GSEB Class 11 Maths Solutions Chapter 6 Linear Inequalities Exercise 6.2

Solve the following inequalities graphically in two dimensional plane:

Question 1. x + y < 5
Answer: Consider the equation \( x + y = 5 \). This line passes through the points \( (5, 0) \) and \( (0, 5) \). This equation is shown as line AB on the graph. Now, let's consider the inequality \( x + y < 5 \). If we put \( x = 0 \) and \( y = 0 \), we get \( 0 + 0 = 0 < 5 \), which is a true statement. Therefore, the origin \( (0, 0) \) is located in the half-plane defined by \( x + y < 5 \). The shaded area in the figure represents this inequality.
In simple words: First, draw the line for \( x + y = 5 \). This line cuts through \( (5,0) \) and \( (0,5) \). Because the inequality is \( x + y < 5 \), we test a point like \( (0,0) \). Since \( 0 < 5 \) is true, we shade the side of the line that includes \( (0,0) \). The line itself is dashed because the inequality is "less than" (not "less than or equal to").

Exam Tip: When dealing with strict inequalities like <, >, always use a dashed line for the boundary. For non-strict inequalities like ≤, ≥, use a solid line.

Question 2. 2x + y ≥ 6
Answer: Consider the equation \( 2x + y = 6 \). This line passes through the points \( (0, 6) \) and \( (3, 0) \). This equation is shown as line AB on the graph. Now, let's consider the inequality \( 2x + y \ge 6 \). If we put \( x = 0 \) and \( y = 0 \), we get \( 0 + 0 = 0 \ge 6 \), which is a false statement. This means the origin \( (0, 0) \) does not lie in the region that satisfies the inequality \( 2x + y \ge 6 \). Hence, the shaded area in the figure represents the inequality \( 2x + y \ge 6 \), which is the region not containing the origin.
In simple words: Draw the line for \( 2x + y = 6 \), going through \( (0,6) \) and \( (3,0) \). Test \( (0,0) \). Since \( 0 \ge 6 \) is false, shade the side of the line that does NOT include \( (0,0) \). Use a solid line because the inequality includes "or equal to".

Exam Tip: When testing for the shaded region, choosing the origin (0,0) as a test point is usually the easiest method, unless the line passes through the origin.

Question 3. 3x + 4y ≤ 12
Answer: We begin by plotting the graph of the equation \( 3x + 4y = 12 \). This line goes through the points \( (4, 0) \) and \( (0, 3) \). We represent this line as AB. Now, let's consider the inequality \( 3x + 4y \le 12 \). If we put \( x = 0 \) and \( y = 0 \), we get \( 0 + 0 = 0 \le 12 \), which is a true statement. Therefore, the origin lies in the region defined by \( 3x + 4y \le 12 \). The shaded area in the figure represents this inequality.
In simple words: Draw the line for \( 3x + 4y = 12 \), crossing at \( (4,0) \) and \( (0,3) \). Since \( 0 \le 12 \) is true, shade the part of the graph that includes the origin. Use a solid line for "less than or equal to".

Exam Tip: Always identify the x and y intercepts to accurately plot the boundary line for two-variable linear inequalities.

Question 4. y + 8 ≥ 2x
Answer: For the given inequality \( y + 8 \ge 2x \), we first draw the graph of the line \( y + 8 = 2x \). Setting \( y = 0 \), we find \( x = 4 \), so the point on the x-axis is \( (4, 0) \). Setting \( x = 0 \), we find \( y = -8 \), so the point on the y-axis is \( (0, -8) \). The line AB represents this equation. To test the inequality \( y + 8 \ge 2x \), we substitute \( x = 0 \) and \( y = 0 \), which gives \( 0 + 8 \ge 0 \) or \( 8 \ge 0 \), which is a true statement. Therefore, the origin lies in the half-plane region that satisfies the inequality. The shaded region, including the line, represents the solution.
In simple words: First, draw the line \( y + 8 = 2x \). It passes through \( (4,0) \) and \( (0,-8) \). Check with \( (0,0) \); since \( 8 \ge 0 \) is true, shade the side of the line that includes the origin. Use a solid line because it's "greater than or equal to".

Exam Tip: For inequalities, always draw the boundary line accurately first, then use a test point to determine which side to shade.

Question 5. x - y ≤ 2
Answer: The given inequality is \( x - y \le 2 \). We will draw the graph of the line \( x - y = 2 \). If we set \( y = 0 \), we find \( x = 2 \), so the point on the x-axis is \( (2, 0) \). If we set \( x = 0 \), we find \( y = -2 \), so the point on the y-axis is \( (0, -2) \). The line AB represents this equation. To test the inequality \( x - y \le 2 \), we substitute \( x = 0 \) and \( y = 0 \), which gives \( 0 - 0 \le 2 \) or \( 0 \le 2 \), which is a true statement. Hence, the origin lies in the half-plane region that satisfies the inequality. The shaded area, including the line, represents the solution.
In simple words: Draw the line \( x - y = 2 \), passing through \( (2,0) \) and \( (0,-2) \). Test \( (0,0) \). Since \( 0 \le 2 \) is true, shade the side of the line that includes the origin. Use a solid line for "less than or equal to".

