Quantitative Aptitude · Linear Inequalities

System of Linear Inequalities: Graphical Method for CA Foundation

Updated 1 October 2026 · Fact-checked

A system of linear inequalities is a set of two or more inequalities in x and y that must hold together. Draw each boundary line, shade the side that satisfies each inequality, and find the overlap. That common area is the feasible region. Find its corner points and test a point to confirm.

Understand System of Linear Inequalities

A single linear inequality in two variables, like x + y ≤ 6, has many solutions. On a graph they fill one side of a line, called a half-plane. The line itself is the boundary line.

A system has two or more such inequalities. A point is a solution only if it satisfies every inequality at once. So you are looking for the overlap of all the half-planes.

That overlap is the feasible region (also called the solution region or common region). It can be a bounded polygon, an unbounded area, a single point, or empty. If the half-planes never overlap, the system has no solution.

The corners of the feasible region are called corner points or vertices. Each one is where two boundary lines meet. You find them by solving two equations together. Many MCQs ask you to identify a point in the region, count corner points, or find a vertex.

Conditions like x ≥ 0 and y ≥ 0 are common. They restrict the region to the first quadrant, which makes the graph much smaller and easier to read.

Key formulas to remember

Boundary line: ax + by ≤ c → boundary line: ax + by = c; Replace the inequality sign with = to get the line you draw.
Line style: ≤ or ≥: solid line. < or >: dashed line; A solid line means points on the line are included. A dashed line means they are not.
Test-point rule: Put (0, 0) into the inequality. True: shade the side with the origin. False: shade the other side.; Works only if the line does not pass through the origin. If it does, test another point such as (1, 0) or (0, 1).
Feasible region: Feasible region = intersection of all the shaded half-planes; A point is in the region only if it satisfies every inequality.
Corner point: Solve the two boundary equations together; Check the result in all other inequalities. A real corner point must satisfy them all.
Axis conditions: x ≥ 0 means right of or on the y-axis. y ≥ 0 means above or on the x-axis.; Together they confine the region to the first quadrant.

How to solve System of Linear Inequalities questions

Use this method for any graphical question on a system of linear inequalities. It works for both identify-the-region and find-the-vertex questions.

1Write each inequality separately. Replace the sign with = to get its boundary line.
2Find two points on each line. Putting x = 0 and then y = 0 gives the intercepts quickly.
3Draw each line. Use solid for ≤ or ≥ and dashed for < or >.
4Pick a test point, usually (0, 0), for each inequality. Shade the side where the inequality is true.
5Mark the area shaded by all the inequalities. This overlap is the feasible region. If there is no overlap, the system has no solution.
6Find the corner points by solving the pairs of boundary equations that meet at the edge of the region.
7Check each corner point in all the other inequalities. Keep only those that satisfy every one.
8Answer what is asked: a point in the region, a vertex, the shape or the area.

Quickest way: Option testing and intercepts

When to use it: Use this in the MCQ paper when options are points or vertices. It is faster than drawing a full graph.

If the options are points, put each point into every inequality. Reject an option as soon as one inequality fails.
Check the simple conditions first, such as x ≥ 0 and y ≥ 0. They remove options with negative coordinates quickly.
If you must find a vertex, solve just the two relevant boundary equations by elimination.
Then test that vertex in the remaining inequalities. A solution that breaks one is not a corner of the region.
For a rough sketch, use intercepts only. Plot (x, 0) and (0, y) for each line and join them.
Watch for contradictions such as x + y ≥ 5 and x + y ≤ 3. If you spot one, the answer is no feasible region. Skip the drawing.
If the question needs a long graph with many lines and you are short of time, mark it and return later. A wrong answer costs 0.25 marks.

Common mistakes in System of Linear Inequalities

Shading the wrong side of a line
Students guess the side from the sign instead of testing a point. This goes wrong when the line has a negative slope or a negative coefficient.
Fix: Always test (0, 0) when it is not on the line. If it makes the inequality true, shade its side. Otherwise shade the opposite side.
Testing a point on the boundary line
If the line passes through the origin, (0, 0) gives 0 on both sides and tells you nothing about the side.
Fix: When the line passes through the origin, use a point such as (1, 0) or (0, 1) instead.
Accepting a vertex without checking all inequalities
Two lines always meet at a point, but that point may lie outside the region set by the other inequalities.
Fix: Put every candidate vertex into all the inequalities. Only points that satisfy all of them are corner points.
Taking the union of the shaded areas
Students count any shaded area as part of the answer after drawing several half-planes.
Fix: The solution must be shaded by every inequality. Take only the overlap.
Ignoring x ≥ 0 and y ≥ 0
These conditions are written in one line and feel minor, so students draw the region in all four quadrants.
Fix: Underline them first. Restrict your graph to the first quadrant unless the question says otherwise.
Using a solid line for strict inequalities
Students forget that < and > do not include the boundary.
Fix: Dashed for < and >, solid for ≤ and ≥. In MCQs this decides whether a boundary point is a valid answer.

