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Systems of Linear Inequalities in Two Variables

6.2 Systems of Linear Inequalities in Two Variables
MATHEMATICS • Form 4 • Chapter 6: Linear Inequalities in Two Variables

6.2 Systems of Linear Inequalities in Two Variables

A system of linear inequalities consists of a combination of two or more linear inequalities involving the same variables. In real-life scenarios such as transportation planning, production, and budgeting, decisions are constrained by multiple simultaneous conditions. The solution is the set of all coordinate points \((x, y)\) that satisfy every inequality in the system simultaneously, represented graphically as an overlapping common region.

Learning Objectives

  • Represent real-world situations and constraints in the form of a system of linear inequalities.
  • Make and verify conjectures about points inside and outside regions that satisfy a system of linear inequalities.
  • Determine, construct boundary lines for, and shade the common region that satisfies a system of linear inequalities.
  • Identify the system of linear inequalities that defines a given shaded region on a Cartesian plane.
  • Solve real-world contextual problems using graphical solutions of systems of linear inequalities.

Identifying Common Feasible Regions for Systems of Linear Inequalities

Overlapping shaded half-planes define the common solution set. Solid lines include boundary points; dashed lines exclude them.

2 4 6 1 2 3 4 0 x y x + y = 4 2y = x x = 0 Common region

The shaded region satisfies \(2y \ge x\), \(x + y < 4\) and \(x \ge 0\). The dashed line \(x + y = 4\) is excluded from the region; the solid lines \(2y = x\) and \(x = 0\) are included.

Brief Explanation

What is a System of Linear Inequalities?

A system of linear inequalities combines several individual inequalities. Instead of having an infinite single half-plane as a solution, the solution is the common region where all constraints are satisfied at the same time.

Non-negativity Constraints

When variables \(x\) and \(y\) represent quantities of physical items or people, non-negativity constraints apply: \(x \ge 0\) and \(y \ge 0\). Graphically, this restricts solutions to the first quadrant.

Example of situation Linear inequality
\(y\) is greater than \(x\)\(y > x\)
\(y\) is less than \(x\)\(y < x\)
\(y\) is not less than \(x\)\(y \ge x\)
\(y\) is not more than \(x\)\(y \le x\)
\(y\) is at least \(k\) times \(x\)\(y \ge kx\)
\(y\) is at most \(k\) times \(x\)\(y \le kx\)
Maximum of \(y\) is \(k\)\(y \le k\)
Minimum of \(y\) is \(k\)\(y \ge k\)
Sum of \(x\) and \(y\) is greater than \(k\)\(x + y > k\)
Difference between \(y\) and \(x\) is less than \(k\)\(y - x < k\)
\(y\) is more than \(x\) by at least \(k\)\(y - x \ge k\)
1
Convert inequalities to linear equations
Write each inequality as \(y = mx + c\), \(x = k\), or \(y = k\) to plot boundary lines.
2
Draw the boundary lines
Use a solid line for \(\le\) or \(\ge\), and a dashed line for \(<\) or \(>\).
3
Mark the region for each inequality
Use distinct markers or letters, such as \(A, B, C\), to identify each satisfying half-plane.
4
Identify the common region
Find the region containing all markers.
5
Shade the common region completely
Ensure shading stays strictly within the boundaries.

Try Answering First

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

1

Write a linear inequality for the statement: “The number of brand \(Y\) shirts, \(y\), is at least twice the number of brand \(X\) shirts, \(x\).”

Answer: \(y \ge 2x\)
2

Given the system of linear inequalities \(x + y < 8\) and \(y \ge 3\), test whether the coordinate point \((4, 2)\) satisfies the system.

Answer: No. Substituting \((4, 2)\): \(4 + 2 = 6 < 8\) (True), but \(y = 2\), so \(2 \ge 3\) is False. Because it does not satisfy \(y \ge 3\), \((4, 2)\) does not satisfy the system.
3

What type of boundary line should be drawn for the inequality \(x + y < 4\), and why?

Answer: A dashed line, because the inequality uses the strict inequality symbol “\(<\)”, which means points on the line \(x + y = 4\) are not included in the solution set.

From Region to System, and Maximum Values

Writing the system of inequalities from a shaded region, and reading a maximum value off the feasible region.

