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Linear Inequalities in Two Variables 

6.1 Linear Inequalities in Two Variables
MATHEMATICS • Form 4 • Chapter 6: Linear Inequalities in Two Variables

6.1 Linear Inequalities in Two Variables

Linear inequalities in two variables are used to describe relationships between two quantities that are not equal, such as budget allocations, maximum capacities, and vehicle dimensions. Unlike a linear equation whose solution lies solely on a line, a linear inequality has solutions that form an entire region on a Cartesian plane.

Learning Objectives

  • Represent real-life situations in the form of linear inequalities using suitable variables.
  • Make and verify conjectures about points lying on a line, in the region above, or in the region below a line and their relationship to linear inequalities.
  • Determine and shade regions on a Cartesian plane that satisfy linear inequalities involving \(y = mx + c\), horizontal lines \(y = h\), and vertical lines \(x = k\).

Regions of Linear Inequalities on the Cartesian Plane

Distinguish between solid and dashed boundary lines, and identify the correct region to shade for each inequality.

Solid line: \(\ge\) or \(\le\)

Points on the line are included in the solution set.

Above: \(y \ge mx + c\)
Below: \(y \le mx + c\)

Dashed line: \(>\) or \(<\)

Points on the line are not included in the solution set.

Above: \(y > mx + c\)
Below: \(y < mx + c\)

Vertical line \(x = k\)

Right: \(x > k\) or \(x \ge k\)
Left: \(x < k\) or \(x \le k\)

Horizontal line \(y = h\)

Above: \(y > h\) or \(y \ge h\)
Below: \(y < h\) or \(y \le h\)

Brief Explanation

What is a linear inequality in two variables?

A linear inequality in two variables defines a set of ordered pairs \((x, y)\) that satisfy an inequality statement. The highest power of each variable is 1.

How do we graph it?

First draw the boundary line corresponding to the linear equation. Use a dashed line for strict inequalities and a solid line for inclusive inequalities. Then identify the correct region to shade.

Line type Equation Region above / right Region below / left
Oblique line \(y = mx + c\) \(y > mx + c\) or \(y \ge mx + c\) \(y < mx + c\) or \(y \le mx + c\)
Horizontal line \(y = h\) \(y > h\) or \(y \ge h\) \(y < h\) or \(y \le h\)
Vertical line \(x = k\) \(x > k\) or \(x \ge k\) \(x < k\) or \(x \le k\)
1
Draw the boundary line
Draw the line for the corresponding equation. Use a dashed line for \(>\) or \(<\), and a solid line for \(\ge\) or \(\le\).
2
Identify the correct region
For \(y = mx + c\) or \(y = h\), shade above for “greater than” and below for “less than”. For \(x = k\), shade right for “greater than” and left for “less than”.
3
Check the \(y\)-coefficient
If the coefficient of \(y\) is negative, rearrange the inequality so that the coefficient of \(y\) is positive before choosing the region.

Try Answering First

Read the question, answer it in your mind, then press “Check Answer”.

1

A chick costs RM2 and a duckling costs RM1.50. Encik Halim has a capital of RM500 to purchase \(x\) chicks and \(y\) ducklings. Write a linear inequality representing this situation.

Answer: \(2x + 1.50y \le 500\), or after multiplying by 2: \(4x + 3y \le 1\,000\).
2

Determine algebraically whether the point \((1, 2)\) satisfies \(y = -3x + 6\), \(y > -3x + 6\), or \(y < -3x + 6\).

Answer: Substitute \(x = 1\): \(-3(1) + 6 = 3\). Since \(y = 2\) and \(2 < 3\), the point \((1, 2)\) satisfies \(y < -3x + 6\).
3

State whether the boundary line for the inequality \(x < 4\) should be drawn as a solid line or a dashed line, and specify which side of the line should be shaded.

Answer: The boundary line is drawn as a dashed line at \(x = 4\), and the region on the left of the line should be shaded.

Shading Inequality Regions

Three complete graphs with test points, including the negative \(y\)-coefficient case.

Shading the region of a linear inequality

Three steps: (1) change the inequality into a linear equation and draw that straight line; (2) choose the line type — a solid line for \(\leqslant\) or \(\geqslant\) (points on the line are included), a dashed line for \(<\) or \(>\) (points on the line are excluded); (3) choose a test point and substitute it into the inequality to decide which region to shade.

-4 -2 2 4 -4 -2 2 4 6 8 O x y (4, 0) y ≤ 2x + 3

Example 1: \(y \leqslant 2x + 3\). The line \(y = 2x + 3\) passes through \((-2, -1)\) and \((2, 7)\). Because the sign is \(\leqslant\), the line is solid. Test the point \((4, 0)\): \(2(4) + 3 = 11\) and \(0 \leqslant 11\) is True — so the region on that side is shaded.