Worked examples

Example 1

Which of the following points lies in the feasible region of the system x ≥ 0, y ≥ 0, x + y ≤ 6, 2x + y ≥ 4? Options: (A) (1, 1) (B) (2, 3) (C) (4, 3) (D) (−1, 2)

Show the solution

Test (1, 1): x + y = 2 ≤ 6 is true. But 2x + y = 3 ≥ 4 is false. Reject A.
Test (2, 3): x ≥ 0 and y ≥ 0 are true. x + y = 5 ≤ 6 is true. 2x + y = 7 ≥ 4 is true. All hold.
Test (4, 3): x + y = 7 ≤ 6 is false. Reject C.
Test (−1, 2): x ≥ 0 is false. Reject D.

Answer: (B) (2, 3)

Example 2

For the system x ≥ 0, y ≥ 0, x + y ≤ 4, x + 2y ≤ 6, which of these is a corner point of the feasible region that does not lie on either axis? Options: (A) (3, 1) (B) (1, 3) (C) (2, 2) (D) (4, 2)

Show the solution

A corner point off the axes is where the lines x + y = 4 and x + 2y = 6 meet.
Subtract the first equation from the second: y = 2.
Then x = 4 − 2 = 2. The point is (2, 2).
Check the other inequalities: x ≥ 0 and y ≥ 0 hold. Both lines pass through (2, 2) exactly, so it satisfies both.
Check the other options. (3, 1): x + 2y = 5, so it lies on x + y = 4 only and is not a meeting point. (1, 3): x + 2y = 7 > 6, so it is outside the region. (4, 2): x + y = 6 > 4, so it is outside the region.

Answer: (C) (2, 2)

Example 3

What does the solution of the system x ≥ 0, y ≥ 0, x + y ≥ 5, x + y ≤ 3 look like on the graph? Options: (A) A bounded triangle (B) An unbounded region (C) No feasible region (D) A single point

Show the solution

The third inequality says x + y must be at least 5.
The fourth says x + y must be at most 3.
No number can be both ≥ 5 and ≤ 3, so no point satisfies both.
On the graph, the half-plane above x + y = 5 and the half-plane below x + y = 3 are separated by parallel lines and never overlap.
Adding x ≥ 0 and y ≥ 0 cannot create an overlap.

Answer: (C) No feasible region

Exam tips

Most questions give points or vertices as options. Test options in the inequalities before you draw anything.
Reject options using x ≥ 0 and y ≥ 0 first. This eliminates negative coordinates in seconds.
Read each sign carefully. A swapped ≤ and ≥ gives a different region and a tempting wrong option.
Look for contradictory inequalities early. They give a quick no-solution answer.
If a graph-based question needs more than three lines, leave it for the end. Attempt only if you have time and are fairly sure, because each wrong answer costs 0.25 marks.

Practice questions from Linear Inequalities

System of Linear Inequalities: frequently asked questions

How do I find the common region of inequalities?

Draw each boundary line and shade the side that satisfies its inequality using a test point. The area shaded by all the inequalities is the common region. If no area is shared, there is no solution.

Which test point should I use for shading?

Use (0, 0) whenever the line does not pass through the origin. If the inequality is true at (0, 0), shade the side with the origin. If the line passes through the origin, test (1, 0) or (0, 1) instead.

How do I find the corner points of the feasible region?

Solve the equations of two boundary lines together to get their meeting point. Then check that point in all the other inequalities. Only points that satisfy every inequality are corner points.

Can a system of linear inequalities have no solution?

Yes. If the half-planes do not overlap, the feasible region is empty. For example, x + y ≥ 5 and x + y ≤ 3 cannot both be true for any point.

Do I need to draw the graph in the exam?

The CA Foundation Quantitative Aptitude paper is objective, so you are not marked on a drawing. Testing options in the inequalities is often faster. Sketch only when the question needs the shape or the corner points.