Writing the system of inequalities from a shaded region

Identify each boundary line, then ask two questions about each: which side the shaded region lies on (above/below, left/right), and whether the line is solid or dashed. The answers to those two questions give one inequality. Remember the \(x\)-axis is \(y = 0\) and the \(y\)-axis is \(x = 0\).

2 4 6 8 2 4 6 8 O x y x + y = 8

The shaded region is bounded by three lines: \(x + y = 8\) (dashed), the \(x\)-axis and the \(y\)-axis (both solid).

1
The line \(x + y = 8\)
The shaded region lies below the line and the line is dashed, so \(x + y < 8\).
2
The \(x\)-axis, that is \(y = 0\)
The shaded region lies above the \(x\)-axis and the axis is solid, so \(y \geqslant 0\).
3
The \(y\)-axis, that is \(x = 0\)
The shaded region lies to the right of the \(y\)-axis and the axis is solid, so \(x \geqslant 0\).
4
Confirm with test points
Take \((2, 2)\) inside the region: \(2 + 2 = 4 < 8\) ✓, \(2 \geqslant 0\) ✓, \(2 \geqslant 0\) ✓. Take \((5, 5)\) outside: \(5 + 5 = 10 < 8\) is False — it fails only the first inequality, exactly as expected. The system is \(x + y < 8\), \(y \geqslant 0\) and \(x \geqslant 0\).

Determining a maximum value from a feasible region

The maximum number of passengers on a train in a theme park is 30. As a safety measure, the number of adults must always be greater than or equal to the number of children. Let \(x\) be the number of children and \(y\) be the number of adults.

From the graph, determine the maximum number of children allowed to board the train. If there are 18 children, can they all board the train at the same time?

ConditionInequality
At most 30 passengers\(x + y \leqslant 30\)
Adults not fewer than children\(y \geqslant x\)
Numbers cannot be negative\(x \geqslant 0\), \(y \geqslant 0\)
10 20 30 10 20 30 O x y (15, 15) x + y = 30 y = x

The feasible region is the triangle with vertices \((0, 0)\), \((0, 30)\) and \((15, 15)\). The vertex \((15, 15)\) is the point with the largest \(x\)-value in the region.

1
Find the vertex where the two lines meet
Solve \(y = x\) and \(x + y = 30\) simultaneously: \(x + x = 30\), so \(x = 15\) and \(y = 15\). The vertex is \((15, 15)\).
2
Read the maximum \(x\)-value off the region
Across the whole feasible region, the largest value of \(x\) occurs at the vertex \((15, 15)\). So the maximum number of children is 15.
3
Check the answer
Test \((15, 15)\): \(15 + 15 = 30 \leqslant 30\) ✓ and \(15 \geqslant 15\) ✓. Test \((16, 15)\): \(16 + 15 = 31 \leqslant 30\) is False, and \(15 \geqslant 16\) is False too.
4
Answer the 18-children question
If \(x = 18\), then \(y \geqslant 18\) and \(y \leqslant 30 - 18 = 12\) at the same time — impossible. No, because the point \(x = 18\) lies outside the shaded region.

Common Mistakes

  • ! Confusing “at least” with \(\le\) and “at most” with \(\ge\). In mathematics, at least means greater than or equal to (\(\ge\)), while at most means less than or equal to (\(\le\)).
  • ! Drawing solid lines for strict inequalities (\(<\) or \(>\)), or drawing dashed lines for inclusive inequalities (\(\le\) or \(\ge\)).
  • ! Forgetting to include the non-negativity constraints \(x \ge 0\) and \(y \ge 0\) when representing real-life quantities such as number of people or items.
  • ! Shading individual regions separately without clearly highlighting or filling only the common intersection region.

Misconceptions

Misconception

Any point lying in the shaded region of at least one inequality is a solution to the system.

Correct Fact

A point is a solution to the system only if it simultaneously satisfies all inequalities in the system; it must lie within the intersection.

Misconception

Points lying directly on any boundary line are always part of the solution.

Correct Fact

Points lying on a boundary line are only included if that line is solid (\(\le\) or \(\ge\)). If the line is dashed (\(<\) or \(>\)), points on the line are excluded.

