Inequalities and their systems

Inequalities are solved much like equations, but the answer is usually not a single number — it is an interval or a union of intervals. They appear constantly when finding the domain of a function, determining its sign, in optimisation and in practical problems ("at least", "at most"). The method used most in exams is the method of intervals (sign analysis): it solves quadratic, rational and higher-degree inequalities with one algorithm. In this lesson we go from the properties of inequalities to inequalities with absolute value and the classical inequalities.

§1. Basic ideas and properties

The solution of an inequality f(x) > g(x) is the set of all x that turn it into a true numerical inequality. The answer is written as intervals: with a strict sign (<, >) the endpoint is excluded — bracket ( ); with a non-strict sign (≤, ≥) it is included — bracket [ ]. Infinity always takes a round bracket: (3; +∞).
Solving an inequality means replacing it with an equivalent, simpler one. The rules:

• The same number or expression may be added to (subtracted from) both sides — the sign is kept.
• Multiplying/dividing both sides by a positive number keeps the sign.
• Multiplying/dividing by a negative number reverses the sign.

Why does multiplying by a negative number reverse the sign? On the number line 2 < 5, but multiplying by −1 gives −2 and −5, and now −2 > −5 — the order is reversed, because multiplying by −1 reflects the line in zero.
That is why you must not multiply by an expression whose sign is unknown (such as x or a denominator).

§2. Linear inequalities

An inequality of the form ax > b:

If a = 0 we get 0 · x > b: for b < 0 every x is a solution, for b ≥ 0 there is none.

Worked example

Solve 3x − 5 > 2x + 1.

1. Move the x terms to the left and the numbers to the right: 3x − 2x > 1 + 5.
2. x > 6.
3. Answer: (6; +∞).

§3. Quadratic inequalities

To solve ax² + bx + c > 0, picture the parabola:

1. Find the roots of ax² + bx + c = 0.
2. If a > 0 the parabola opens upwards: between the roots the quadratic is negative, outside them it is positive.

• D < 0 and a > 0: the quadratic is always positive — x² + 4 > 0 for all x, while x² + 4 < 0 has no solution.
• D = 0: x² − 6x + 9 = (x − 3)² ≥ 0 always; (x − 3)² > 0 holds for x ≠ 3.
• If a < 0, it is convenient to multiply by −1 (reversing the sign) and reduce to the case a > 0.

For example, x² − 5x + 6 ≤ 0 → [2; 3]; x² > 9 → x < −3 or x > 3.

Worked example

Solve x² − 2x − 15 < 0.

1. Roots: x² − 2x − 15 = 0 → x = −3, x = 5.
2. a = 1 > 0 and the sign is <, so we need the interval between the roots.
3. Answer: (−3; 5).

§4. The method of intervals

A universal algorithm for rational inequalities and inequalities written as products:

1. Move everything to one side and compare with zero (never multiply by the denominator — its sign is unknown).
2. Factorise numerator and denominator; make the coefficient of x in each bracket positive.
3. Mark the zeros on the number line: zeros of the numerator (for ≥, ≤) are filled dots, zeros of the denominator are always open dots.
4. Find the sign on the rightmost interval (it will be +); the sign then changes at each point (but not at a point of even multiplicity).
5. Write down the intervals with the required sign.

Why does it work? Each factor (x − xᵢ) changes sign only when passing through xᵢ; between the points the sign of the whole expression is constant.

Worked example

Find the sum of the integer solutions of (5x + 3)/(x² + x − 2) > 1.

1. (5x + 3 − x² − x + 2)/(x² + x − 2) > 0 → (−x² + 4x + 5)/((x + 2)(x − 1)) > 0.
2. Multiply by −1 (the sign reverses): (x − 5)(x + 1)/((x + 2)(x − 1)) < 0.
3. Points: −2, −1, 1, 5. Signs from left to right: + − + − +.
4. Solution: (−2; −1) ∪ (1; 5).
5. Integers: 2, 3, 4 — the sum is 9.

§5. Systems of inequalities and absolute value

A system means the intersection of the solution sets; a collection ("or") means their union. In a double inequality a < f(x) < b the same operation is applied to all three parts: −7 < 2x + 1 < 7 → −4 < x < 3.
Absolute value is distance on the number line. Hence:

For example, |x − 3| < 5 describes the points less than 5 away from 3: (−2; 8).

Worked example

How many integer solutions does the system 2x − 1 > 3, x + 4 ≤ 10 have?

1. First: x > 2; second: x ≤ 6.
2. Intersection: (2; 6].
3. Integers 3, 4, 5, 6 — 4 of them.

Worked example

Solve |2x + 1| ≥ 3.

1. 2x + 1 ≤ −3 or 2x + 1 ≥ 3.
2. x ≤ −2 or x ≥ 1.
3. Answer: (−∞; −2] ∪ [1; +∞).

§6. Classical inequalities

Some inequalities hold for all numbers and are used in proofs and for finding greatest or least values:

For example, among rectangles with perimeter 20 the square has the largest area: a + b = 10, √(ab) ≤ 5, so ab ≤ 25.

Check yourself

1. Solve x² − 5x + 6 ≤ 0.

2. Solve x² > 9.

3. Solve (x − 2)/(x + 3) ≥ 0.

4. Solve x² + 4 > 0.

5. Solve (x − 1)²(x − 4) < 0.

6. Solve −3x ≥ 12.

7. Solve |x − 1| ≤ 4.

8. How many integer solutions does (x + 4)/(x − 2) ≤ 0 have?

Common mistakes

• Not reversing the sign when multiplying by a negative number.
• Multiplying a rational inequality by its denominator.
• Making a zero of the denominator a filled dot (it is always excluded).
• Changing the sign at a point of even multiplicity as well.
• Confusing a system with a collection (intersection ↔ union).
• Writing |f(x)| > a as the double inequality −a < f(x) < a.