Exam Tip: Remember that \( x - y \le 2 \) is equivalent to \( y \ge x - 2 \). When \( y \) is greater than or equal to the expression, shade above the line.

Question 6. 2x - 3y > 6
Answer: We begin by drawing the graph of \( 2x - 3y = 6 \). This line passes through the points \( (3, 0) \) and \( (0, -2) \). This equation is represented by AB. Now, consider the inequality \( 2x - 3y > 6 \). If we put \( x = 0 \) and \( y = 0 \), we get \( 0 - 0 = 0 > 6 \), which is a false statement. Therefore, the origin does not lie in the region that satisfies \( 2x - 3y > 6 \). The shaded region represents this inequality.
In simple words: Draw the line for \( 2x - 3y = 6 \), passing through \( (3,0) \) and \( (0,-2) \). Test \( (0,0) \). Since \( 0 > 6 \) is false, shade the side of the line that does NOT include the origin. Use a dashed line because it's "greater than" (not "greater than or equal to").

Exam Tip: For strict inequalities, the boundary line itself is not part of the solution set, hence it's represented as a dashed line.

Question 7. - 3x + 2y ≥ -6
Answer: Let's draw the line \( -3x + 2y = -6 \). This line passes through the points \( (2, 0) \) and \( (0, -3) \). This equation is represented by AB. Now, consider the inequality \( -3x + 2y \ge -6 \). If we put \( x = 0 \) and \( y = 0 \), we get \( 0 + 0 = 0 \ge -6 \), which is a true statement. Thus, the origin lies in the region that satisfies \( -3x + 2y \ge -6 \). The shaded region represents this inequality.
In simple words: Draw the line for \( -3x + 2y = -6 \), passing through \( (2,0) \) and \( (0,-3) \). Test \( (0,0) \). Since \( 0 \ge -6 \) is true, shade the side of the line that includes the origin. Use a solid line for "greater than or equal to".

Exam Tip: For inequalities, always choose a test point that is not on the line itself to determine the correct half-plane for shading.

Question 8. 3y - 5x < 30
Answer: The given inequality is \( 3y - 5x < 30 \). We draw the graph of the equation \( 3y - 5x = 30 \). Setting \( y = 0 \), we find \( x = -6 \), so the point on the x-axis is \( (-6, 0) \). Setting \( x = 0 \), we find \( y = 10 \), so the point on the y-axis is \( (0, 10) \). The line AB represents this equation. To test the inequality, we substitute \( x = 0 \) and \( y = 0 \) into \( 3y - 5x < 30 \), which gives \( 0 - 0 < 30 \) or \( 0 < 30 \), which is a true statement. Therefore, the origin lies in the half-plane region that satisfies the inequality. The shaded area, excluding the line AB, represents the solution region.
In simple words: Draw the line for \( 3y - 5x = 30 \), crossing at \( (-6,0) \) and \( (0,10) \). Test \( (0,0) \). Since \( 0 < 30 \) is true, shade the side of the line that includes the origin. Use a dashed line because it's "less than" (not "less than or equal to").

Exam Tip: Always make sure your chosen test point clearly falls into one of the half-planes, not on the boundary line itself.

Question 9. y < - 2
Answer: The given inequality is \( y < -2 \). We draw the graph of the line \( y = -2 \). This is a horizontal line that passes through \( (0, -2) \). This line is represented by AB. To test the inequality, we substitute \( y = 0 \) (since \( x \) is not in the inequality, any \( x \) value works for \( (0,0) \)) into \( y < -2 \), which gives \( 0 < -2 \), which is a false statement. Therefore, the solution region is the shaded area below the line AB, excluding the line itself. Hence, every point below the line AB is part of the solution area.
In simple words: Draw a dashed horizontal line at \( y = -2 \). Test \( (0,0) \). Since \( 0 < -2 \) is false, shade the area below the line. The line is dashed because it's strictly "less than".

Exam Tip: For inequalities involving only one variable (e.g., \( y < k \) or \( x > k \)), the boundary line will be horizontal or vertical, respectively.

Question 10. x > - 3
Answer: Let's draw the graph of the line \( x = -3 \). This is a vertical line passing through \( (-3, 0) \) and \( (-3, 2) \), among other points. This line is represented by AB. By substituting \( x = 0 \) into the inequality \( x > -3 \), we get \( 0 > -3 \), which is a true statement. Therefore, the origin lies in the region defined by \( x > -3 \). The graph of the inequality \( x > -3 \) is shown as the shaded area to the right of the line, excluding the line itself.
In simple words: Draw a dashed vertical line at \( x = -3 \). Test \( (0,0) \). Since \( 0 > -3 \) is true, shade the area to the right of the line, which includes the origin. The line is dashed because it's strictly "greater than".

Exam Tip: For inequalities involving only \( x \), shade to the right for \( x > k \) or \( x \ge k \) and to the left for \( x < k \) or \( x \le k \).