Cheatsheet: Linear Inequalities | Quantitative Aptitude for CA Foundation PDF Download

What is a Linear Inequality?

A linear inequality is similar to a linear equation, but instead of an equals sign (=), it uses inequality signs, such as <, >, <=, or >=. It represents a region of solutions rather than exact points.

Examples of Linear Inequalities

• One variable: x > 0, x ≤ 5
• Two variables: 3x + y ≤ 6, x - y ≥ -2

Graphing a Linear Inequality

1. Replace the inequality with an equality and draw the line.
2. Use a solid line for ≤ or ≥, and dashed line for < or >.
3. Test a point (usually the origin) to determine which side to shade.
4. The shaded area represents the solution space.

Example: Graph 3x + y < 6
Convert to equality: y = 6 - 3x
Plot (0,6) and (2,0); then shade below the line since y < 6 - 3x

Step 1: Convert to equality → 3x + y = 6 → y = 6 - 3x
Step 2: Plot points (0,6) and (2,0) to draw the dashed boundary line
Step 3: Since it’s ‘<’, shade below the line (strict inequality)
Note: Dashed line indicates the boundary is not included

System of Inequalities

A system of inequalities includes two or more inequalities. The solution set is the region where all shaded areas overlap.

Step 1: Start with the inequality: 3x + y < 6
Step 2: Convert it to equality to find the boundary line:
        3x + y = 6 → y = 6 - 3x
Step 3: Find two points on the line:
        When x = 0, y = 6 → (0, 6)
        When x = 2, y = 0 → (2, 0)
Step 4: Plot these points and draw a dashed line through them
        (dashed because the inequality is ‘<’, not ‘≤’)
Step 5: Shade the region below the line since the inequality is y < 6 - 3x
Final: The shaded area represents all solutions to the inequality.

Feasible Region & Optimal Solution

• The region satisfying all constraints is the feasible region.
• It is bounded if enclosed, and unbounded if open-ended.
• Extreme points (corners) of this region help find max/min of an objective function.
• To maximise Z = ax + by, evaluate Z at all corner points.

Objective Function & Matrix Method

Let matrix E contain corner points, and C be the coefficient vector of the objective function.

Multiply: EC to get function values. The max/Min value of EC gives the optimal point.

Solution: 

Goal: Maximize Z = x + 2y
Coefficient Vector: C = [1, 2]
Corner Points Matrix E:
 E = [[0, 0],
    [6, 0],
    [5, 7],
    [0, 7]]

Step 1: Multiply E and C to get Z values:
 Z = E × C = [0, 6, 19, 14]

Step 2: Identify the optimal value:
 Maximum Z = 19 at point (5, 7)

Optimal Solution: Max Z = 19 at (5, 7)

Word Problems to Inequalities

• Translate constraints into inequalities.
• Identify objective (maximise profit, minimise cost).
• Examples: resource limits, time restrictions, production mix.

Step 1: Define variables:
 Let x = units of A, y = units of B

Step 2: Write constraints from time usage:
 Machine M1: 2x + 6y ≤ 24
 Machine M2: 6x + 2y ≤ 24
 Also: x ≥ 0, y ≥ 0 (non-negativity)

Step 3: Objective Function:
 Maximize Z = 5x + 2y (profit)

Step 4: Graph the inequalities and identify feasible region
 (Area that satisfies all constraints)