Misconception

The inequality \(y - x \le 6\) means the difference must be strictly 6.

Correct Fact

It defines an entire half-plane where the value of \(y - x\) can be any real number less than or equal to 6.

Summary

  • A system of linear inequalities combines two or more linear inequalities representing simultaneous conditions.
  • Its solution is the intersection of all satisfying regions on a Cartesian plane.
  • Boundary lines are drawn solid for \(\le\) and \(\ge\), and dashed for \(<\) and \(>\).
  • In real-world applications, non-negativity conditions (\(x \ge 0, y \ge 0\)) restrict solutions to the first quadrant.
  • The shaded feasible region allows students to determine maximum, minimum, and allowable values.

Short Activity

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

A. Objective Quiz

1 A factory produces \(x\) units of toy cars and \(y\) units of toy boats. The total production is at most 100 units, and the number of toy boats is not less than the number of toy cars. Which system of linear inequalities represents these conditions?

2 Which of the following points satisfies the system of inequalities \(y \le 2x\) and \(y \ge 4\)?

B. Fill in the Blanks

3 A boundary line on a Cartesian plane is drawn with a line when the linear inequality uses the symbol \(<\) or \(>\).

4 The mathematical inequality that represents the statement “the minimum value of \(y\) is 15” is .

C. Drag and Drop: Match Verbal Constraints with Linear Inequalities

Drag each verbal constraint to its matching linear inequality. On a phone, tap the answer choice first, then tap the matching box.

Verbal Constraint
Maximum of \(y\) is \(k\)
Minimum of \(y\) is \(k\)
\(y\) is at least \(k\) times \(x\)
\(y\) is at most \(k\) times \(x\)
Linear Inequality
\(y \le k\)CorrectTry againEmpty
\(y \ge k\)CorrectTry againEmpty
\(y \ge kx\)CorrectTry againEmpty
\(y \le kx\)CorrectTry againEmpty
 

Key Terms

System of linear inequalities Common region Solid line Dashed line Non-negativity constraint At least At most
The views expressed in the notes here are those of the contributor and don’t necessarily represent the views of Pandai.

Systems of Linear Inequalities in Two Variables

6.2 Systems of Linear Inequalities in Two Variables
MATHEMATICS • Form 4 • Chapter 6: Linear Inequalities in Two Variables

6.2 Systems of Linear Inequalities in Two Variables

A system of linear inequalities consists of a combination of two or more linear inequalities involving the same variables. In real-life scenarios such as transportation planning, production, and budgeting, decisions are constrained by multiple simultaneous conditions. The solution is the set of all coordinate points \((x, y)\) that satisfy every inequality in the system simultaneously, represented graphically as an overlapping common region.

Learning Objectives

  • Represent real-world situations and constraints in the form of a system of linear inequalities.
  • Make and verify conjectures about points inside and outside regions that satisfy a system of linear inequalities.
  • Determine, construct boundary lines for, and shade the common region that satisfies a system of linear inequalities.
  • Identify the system of linear inequalities that defines a given shaded region on a Cartesian plane.
  • Solve real-world contextual problems using graphical solutions of systems of linear inequalities.

Identifying Common Feasible Regions for Systems of Linear Inequalities

Overlapping shaded half-planes define the common solution set. Solid lines include boundary points; dashed lines exclude them.

2 4 6 1 2 3 4 0 x y x + y = 4 2y = x x = 0 Common region

The shaded region satisfies \(2y \ge x\), \(x + y < 4\) and \(x \ge 0\). The dashed line \(x + y = 4\) is excluded from the region; the solid lines \(2y = x\) and \(x = 0\) are included.

Brief Explanation

What is a System of Linear Inequalities?

A system of linear inequalities combines several individual inequalities. Instead of having an infinite single half-plane as a solution, the solution is the common region where all constraints are satisfied at the same time.

Non-negativity Constraints

When variables \(x\) and \(y\) represent quantities of physical items or people, non-negativity constraints apply: \(x \ge 0\) and \(y \ge 0\). Graphically, this restricts solutions to the first quadrant.

