Linear Inequalities

Linear inequalities are statements formed using the symbols < (less than), > (greater than), ≤ (less than or equal to) and ≥ (greater than or equal to) instead of the equality sign, where the expressions involved are of degree one (linear). For example, 3x - 5 < 7 and 2x + 3y ≤ 12 are linear inequalities in one and two variables respectively, while 5 < 7 is called a numerical inequality.

Inequalities are solved almost like equations, but two special rules must be remembered because they can change the direction (sense) of the inequality sign.

Rule 1 (Addition/Subtraction): Adding or subtracting the same number from both sides of an inequality does not change its sign. Example: if x > 3, then x + 5 > 8 (5 added to both sides).

Rule 2 (Multiplication/Division by a positive number): Multiplying or dividing both sides by the same positive number keeps the sign unchanged. Example: if x > 3, multiplying by 2 gives 2x > 6.

Rule 3 (Multiplication/Division by a negative number): Multiplying or dividing both sides by the same negative number reverses the sign.

Worked Example: Given x > 3, multiply both sides by -2. Since we multiply by a negative number, the sign flips: -2x < -6. Check with x = 4 (which satisfies x > 3): -2(4) = -8, and indeed -8 < -6, confirming the rule.

Because exactly one of the relations a < b, a = b, a > b holds for any two real numbers a and b, this is called the Law of Trichotomy, and it is why every linear inequality has a well-defined solution set.

An algebraic solution of a linear inequality in one variable follows the same steps as solving a linear equation — collect the variable terms on one side and the constants on the other — except that the inequality sign must be reversed whenever both sides are multiplied or divided by a negative number.

Worked Example: Solve 5x - 3 < 3x + 1 for (i) x a real number and (ii) x an integer.

• 5x - 3 < 3x + 1
• 5x - 3x < 1 + 3 (collect x-terms on the left, constants on the right)
• 2x < 4
• x < 2 (dividing both sides by the positive number 2, sign unchanged)

(i) If x is real, the solution set is {x : x < 2} = (-∞, 2).

(ii) If x is an integer, the solution set is {..., -3, -2, -1, 0, 1}, i.e. every integer strictly less than 2.

This shows the same inequality can have infinitely many real solutions but only a restricted (discrete) list of solutions when the domain is limited to integers or natural numbers.

The solution set of a linear inequality in one variable can be shown visually on a number line. A hollow (open) circle is used at a boundary point when it is not included in the solution (strict inequalities < or >), and a solid (filled) circle is used when the boundary point is included (≤ or ≥). An arrow is then drawn from the circle in the direction of all the numbers that satisfy the inequality.

Worked Example: Represent the solution of 3x + 6 ≥ 0, x ∈ R, on a number line.

• 3x + 6 ≥ 0
• 3x ≥ -6
• x ≥ -2 (divided by the positive number 3)

Since the inequality is ≥ (not strict), a solid circle is drawn at x = -2, and the arrow points to the right (towards larger values), because every real number greater than or equal to -2 satisfies the inequality.

If instead the inequality had been strict, x > -2, an open circle would be used at -2, with the arrow still pointing right, showing -2 itself is excluded.

When two (or more) linear inequalities in the same variable must hold simultaneously, they form a system of linear inequalities. Each inequality is solved separately, and the final solution is the intersection (common part) of all the individual solution sets — found by marking each solution on the same number line and reading off the overlap.

Worked Example: Solve the system 2x - 3 < 7 and 2x + 5 > 1, x ∈ R.

• First inequality: 2x - 3 < 7 ⟹ 2x < 10 ⟹ x < 5.
• Second inequality: 2x + 5 > 1 ⟹ 2x > -4 ⟹ x > -2.

Combining both, x must be greater than -2 and less than 5 at the same time, so the solution set is {x : -2 < x < 5} = (-2, 5). On the number line this is shown with open circles at -2 and 5 and shading between them.

Applied Worked Example: A pool's water temperature C (in °C) must satisfy 25 < C < 30. Using F = (9/5)C + 32, find the corresponding range of Fahrenheit temperature F.

• 25 < C < 30
• Multiply throughout by 9/5: 45 < (9/5)C < 54
• Add 32 throughout: 77 < (9/5)C + 32 < 86
• So 77 < F < 86, i.e. the Fahrenheit temperature lies strictly between 77°F and 86°F.

A linear inequality in two variables, such as ax + by ≤ c, is solved graphically. The equation ax + by = c is first drawn as a straight line; this line splits the entire coordinate plane into two half-planes. Exactly one of these half-planes (sometimes together with the line itself) forms the solution region.

Rule for the boundary line: draw it as a solid (unbroken) line if the inequality is ≤ or ≥ (points on the line satisfy it), and as a dashed (broken) line if the inequality is strict, < or > (points on the line do not satisfy it).

Rule for shading (test-point method): pick any convenient point not on the line — the origin (0, 0) is easiest when the line does not pass through it. Substitute this point into the inequality; if it is satisfied, shade the half-plane containing that point, otherwise shade the other half-plane.

Worked Example: Represent x + 2y ≤ 8 graphically.

Boundary line: x + 2y = 8, meeting the x-axis at (8, 0) [put y = 0] and the y-axis at (0, 4) [put x = 0]. Since the inequality is ≤, the line is drawn solid.

Test point (0, 0): substituting gives 0 + 2(0) = 0 ≤ 8, which is true. So the half-plane containing the origin is shaded — this shaded region together with the solid boundary line is the complete graph of the solution.

To solve a system of linear inequalities in two variables graphically, draw the boundary line for each inequality on the same coordinate axes and shade the correct half-plane for each (using the boundary-line and test-point rules). The region where all the shadings overlap is called the feasible region (or solution region) of the system.

Worked Example: Solve graphically: x + y ≤ 6, x ≥ 2, y ≥ 0.

• Step 1: x + y ≤ 6 → solid line through (6, 0) and (0, 6); test (0,0): 0 ≤ 6 true, so shade towards the origin.
• Step 2: x ≥ 2 → solid vertical line at x = 2; shade the region to its right.
• Step 3: y ≥ 0 → the region on or above the x-axis.

Step 4 (corner points of the feasible region):

• Intersection of x = 2 and y = 0 ⟹ (2, 0).
• Intersection of x = 2 and x + y = 6 ⟹ y = 4 ⟹ (2, 4).
• Intersection of y = 0 and x + y = 6 ⟹ (6, 0).

The feasible region is the triangular region with vertices (2, 0), (2, 4) and (6, 0). Checking an interior point, (3, 1): x ≥ 2 ✓, y ≥ 0 ✓, x + y = 4 ≤ 6 ✓, confirming it lies inside the feasible region.