-4 -2 2 4 -4 -2 2 4 6 8 O x y (−4, 5) y > x + 5

Example 2: \(y > x + 5\). The line \(y = x + 5\) passes through \((0, 5)\) and \((-5, 0)\). Because the sign is \(>\), the line is dashed. Test the point \((-4, 5)\): \(-4 + 5 = 1\) and \(5 > 1\) is True — the upper region is shaded.

When the coefficient of \(y\) is negative

If the coefficient of \(y\) is negative, make it positive first so that choosing the region is easier. Remember: when both sides are multiplied by a negative number, the inequality sign reverses.

1
Start from the original inequality
\(x - y > 4\)
2
Multiply both sides by \(-1\) and reverse the sign
\(-x + y < -4\)
3
Make \(y\) the subject
\(y < x - 4\). Because the sign is \(<\), the line \(y = x - 4\) is drawn dashed and the region below the line is shaded.
4
Confirm with a test point
Test \((4, -4)\) in the original form: \(x - y = 4 - (-4) = 8\) and \(8 > 4\) is True. Test it in the new form too: \(4 - 4 = 0\) and \(-4 < 0\) is True. Both forms give the same answer.
-4 -2 2 4 -4 -2 2 4 6 8 O x y (4, −4) y < x − 4

Example 3: \(x - y > 4\), that is \(y < x - 4\). The dashed line passes through \((0, -4)\) and \((4, 0)\); the region below and to the right is shaded.

Common Mistakes

  • ! Using a solid line instead of a dashed line for strict inequalities (\(>\) and \(<\)).
  • ! Forgetting to reverse the inequality sign when multiplying or dividing both sides by a negative number.
  • ! Misidentifying the “above” and “below” regions when a straight line has a negative gradient (\(m < 0\)).
  • ! Confusing vertical line regions (\(x = k\)) by attempting to shade “above” or “below” instead of right (\(>\)) or left (\(<\)).

Misconceptions

Misconception

A point lying on a dashed boundary line is part of the inequality's solution.

Correct Fact

A dashed line indicates that points lying on the line do not satisfy the strict inequality (\(>\) or \(<\)).

Misconception

The gradient of a vertical line \(x = k\) is zero.

Correct Fact

The gradient of a horizontal line \(y = h\) is zero, whereas the gradient of a vertical line \(x = k\) is undefined.

Misconception

The inequality sign “\(\le\)” means strictly “less than”.

Correct Fact

The symbol “\(\le\)” means “less than or equal to” (or “at most”), meaning points on the boundary line are included in the solution.

Summary

  • A linear inequality in two variables models constraints where quantities are not strictly equal, with variable powers equal to 1.
  • A solid boundary line represents \(\le\) or \(\ge\), including points on the line.
  • A dashed boundary line represents \(<\) or \(>\), excluding points on the line.
  • For \(y = mx + c\) and \(y = h\): the region above satisfies \(>\) or \(\ge\); the region below satisfies \(<\) or \(\le\).
  • For \(x = k\): the region to the right satisfies \(>\) or \(\ge\); the region to the left satisfies \(<\) or \(\le\).
  • To determine the region algebraically or graphically for forms like \(ax + by \le c\), ensure the \(y\)-coefficient is positive before choosing above or below.

Short Activity

Complete all parts first, then press the check answer button.

A. Objective Quiz

1 Which type of boundary line must be used when graphing the inequality \(y > 2x + 3\) on a Cartesian plane?

2 Given the linear equation \(y = -2x + 6\), which inequality does the point \((2, 6)\) satisfy?

B. Fill in the Blanks

3 For a vertical line \(x = k\), points located in the region on the of the straight line satisfy the inequality \(x > k\).

4 The gradient of the horizontal straight line \(y = h\) is equal to .

C. Matching: Inequality Statement to Graphical Property

Match each inequality statement to its correct graphical representation or boundary property. On a phone, tap the answer choice first, then tap the matching box.

Inequality Statement
\(y \ge mx + c\)
\(y < mx + c\)
\(x < k\)
\(x \ge k\)
Graphical Property
Region above a solid straight lineCorrectTry againEmpty
Region below a dashed straight lineCorrectTry againEmpty
Region to the left of a dashed vertical lineCorrectTry againEmpty
Region to the right of a solid vertical lineCorrectTry againEmpty
 

Key Terms

Linear inequality Dashed line Solid line Region above Region below Region on the right Region on the left
The views expressed in the notes here are those of the contributor and don’t necessarily represent the views of Pandai.

