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Quantitative Aptitude · Linear Inequalities

Linear Inequalities in Two Variables: Graphing and Shading the Region

Updated 1 October 2026 · Fact-checked

A linear inequality in two variables, like ax + by < c, is solved by drawing the line ax + by = c and shading the half-plane that satisfies it. Draw the line solid for ≤ or ≥ and dashed for < or >. Pick a test point, usually (0, 0), to decide which side to shade.

Understand Linear Inequalities in Two Variables

An equation like x + y = 4 gives a straight line. Every point on that line satisfies the equation. An inequality like x + y < 4 is different. It is satisfied by a whole region of points, not just a line.

The line ax + by = c is called the boundary line. It splits the plane into two half-planes. All points on one side satisfy ax + by < c. All points on the other side satisfy ax + by > c. The line itself satisfies ax + by = c.

So your job has two parts. First, draw the boundary line. Second, decide which side to shade. The test point method does the second part. Pick any point not on the line, put its coordinates in the inequality, and check if it is true. If true, shade the side containing that point. If false, shade the other side.

The type of inequality sign decides the line style. With ≤ or ≥, points on the line are included, so the line is solid. With < or >, points on the line are not included, so the line is dashed.

In MCQs you rarely need to draw anything. You usually check which option matches a given point or region. The same test-point idea does all the work.

Key formulas to remember

Boundary line
Replace the sign in ax + by (<, >, ≤, ≥) c with = to get ax + by = c
Draw this line first. It separates the two half-planes.
Line style
≤ or ≥: solid line. < or >: dashed line
Solid means points on the line are solutions. Dashed means they are not.
Test point rule
Put (x₀, y₀) not on the line into the inequality. True: shade its side. False: shade the opposite side.
If c ≠ 0, (0, 0) is the easiest test point. If c = 0, the line passes through the origin, so do not use (0, 0).
Intercepts of the line
x-intercept = c ÷ a (put y = 0). y-intercept = c ÷ b (put x = 0).
Use for a ≠ 0, b ≠ 0 and c ≠ 0. Then two intercepts are enough to draw the line. If c = 0, both intercepts are the origin, so find another point, such as the one at x = 1.
Special lines
x = k is a vertical line. y = k is a horizontal line.
x > k shades to the right of x = k. y > k shades above y = k.

How to solve Linear Inequalities in Two Variables questions

Use this method for any question that asks you to graph, shade or identify the solution region of a linear inequality in two variables.

  1. 1Replace the inequality sign with = to get the boundary line.
  2. 2Find two points on the line. If c ≠ 0, use the x-intercept (put y = 0) and the y-intercept (put x = 0). If c = 0, the line passes through the origin, so use (0, 0) and one more point, such as the one at x = 1.
  3. 3Decide the line style: solid for ≤ or ≥, dashed for < or >.
  4. 4Choose a test point not on the line. Use (0, 0) if c ≠ 0, that is, if the line does not pass through the origin.
  5. 5Substitute the test point into the original inequality.
  6. 6If the statement is true, shade the side containing the test point. If false, shade the other side.
  7. 7For an MCQ, check the given point or region against the inequality, and verify with one more point if unsure.

Quickest way: Test-point elimination for MCQs

When to use it: Use when options list points, regions or inequalities and you must pick the correct one. No drawing is needed.

  1. Substitute each given point directly into the inequality. Keep only points that make it true.
  2. Check boundary points carefully. A point on the line satisfies ≤ and ≥ but not < or >.
  3. For region questions, test (0, 0) first. Its truth value tells you whether the origin is in the shaded region.
  4. For 'which inequality matches this graph' questions, test one point in the shaded area and one on each side of the line.
  5. If the line passes through the origin, test a point like (1, 0) or (0, 1) instead.
  6. Skip a question if the graph is hard to read after one attempt. Wrong answers cost 0.25 marks.

Common mistakes in Linear Inequalities in Two Variables

  • Using a solid line for a strict inequality, or a dashed line for ≤ or ≥.

    Students forget that the line style shows whether boundary points are included.

    Fix: Remember: the 'equal to' bar in ≤ and ≥ means the line is included, so draw it solid.

