GSEB Class 11 Maths Solutions Chapter 6 Linear Inequalities Exercise 6.3

Solve the following systems of inequalities graphically:

Question 1. \( x \ge 3, y \ge 2 \)
Answer:
(i) The line \( x = 3 \) goes through the points \( (3, 0) \) and \( (3, 2) \). This line is called AB.
When we put \( x = 0 \) into \( x \ge 3 \), we get \( 0 \ge 3 \), which is false. This shows that the origin does not lie in the region for \( x \ge 3 \). The graph for this region is shaded with horizontal lines.
(ii) The line \( y = 2 \) goes through the points \( (0, 2) \) and \( (2, 2) \). This line is called CD.
When we put \( y = 0 \) into \( y \ge 2 \), we get \( 0 \ge 2 \), which is also false. This means the origin does not lie in the region for \( y \ge 2 \). The graph for \( y \ge 2 \) is shown with vertical lines.
The final solution for both \( x \ge 3 \) and \( y \ge 2 \) is the area that is common to both the horizontal and vertical line shadings in the region called PBQ.
In simple words: First, draw the lines for \( x=3 \) and \( y=2 \). Then, for \( x \ge 3 \), shade to the right of \( x=3 \). For \( y \ge 2 \), shade above \( y=2 \). The area where both shaded regions overlap is the solution.

Exam Tip: Remember that for inequalities like \( x \ge a \), the region is to the right of the line \( x=a \), and for \( y \ge b \), the region is above the line \( y=b \). Always check a test point like the origin \( (0,0) \) to decide which side to shade.

Question 2. \( 3x + 2y \le 12, x \ge 1, y \ge 2 \)
Answer: The given inequalities are \( 3x + 2y \le 12 \), \( x \ge 1 \), and \( y \ge 2 \).
(i) The line \( 3x + 2y = 12 \) passes through the points \( (4, 0) \) and \( (0, 6) \). This line is represented by AB.
If we put \( x = 0, y = 0 \) into \( 3x + 2y \le 12 \), we get \( 0 + 0 \le 12 \), which is true. This indicates that the origin lies within the region defined by \( 3x + 2y \le 12 \). So, \( 3x + 2y \le 12 \) represents the area below the line \( 3x + 2y = 12 \) and includes all points lying on that line.
(ii) The line \( x = 1 \) passes through the points \( (1, 0) \) and \( (1, 2) \). This line is represented by EF.
Considering the inequality \( x \ge 1 \), if we put \( x = 0 \), we get \( 0 \ge 1 \), which is false. This means the origin does not lie in the region for \( x \ge 1 \). The region on the right side of line EF and all points on EF satisfy the inequality \( x \ge 1 \).
(iii) The line \( y = 2 \) passes through the points \( (1, 2) \) and \( (2, 2) \). This line is represented by BC.
If we put \( y = 0 \) into \( y \ge 2 \), we get \( 0 \ge 2 \), which is false. Therefore, the origin does not lie in this region. The inequality \( y \ge 2 \) is represented by the area above the line \( y = 2 \) and includes all points on this line.
The region that satisfies all the inequalities \( 3x + 2y \le 12 \), \( x \ge 1 \), and \( y \ge 2 \) is the triangle PQR.
In simple words: Draw three lines: \( 3x+2y=12 \), \( x=1 \), and \( y=2 \). For \( 3x+2y \le 12 \), shade towards the origin. For \( x \ge 1 \), shade to the right. For \( y \ge 2 \), shade upwards. The triangular area where all three shadings meet is the answer.

Exam Tip: When dealing with multiple inequalities, graph each line separately, then determine the correct shading region for each using a test point (like the origin if it's not on the line). The solution region is where all shaded areas overlap.