Example of situation Linear inequality
\(y\) is greater than \(x\)\(y > x\)
\(y\) is less than \(x\)\(y < x\)
\(y\) is not less than \(x\)\(y \ge x\)
\(y\) is not more than \(x\)\(y \le x\)
\(y\) is at least \(k\) times \(x\)\(y \ge kx\)
\(y\) is at most \(k\) times \(x\)\(y \le kx\)
Maximum of \(y\) is \(k\)\(y \le k\)
Minimum of \(y\) is \(k\)\(y \ge k\)
Sum of \(x\) and \(y\) is greater than \(k\)\(x + y > k\)
Difference between \(y\) and \(x\) is less than \(k\)\(y - x < k\)
\(y\) is more than \(x\) by at least \(k\)\(y - x \ge k\)
1
Convert inequalities to linear equations
Write each inequality as \(y = mx + c\), \(x = k\), or \(y = k\) to plot boundary lines.
2
Draw the boundary lines
Use a solid line for \(\le\) or \(\ge\), and a dashed line for \(<\) or \(>\).
3
Mark the region for each inequality
Use distinct markers or letters, such as \(A, B, C\), to identify each satisfying half-plane.
4
Identify the common region
Find the region containing all markers.
5
Shade the common region completely
Ensure shading stays strictly within the boundaries.

Try Answering First

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

1

Write a linear inequality for the statement: “The number of brand \(Y\) shirts, \(y\), is at least twice the number of brand \(X\) shirts, \(x\).”

Answer: \(y \ge 2x\)
2

Given the system of linear inequalities \(x + y < 8\) and \(y \ge 3\), test whether the coordinate point \((4, 2)\) satisfies the system.

Answer: No. Substituting \((4, 2)\): \(4 + 2 = 6 < 8\) (True), but \(y = 2\), so \(2 \ge 3\) is False. Because it does not satisfy \(y \ge 3\), \((4, 2)\) does not satisfy the system.
3

What type of boundary line should be drawn for the inequality \(x + y < 4\), and why?

Answer: A dashed line, because the inequality uses the strict inequality symbol “\(<\)”, which means points on the line \(x + y = 4\) are not included in the solution set.

From Region to System, and Maximum Values

Writing the system of inequalities from a shaded region, and reading a maximum value off the feasible region.

Writing the system of inequalities from a shaded region

Identify each boundary line, then ask two questions about each: which side the shaded region lies on (above/below, left/right), and whether the line is solid or dashed. The answers to those two questions give one inequality. Remember the \(x\)-axis is \(y = 0\) and the \(y\)-axis is \(x = 0\).

2 4 6 8 2 4 6 8 O x y x + y = 8

The shaded region is bounded by three lines: \(x + y = 8\) (dashed), the \(x\)-axis and the \(y\)-axis (both solid).

1
The line \(x + y = 8\)
The shaded region lies below the line and the line is dashed, so \(x + y < 8\).
2
The \(x\)-axis, that is \(y = 0\)
The shaded region lies above the \(x\)-axis and the axis is solid, so \(y \geqslant 0\).
3
The \(y\)-axis, that is \(x = 0\)
The shaded region lies to the right of the \(y\)-axis and the axis is solid, so \(x \geqslant 0\).
4
Confirm with test points
Take \((2, 2)\) inside the region: \(2 + 2 = 4 < 8\) ✓, \(2 \geqslant 0\) ✓, \(2 \geqslant 0\) ✓. Take \((5, 5)\) outside: \(5 + 5 = 10 < 8\) is False — it fails only the first inequality, exactly as expected. The system is \(x + y < 8\), \(y \geqslant 0\) and \(x \geqslant 0\).

Determining a maximum value from a feasible region

The maximum number of passengers on a train in a theme park is 30. As a safety measure, the number of adults must always be greater than or equal to the number of children. Let \(x\) be the number of children and \(y\) be the number of adults.

From the graph, determine the maximum number of children allowed to board the train. If there are 18 children, can they all board the train at the same time?

ConditionInequality
At most 30 passengers\(x + y \leqslant 30\)
Adults not fewer than children\(y \geqslant x\)
Numbers cannot be negative\(x \geqslant 0\), \(y \geqslant 0\)
10 20 30 10 20 30 O x y (15, 15) x + y = 30 y = x

The feasible region is the triangle with vertices \((0, 0)\), \((0, 30)\) and \((15, 15)\). The vertex \((15, 15)\) is the point with the largest \(x\)-value in the region.