  • Using (0, 0) as the test point when the line passes through the origin.

    Students use (0, 0) by habit without checking the line.

    Fix: If c = 0, the origin is on the line. Use another point such as (1, 0) or (0, 1).

  • Shading the wrong side after substituting the test point.

    Students rush and mix up 'true' and 'false'.

    Fix: Write the result: 'true, shade the origin side' or 'false, shade the other side' before shading.

  • Treating x > k as a horizontal line and y > k as a vertical line.

    Students mix up which variable is fixed.

    Fix: x = k is vertical because every point has the same x. y = k is horizontal.

  • Making arithmetic errors in intercepts, such as using c ÷ a for the y-intercept.

    Students memorise a formula without setting the other variable to zero.

    Fix: Always set one variable to zero and solve. For 3x + 2y = 6, x = 0 gives y = 3, and y = 0 gives x = 2.

Worked examples

Example 1

Which of the following points lies in the solution region of 2x + 3y ≤ 12? (A) (6, 2) (B) (3, 3) (C) (3, 1) (D) (4, 3)

Show the solution
  1. Test each point in 2x + 3y and compare with 12.
  2. (A) (6, 2): 12 + 6 = 18, which is greater than 12. Not valid.
  3. (B) (3, 3): 6 + 9 = 15, which is greater than 12. Not valid.
  4. (C) (3, 1): 6 + 3 = 9, which is at most 12. Valid.
  5. (D) (4, 3): 8 + 9 = 17, which is greater than 12. Not valid.

Answer: (C) (3, 1)

Example 2

The origin (0, 0) lies in the solution region of which inequality? (A) x + y > 5 (B) 2x − y ≥ 3 (C) 3x + 4y < 12 (D) x − 2y > 1

Show the solution
  1. Substitute x = 0, y = 0 into each inequality.
  2. (A) 0 > 5 is false.
  3. (B) 0 ≥ 3 is false.
  4. (C) 0 < 12 is true.
  5. (D) 0 > 1 is false.

Answer: (C) 3x + 4y < 12

Example 3

The solution region of x + 2y ≥ 6 is shaded. Which of the following points is in the region? (A) (0, 0) (B) (1, 1) (C) (2, 3) (D) (1, 2)

Show the solution
  1. The boundary line is x + 2y = 6, with intercepts (6, 0) and (0, 3).
  2. Test (0, 0): 0 ≥ 6 is false. So the region is on the side away from the origin.
  3. (1, 1): 1 + 2 = 3, which is less than 6. Not in region.
  4. (2, 3): 2 + 6 = 8, which is at least 6. In region.
  5. (1, 2): 1 + 4 = 5, which is less than 6. Not in region.

Answer: (C) (2, 3)

Exam tips

  • Most MCQs ask which point or inequality fits a region. Substitute values instead of drawing the graph.
  • Watch the sign carefully. A point on the boundary line is valid for ≤ and ≥ but not for < or >.
  • Check all four options quickly with simple arithmetic. Two wrong options usually fail on the first test.
  • If the line passes through the origin, do not trust (0, 0) as a test point.
  • Combine this topic with systems of inequalities. The feasible region is where all shaded regions overlap.

Practice questions from Linear Inequalities

Linear Inequalities in Two Variables: frequently asked questions

How do I know which side to shade for ax + by < c?

Use a test point. Put (0, 0) in the inequality if c ≠ 0, that is, if the line does not pass through the origin. If the statement is true, shade the side with the origin. If false, shade the other side.

When is the boundary line dashed?

Draw it dashed for strict inequalities, which are < and >. Points on the line do not satisfy these. For ≤ and ≥, draw it solid.

What if the line passes through the origin?

Then c = 0 and (0, 0) lies on the line, so it cannot be your test point. Choose another point, such as (1, 0) or (0, 1), that is clearly off the line. To draw the line, the two-intercept method fails because both intercepts are the origin. Use (0, 0) and one more point, such as the one at x = 1.

Do I need to draw the graph in the MCQ exam?

Often not. You can substitute the given points or test points directly. Rough sketching helps only for region-identification questions, and it should be quick.