Question 3. \( 2x + y \ge 6, 3x + 4y \le 12 \)
Answer: The given inequalities are \( 2x + y \ge 6 \) and \( 3x + 4y \le 12 \).
(i) The line \( 2x + y = 6 \) passes through points \( (3, 0) \) and \( (0, 6) \). This line is represented by AB.
If we put \( x = 0, y = 0 \) into \( 2x + y \ge 6 \), we get \( 0 \ge 6 \), which is false. This shows that the origin does not lie in this region. Thus, the region lying above line AB and including all points on AB represents the inequality \( 2x + y \ge 6 \).
(ii) The line \( 3x + 4y = 12 \) passes through points \( (4, 0) \) and \( (0, 3) \). This line is represented by CD.
If we put \( x = 0, y = 0 \) into \( 3x + 4y \le 12 \), we get \( 0 \le 12 \), which is true. This means the origin lies in this region. Therefore, \( 3x + 4y \le 12 \) represents the area below line CD (towards the origin) and includes all points lying on it.
The common region is the solution to \( 2x + y \ge 6 \) and \( 3x + 4y \le 12 \), which is the area between these two lines.
In simple words: Draw the lines \( 2x+y=6 \) and \( 3x+4y=12 \). For \( 2x+y \ge 6 \), shade away from the origin. For \( 3x+4y \le 12 \), shade towards the origin. The area that is shaded by both is the solution.

Exam Tip: Always pay attention to the inequality sign: \( \ge \) or \( > \) usually means shading away from the origin (or above the line), while \( \le \) or \( < \) usually means shading towards the origin (or below the line), assuming the origin is a valid test point.

Question 4. \( x + y > 4, 2x - y > 0 \)
Answer: The given inequalities are \( x + y > 4 \) and \( 2x - y > 0 \).
(i) The line \( x + y = 4 \) passes through the points \( (4, 0) \) and \( (0, 4) \). This line is represented by AB.
If we put \( x = 0, y = 0 \) into \( x + y > 4 \), we get \( 0 > 4 \), which is false. This means the origin does not lie in this region. Thus, \( x + y > 4 \) is represented by the region above the line \( x + y = 4 \). Since the inequality is strictly \( > \), the line itself is not included in the solution.
(ii) The line \( 2x - y = 0 \) passes through the points \( (0, 0) \) and \( (1, 2) \). This line is represented by CD.
If we put a test point like \( (1, 0) \) into \( 2x - y > 0 \), we get \( 2(1) - 0 > 0 \), which is \( 2 > 0 \), and this is true. This shows that the point \( (1, 0) \) lies in the region, meaning the region lying below the line \( 2x - y = 0 \) represents \( 2x - y > 0 \). Since the inequality is strictly \( > \), the line itself is not included in the solution.
The common region satisfying both inequalities is the area where the shaded parts for \( x + y > 4 \) and \( 2x - y > 0 \) overlap.
In simple words: Draw \( x+y=4 \) and \( 2x-y=0 \). For \( x+y > 4 \), shade above the line (not including the line). For \( 2x-y > 0 \), shade below the line (not including the line). The overlapping area is the solution.

Exam Tip: For strict inequalities (with \( < \) or \( > \)), the boundary line should be drawn as a dashed line to show it is not part of the solution. For non-strict inequalities (with \( \le \) or \( \ge \)), a solid line is used.

Question 5. \( 2x - y > 1, x - 2y < -1 \)
Answer: The given inequalities are \( 2x - y > 1 \) and \( x - 2y < -1 \).
Graph for inequality (1): Let us draw the graph of the line \( 2x - y = 1 \).
When \( y = 0 \), \( x = \frac{1}{2} \), so point A is \( (\frac{1}{2}, 0) \). When \( x = 0 \), \( y = -1 \), so point B is \( (0, -1) \).
If we put \( x = 0, y = 0 \) into \( 2x - y > 1 \), we get \( 0 > 1 \), which is false. Therefore, the half-plane region that lies below line AB is the solution region, which is the shaded area (not including the line AB because of the strict inequality).
Graph for inequality (2): Let us draw the graph of the line \( x - 2y = -1 \).
When \( y = 0 \), \( x = -1 \), so point C is \( (-1, 0) \). When \( x = 0 \), \( y = \frac{1}{2} \), so point D is \( (0, \frac{1}{2}) \). The graph of this line is CD.
If we put \( x = 0, y = 0 \) into \( x - 2y < -1 \), we get \( 0 < -1 \), which is false. Therefore, the half-plane region that lies above line CD is the solution area. The double shaded area in the figure is the solution region for the given inequalities.
In simple words: Draw the lines \( 2x-y=1 \) and \( x-2y=-1 \). For \( 2x-y > 1 \), shade away from the origin. For \( x-2y < -1 \), shade away from the origin. The area where both shadings overlap is the solution, and the lines themselves are not included.