Linear Inequalities in Two Variables 

6.1 Linear Inequalities in Two Variables
MATHEMATICS • Form 4 • Chapter 6: Linear Inequalities in Two Variables

6.1 Linear Inequalities in Two Variables

Linear inequalities in two variables are used to describe relationships between two quantities that are not equal, such as budget allocations, maximum capacities, and vehicle dimensions. Unlike a linear equation whose solution lies solely on a line, a linear inequality has solutions that form an entire region on a Cartesian plane.

Learning Objectives

  • Represent real-life situations in the form of linear inequalities using suitable variables.
  • Make and verify conjectures about points lying on a line, in the region above, or in the region below a line and their relationship to linear inequalities.
  • Determine and shade regions on a Cartesian plane that satisfy linear inequalities involving \(y = mx + c\), horizontal lines \(y = h\), and vertical lines \(x = k\).

Regions of Linear Inequalities on the Cartesian Plane

Distinguish between solid and dashed boundary lines, and identify the correct region to shade for each inequality.

Solid line: \(\ge\) or \(\le\)

Points on the line are included in the solution set.

Above: \(y \ge mx + c\)
Below: \(y \le mx + c\)

Dashed line: \(>\) or \(<\)

Points on the line are not included in the solution set.

Above: \(y > mx + c\)
Below: \(y < mx + c\)

Vertical line \(x = k\)

Right: \(x > k\) or \(x \ge k\)
Left: \(x < k\) or \(x \le k\)

Horizontal line \(y = h\)

Above: \(y > h\) or \(y \ge h\)
Below: \(y < h\) or \(y \le h\)

Brief Explanation

What is a linear inequality in two variables?

A linear inequality in two variables defines a set of ordered pairs \((x, y)\) that satisfy an inequality statement. The highest power of each variable is 1.

How do we graph it?

First draw the boundary line corresponding to the linear equation. Use a dashed line for strict inequalities and a solid line for inclusive inequalities. Then identify the correct region to shade.

Line type Equation Region above / right Region below / left
Oblique line \(y = mx + c\) \(y > mx + c\) or \(y \ge mx + c\) \(y < mx + c\) or \(y \le mx + c\)
Horizontal line \(y = h\) \(y > h\) or \(y \ge h\) \(y < h\) or \(y \le h\)
Vertical line \(x = k\) \(x > k\) or \(x \ge k\) \(x < k\) or \(x \le k\)
1
Draw the boundary line
Draw the line for the corresponding equation. Use a dashed line for \(>\) or \(<\), and a solid line for \(\ge\) or \(\le\).
2
Identify the correct region
For \(y = mx + c\) or \(y = h\), shade above for “greater than” and below for “less than”. For \(x = k\), shade right for “greater than” and left for “less than”.
3
Check the \(y\)-coefficient
If the coefficient of \(y\) is negative, rearrange the inequality so that the coefficient of \(y\) is positive before choosing the region.

Try Answering First

Read the question, answer it in your mind, then press “Check Answer”.

1

A chick costs RM2 and a duckling costs RM1.50. Encik Halim has a capital of RM500 to purchase \(x\) chicks and \(y\) ducklings. Write a linear inequality representing this situation.

Answer: \(2x + 1.50y \le 500\), or after multiplying by 2: \(4x + 3y \le 1\,000\).
2

Determine algebraically whether the point \((1, 2)\) satisfies \(y = -3x + 6\), \(y > -3x + 6\), or \(y < -3x + 6\).

Answer: Substitute \(x = 1\): \(-3(1) + 6 = 3\). Since \(y = 2\) and \(2 < 3\), the point \((1, 2)\) satisfies \(y < -3x + 6\).
3

State whether the boundary line for the inequality \(x < 4\) should be drawn as a solid line or a dashed line, and specify which side of the line should be shaded.

Answer: The boundary line is drawn as a dashed line at \(x = 4\), and the region on the left of the line should be shaded.

Shading Inequality Regions

Three complete graphs with test points, including the negative \(y\)-coefficient case.

Shading the region of a linear inequality

Three steps: (1) change the inequality into a linear equation and draw that straight line; (2) choose the line type — a solid line for \(\leqslant\) or \(\geqslant\) (points on the line are included), a dashed line for \(<\) or \(>\) (points on the line are excluded); (3) choose a test point and substitute it into the inequality to decide which region to shade.

-4 -2 2 4 -4 -2 2 4 6 8 O x y (4, 0) y ≤ 2x + 3

Example 1: \(y \leqslant 2x + 3\). The line \(y = 2x + 3\) passes through \((-2, -1)\) and \((2, 7)\). Because the sign is \(\leqslant\), the line is solid. Test the point \((4, 0)\): \(2(4) + 3 = 11\) and \(0 \leqslant 11\) is True — so the region on that side is shaded.

