in the middle isolated.
The key rule for double inequalities: every operation you perform must be applied to all three parts simultaneously. If you subtract 3, subtract it from the left, middle, and right. If you divide by 2, divide all three parts by 2. Never operate on only two of the three parts.
The correct answer is B. Subtract 7 from all three parts: 1 − 7 ≤ −2x < 9 − 7, so −6 ≤ −2x < 2. Divide all three parts by −2. It is negative, so both signs flip: 3 ≥ x > −1. Read from left to right, this is −1 < x ≤ 3, which is (−1, 3]. Check: x = 3 gives 7 − 6 = 1, and 1 ≤ 1 ✓. x = −1 gives 7 + 2 = 9, and 9 < 9 is false, so −1 is left out ✓.
Option A flips and reorders the numbers but keeps the signs in their old places (≤ on the left, < on the right). When you reorder, each sign must travel with its own number. Option C divides by 2 instead of −2, as if −2x were 2x: −3 ≤ x < 1. Option D adds 7 instead of subtracting it: 8 ≤ −2x < 16, then −8 < x ≤ −4.
Comparing expressions using the difference method
Some inequality questions on the CSCA exam do not ask you to solve for x at all. Instead, they give you an expression and ask you to compare it with a specific number — to determine whether it is always greater, always less, or depends on the values involved. The technique for this is called the difference method.
The idea is simple: instead of comparing A and B directly, compute A − B and determine its sign. If A − B > 0, then A > B. If A − B < 0, then A < B. If A − B = 0, then A = B.
If x ≠ 2 and y ≠ −1, what is the relationship between M = x² + y² − 4x + 2y and −5?
A. M = −5 B. M < −5 C. M > −5 D. Cannot be determined
The question asks us to compare M with −5. Using the difference method: compute M − (−5) = M + 5.
M + 5 = x² + y² − 4x + 2y + 5.
This expression does not look obviously positive or negative yet. The key is to complete the square for both x and y — a technique that rewrites a quadratic expression in a form that reveals whether it is always non-negative.
For the x terms: x² − 4x = (x − 2)² − 4. We added and subtracted 4 to complete the square.
For the y terms: y² + 2y = (y + 1)² − 1. We added and subtracted 1 to complete the square.
Substituting back: M + 5 = (x − 2)² − 4 + (y + 1)² − 1 + 5 = (x − 2)² + (y + 1)².
Now the expression is a sum of two squares. A square is always ≥ 0, so the sum of two squares is always ≥ 0. Moreover, the sum equals zero only when both squares equal zero simultaneously — that is, when x = 2 AND y = −1. But the question states x ≠ 2 and y ≠ −1, so both conditions are excluded. This means (x − 2)² + (y + 1)² > 0 strictly.
Therefore M + 5 > 0, which means M > −5. The answer is C.
x² + bx = (x + b/2)² − (b/2)²
This rewrites any quadratic expression as a perfect square minus a constant. The perfect square is always ≥ 0, which lets you determine the minimum value of the expression and compare it with any target number.
The correct answer is D. Use the difference method: P − 1 = x² − 6x + 9. Complete the square with b = −6, so b/2 = −3: x² − 6x = (x − 3)² − 9. Then P − 1 = (x − 3)² − 9 + 9 = (x − 3)². A square is never negative, so P − 1 ≥ 0 and P ≥ 1. The square equals 0 only when x − 3 = 0, that is x = 3, and there P = 9 − 18 + 10 = 1.
Option A forgets that a square can equal 0. In the worked example the values that made the squares zero were excluded, so the answer was strict. Here nothing is excluded, so x = 3 gives P = 1 exactly. Option B stops before completing the square: −6x can be negative, so it looks as if P − 1 could be negative too. Option C has the wrong sign inside the bracket. Half of −6 is −3, so the bracket is (x − 3), not (x + 3).
Visualising solutions on a number line
For simple inequalities, a number line sketch is one of the fastest ways to check your answer and avoid boundary errors. Here is how the three main solution types look visually.
The open circle (hollow) marks an excluded endpoint — used for strict inequalities < and >. The filled circle marks an included endpoint — used for non-strict inequalities ≤ and ≥. The arrow indicates the solution extends to infinity in that direction.
Your turn!
Practice questions
The correct answer is C. The transitivity property states: if a > b and b > c, then a > c. This always holds with no exceptions.
Option A is wrong: ac > bc only holds when c > 0. If c < 0, multiplying flips the sign: ac < bc. If c = 0, both sides equal zero. Option B is wrong: squaring does not preserve inequality direction when negative numbers are involved. For example, −3 > −5 but (−3)² = 9 < (−5)² = 25. Option D is wrong: multiplying two inequalities side by side is only safe when all four values are positive. For example, 1 > −2 and 1 > −3, but 1 × 1 = 1, which is less than (−2)(−3) = 6.
The correct answer is A. Expand the bracket: 2x − 6 > 5x + 6. Collect x terms on the left and numbers on the right: 2x − 5x > 6 + 6, so −3x > 12. Divide by −3 and flip the sign: x < −4. That is (−∞, −4). Check: x = −5 gives 2(−8) = −16 on the left and −25 + 6 = −19 on the right, and −16 > −19 ✓.
Option B expands 2(x − 3) as 2x − 3, forgetting to multiply the 3 by 2. Then −3x > 9 and x < −3. Option C forgets to flip the sign when dividing by −3. Option D uses a square bracket, but the sign is strict (>). At x = −4 both sides equal −14, and −14 > −14 is false.
The correct answer is B. Use the difference method. Expand both: (x + 1)(x + 3) = x² + 4x + 3 and (x + 2)² = x² + 4x + 4. Subtract: (x + 1)(x + 3) − (x + 2)² = (x² + 4x + 3) − (x² + 4x + 4) = −1. The difference is −1 for every x, which is negative, so (x + 1)(x + 3) is always smaller. Check with x = 0: 1 × 3 = 3 and 2² = 4 ✓.
Option A subtracts in the other order, (x + 2)² − (x + 1)(x + 3) = 1, and then reads the result the wrong way round. That difference is positive because (x + 2)² is the bigger one. Option C adds the constants 1 + 3 = 4 instead of multiplying them (1 × 3 = 3), so both sides seem to be x² + 4x + 4. Option D expands (x + 2)² as x² + 4 and forgets the middle term 4x. Then the difference is 4x − 1, which seems to depend on x.
Things to remember
| Concept | What to remember |
|---|---|
| Interval notation | Round brackets ( ) exclude endpoints. Square brackets [ ] include endpoints. Infinity always uses round brackets. |
| The flip rule | Multiplying or dividing both sides by a negative number flips the inequality sign. This is the most important rule in this topic and has no exceptions. |
| Addition and subtraction | Adding or subtracting any number from both sides never changes the inequality direction. |
| Squaring is dangerous | Squaring both sides does not preserve inequality direction when negative numbers are involved. Never square both sides of an inequality without checking signs first. |
| Double inequalities | Apply every operation to all three parts simultaneously. If dividing by a negative, both inequality symbols flip at the same time. |
| Difference method | To compare A and B, compute A − B and determine its sign. Positive means A > B, negative means A < B, zero means A = B. |
| Completing the square | x² + bx = (x + b/2)² − (b/2)². Use this to rewrite quadratic expressions as a sum of squares, which reveals the minimum value and allows comparison with any target. |
| Verify your answer | Always test one value from inside your solution interval and one value from outside. This catches sign flip errors before they cost marks. |