1
Find the vertex where the two lines meet
Solve \(y = x\) and \(x + y = 30\) simultaneously: \(x + x = 30\), so \(x = 15\) and \(y = 15\). The vertex is \((15, 15)\).
2
Read the maximum \(x\)-value off the region
Across the whole feasible region, the largest value of \(x\) occurs at the vertex \((15, 15)\). So the maximum number of children is 15.
3
Check the answer
Test \((15, 15)\): \(15 + 15 = 30 \leqslant 30\) ✓ and \(15 \geqslant 15\) ✓. Test \((16, 15)\): \(16 + 15 = 31 \leqslant 30\) is False, and \(15 \geqslant 16\) is False too.
4
Answer the 18-children question
If \(x = 18\), then \(y \geqslant 18\) and \(y \leqslant 30 - 18 = 12\) at the same time — impossible. No, because the point \(x = 18\) lies outside the shaded region.

Common Mistakes

  • ! Confusing “at least” with \(\le\) and “at most” with \(\ge\). In mathematics, at least means greater than or equal to (\(\ge\)), while at most means less than or equal to (\(\le\)).
  • ! Drawing solid lines for strict inequalities (\(<\) or \(>\)), or drawing dashed lines for inclusive inequalities (\(\le\) or \(\ge\)).
  • ! Forgetting to include the non-negativity constraints \(x \ge 0\) and \(y \ge 0\) when representing real-life quantities such as number of people or items.
  • ! Shading individual regions separately without clearly highlighting or filling only the common intersection region.

Misconceptions

Misconception

Any point lying in the shaded region of at least one inequality is a solution to the system.

Correct Fact

A point is a solution to the system only if it simultaneously satisfies all inequalities in the system; it must lie within the intersection.

Misconception

Points lying directly on any boundary line are always part of the solution.

Correct Fact

Points lying on a boundary line are only included if that line is solid (\(\le\) or \(\ge\)). If the line is dashed (\(<\) or \(>\)), points on the line are excluded.

Misconception

The inequality \(y - x \le 6\) means the difference must be strictly 6.

Correct Fact

It defines an entire half-plane where the value of \(y - x\) can be any real number less than or equal to 6.

Summary

  • A system of linear inequalities combines two or more linear inequalities representing simultaneous conditions.
  • Its solution is the intersection of all satisfying regions on a Cartesian plane.
  • Boundary lines are drawn solid for \(\le\) and \(\ge\), and dashed for \(<\) and \(>\).
  • In real-world applications, non-negativity conditions (\(x \ge 0, y \ge 0\)) restrict solutions to the first quadrant.
  • The shaded feasible region allows students to determine maximum, minimum, and allowable values.

Short Activity

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

A. Objective Quiz

1 A factory produces \(x\) units of toy cars and \(y\) units of toy boats. The total production is at most 100 units, and the number of toy boats is not less than the number of toy cars. Which system of linear inequalities represents these conditions?

2 Which of the following points satisfies the system of inequalities \(y \le 2x\) and \(y \ge 4\)?

B. Fill in the Blanks

3 A boundary line on a Cartesian plane is drawn with a line when the linear inequality uses the symbol \(<\) or \(>\).

4 The mathematical inequality that represents the statement “the minimum value of \(y\) is 15” is .

C. Drag and Drop: Match Verbal Constraints with Linear Inequalities

Drag each verbal constraint to its matching linear inequality. On a phone, tap the answer choice first, then tap the matching box.

Verbal Constraint
Maximum of \(y\) is \(k\)
Minimum of \(y\) is \(k\)
\(y\) is at least \(k\) times \(x\)
\(y\) is at most \(k\) times \(x\)
Linear Inequality
\(y \le k\)CorrectTry againEmpty
\(y \ge k\)CorrectTry againEmpty
\(y \ge kx\)CorrectTry againEmpty
\(y \le kx\)CorrectTry againEmpty
 

Key Terms

System of linear inequalities Common region Solid line Dashed line Non-negativity constraint At least At most
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 : System of Linear Inequalities in Two Variables
Form 4 Mathematics
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