-4 -2 2 4 -4 -2 2 4 6 8 O x y (−4, 5) y > x + 5

Example 2: \(y > x + 5\). The line \(y = x + 5\) passes through \((0, 5)\) and \((-5, 0)\). Because the sign is \(>\), the line is dashed. Test the point \((-4, 5)\): \(-4 + 5 = 1\) and \(5 > 1\) is True — the upper region is shaded.

When the coefficient of \(y\) is negative

If the coefficient of \(y\) is negative, make it positive first so that choosing the region is easier. Remember: when both sides are multiplied by a negative number, the inequality sign reverses.

1
Start from the original inequality
\(x - y > 4\)
2
Multiply both sides by \(-1\) and reverse the sign
\(-x + y < -4\)
3
Make \(y\) the subject
\(y < x - 4\). Because the sign is \(<\), the line \(y = x - 4\) is drawn dashed and the region below the line is shaded.
4
Confirm with a test point
Test \((4, -4)\) in the original form: \(x - y = 4 - (-4) = 8\) and \(8 > 4\) is True. Test it in the new form too: \(4 - 4 = 0\) and \(-4 < 0\) is True. Both forms give the same answer.
-4 -2 2 4 -4 -2 2 4 6 8 O x y (4, −4) y < x − 4

Example 3: \(x - y > 4\), that is \(y < x - 4\). The dashed line passes through \((0, -4)\) and \((4, 0)\); the region below and to the right is shaded.

Common Mistakes

  • ! Using a solid line instead of a dashed line for strict inequalities (\(>\) and \(<\)).
  • ! Forgetting to reverse the inequality sign when multiplying or dividing both sides by a negative number.
  • ! Misidentifying the “above” and “below” regions when a straight line has a negative gradient (\(m < 0\)).
  • ! Confusing vertical line regions (\(x = k\)) by attempting to shade “above” or “below” instead of right (\(>\)) or left (\(<\)).

Misconceptions

Misconception

A point lying on a dashed boundary line is part of the inequality's solution.

Correct Fact

A dashed line indicates that points lying on the line do not satisfy the strict inequality (\(>\) or \(<\)).

Misconception

The gradient of a vertical line \(x = k\) is zero.

Correct Fact

The gradient of a horizontal line \(y = h\) is zero, whereas the gradient of a vertical line \(x = k\) is undefined.

Misconception

The inequality sign “\(\le\)” means strictly “less than”.

Correct Fact

The symbol “\(\le\)” means “less than or equal to” (or “at most”), meaning points on the boundary line are included in the solution.

Summary

  • A linear inequality in two variables models constraints where quantities are not strictly equal, with variable powers equal to 1.
  • A solid boundary line represents \(\le\) or \(\ge\), including points on the line.
  • A dashed boundary line represents \(<\) or \(>\), excluding points on the line.
  • For \(y = mx + c\) and \(y = h\): the region above satisfies \(>\) or \(\ge\); the region below satisfies \(<\) or \(\le\).
  • For \(x = k\): the region to the right satisfies \(>\) or \(\ge\); the region to the left satisfies \(<\) or \(\le\).
  • To determine the region algebraically or graphically for forms like \(ax + by \le c\), ensure the \(y\)-coefficient is positive before choosing above or below.

Short Activity

Complete all parts first, then press the check answer button.

A. Objective Quiz

1 Which type of boundary line must be used when graphing the inequality \(y > 2x + 3\) on a Cartesian plane?

2 Given the linear equation \(y = -2x + 6\), which inequality does the point \((2, 6)\) satisfy?

B. Fill in the Blanks

3 For a vertical line \(x = k\), points located in the region on the of the straight line satisfy the inequality \(x > k\).

4 The gradient of the horizontal straight line \(y = h\) is equal to .

C. Matching: Inequality Statement to Graphical Property

Match each inequality statement to its correct graphical representation or boundary property. On a phone, tap the answer choice first, then tap the matching box.

Inequality Statement
\(y \ge mx + c\)
\(y < mx + c\)
\(x < k\)
\(x \ge k\)
Graphical Property
Region above a solid straight lineCorrectTry againEmpty
Region below a dashed straight lineCorrectTry againEmpty
Region to the left of a dashed vertical lineCorrectTry againEmpty
Region to the right of a solid vertical lineCorrectTry againEmpty
 

Key Terms

Linear inequality Dashed line Solid line Region above Region below Region on the right Region on the left
The views expressed in the notes here are those of the contributor and don’t necessarily represent the views of Pandai.
Chapter : Linear Inequalities in Two Variables
Topic : Linear Inequalities in Two Variables
Form 4 Mathematics
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