Linear Inequalities for CA Foundation
Solving linear inequalities in one variable, graphing inequalities in two variables, feasible regions and the corner-point idea of linear programming.
Quantitative AptitudePaper 3, Business Mathematics12 MCQs2 lessons2 flashcards3 memory tricks
Practise Linear Inequalities free
What you'll learn
- Inequalities in one variable
Solve linear and double inequalities and show the solution on a number line. - Inequalities in two variables
Graph inequalities, find the feasible region and optimise a linear function.
Key formulas
| |x| < a | −a < x < a |
|---|---|
| |x| > a | x < −a or x > a |
Concepts explained
Solving linear inequalities
Solve like an equation, but reverse the sign whenever you multiply or divide both sides by a negative number.
What it means
- Adding/subtracting: sign unchanged.
- Multiplying/dividing by negative: sign reversed.
- Double inequality: same operation on all three parts.
- |x| < a ⇔ −a < x < a; |x| > a ⇔ x < −a or x > a.
- −2x > 6
Answer: x < −3
- 2 ≤ x + 1 < 5
Answer: 1 ≤ x < 4
Related terms
- Number line: open circle for <, filled for ≤
- Interval notation: [1, 4) for 1 ≤ x < 4
Exam tip: Check your answer by substituting one value from your solution set.
Graphing inequalities in two variables
Draw the boundary line, test a point to decide which side to shade, and take the overlap of all shadings as the feasible region.
What it means
- Solid line for ≤ or ≥; dashed for < or >.
- (0, 0) is the easiest test point if it isn't on the line.
- x ≥ 0, y ≥ 0 restrict to the first quadrant.
- Feasible region: all constraints satisfied together.
- 2x + 3y ≤ 6; test (0, 0): 0 ≤ 6 true
Answer: Shade the side containing the origin
- x ≥ 3
Answer: Line x = 3 and the right side
Related terms
- Half-plane: each linear inequality covers one side of a line
- Bounded region: enclosed feasible region
Exam tip: If the boundary passes through the origin, test (1, 0) or (0, 1) instead.
Linear programming by corner points
The maximum or minimum of a linear objective function over a bounded feasible region occurs at a corner point.
What it means
- List the corner points (intersections of boundary lines).
- Evaluate Z at each.
- Largest value is the maximum; smallest is the minimum.
- If two corners tie, every point on the edge between them is optimal.
- Z = 3x + 2y at (0,0), (4,0), (2,2), (0,3)
Answer: Maximum 12 at (4, 0)
- Maximise 5x + 4y; x + y ≤ 6, x ≤ 4
Answer: 28 at (4, 2)
Related terms
- Objective function: the quantity to maximise or minimise
- Constraints: the inequalities that limit choices
Exam tip: Find corners by solving the pairs of boundary equations simultaneously.
Memory tricks
Multiplying or dividing by a negative reverses the inequality.
≤, ≥ solid boundary; <, > dashed boundary.
Optimum of a linear objective occurs at a vertex of the feasible region.
Try 3 CA Foundation-style questions
Q1. The boundary of a strict inequality such as 2x + y < 6 is drawn as
- a solid line
- a shaded strip
- a single point
- a dashed (broken) line, because points on it are not included
Show answer
Answer: (4) a dashed (broken) line, because points on it are not included
For < or >, the boundary isn't part of the solution, so it's drawn dashed; for ≤ or ≥ it's drawn solid.
Q2. The solution of |x| < 3 is
- x < 3
- −3 < x < 3
- x > 3 or x < −3
- x > −3
Show answer
Answer: (2) −3 < x < 3
|x| < a means −a < x < a, so −3 < x < 3.
Q3. Which inequality is satisfied by the point (1, 2)?
- 2x + y ≥ 5
- x − y > 0
- x + y ≤ 4
- y < 1
Show answer
Answer: (3) x + y ≤ 4
Substitute: 1 + 2 = 3 ≤ 4 ✓; 2 + 2 = 4 is not ≥ 5; 1 − 2 = −1 is not > 0; 2 is not < 1.
Practise all 12 questions Free · explanations for every option · see where you go wrong