Exam Tip: When shading regions for inequalities, if the origin \( (0,0) \) is not on the boundary line, test it. If \( (0,0) \) satisfies the inequality, shade the region containing the origin. If it doesn't, shade the region opposite to the origin. If \( (0,0) \) is on the line, choose another test point.

Question 6. \( x + y \le 6, x + y \ge 4 \)
Answer: The given inequalities are \( x + y \le 6 \) and \( x + y \ge 4 \).
(i) The line \( x + y = 6 \) passes through points \( (6, 0) \) and \( (0, 6) \). This line is represented by AB.
If we put \( x = 0, y = 0 \) into \( x + y \le 6 \), we get \( 0 < 6 \), which is true. This means the origin lies in the region for \( x + y \le 6 \). So, \( x + y \le 6 \) represents the region below the line \( x + y = 6 \) and includes all points on that line.
(ii) The line \( x + y = 4 \) passes through points \( (4, 0) \) and \( (0, 4) \). This line is represented by CD.
If we put \( x = 0, y = 0 \) into \( x + y \ge 4 \), we get \( 0 + 0 \ge 4 \), which is false. This shows that the origin does not lie in this region. Therefore, the points lying above this line and on this line represent the inequality \( x + y \ge 4 \).
The solution region is the area between lines AB and CD, including all the points lying on these lines.
In simple words: Draw two parallel lines: \( x+y=6 \) and \( x+y=4 \). For \( x+y \le 6 \), shade below the line. For \( x+y \ge 4 \), shade above the line. The solution is the strip between these two lines, including the lines themselves.

Exam Tip: For inequalities involving parallel lines, the solution often forms a band or strip between them. Always draw solid lines for \( \le \) or \( \ge \) to show the boundary is included.

Question 7. \( 2x + y \ge 8, x + 2y \ge 10 \)
Answer: The given inequalities are \( 2x + y \ge 8 \) and \( x + 2y \ge 10 \).
Graph for inequality (1): Let us draw the graph of the line \( 2x + y = 8 \).
When \( y = 0 \), \( x = 4 \), so point A is \( (4, 0) \). When \( x = 0 \), \( y = 8 \), so point B is \( (0, 8) \).
If we put \( x = 0, y = 0 \) into inequality (1), we get \( 0 \ge 8 \), which is false. Hence, the half-plane region that lies above line AB and includes the line AB is the solution area.
Graph for inequality (2): Let us draw the graph of the line \( x + 2y = 10 \).
When \( y = 0 \), \( x = 10 \), so point C is \( (10, 0) \). When \( x = 0 \), \( y = 5 \), so point D is \( (0, 5) \).
If we put \( x = 0, y = 0 \) into inequality (2), we get \( 0 \ge 10 \), which is false. Hence, the upper portion of line CD, including line CD, is the shaded area. The double shaded area of the figure is the overall solution region.
In simple words: Draw the lines \( 2x+y=8 \) and \( x+2y=10 \). For both inequalities, test the origin. Since \( 0 \ge 8 \) and \( 0 \ge 10 \) are both false, shade the region away from the origin for both lines. The area where these two shadings overlap is the solution.

Exam Tip: When all inequalities are \( \ge \), the solution region is typically unbounded and lies "away" from the origin, forming an open region. Ensure you correctly identify the intersection point(s) if necessary.

Question 8. \( x + y \le 9, y > x, x \ge 0 \)
Answer: The given inequalities are \( x + y \le 9 \), \( y > x \), and \( x \ge 0 \).
(i) The line \( x + y = 9 \) passes through points \( (9, 0) \) and \( (0, 9) \). This line is represented by AB.
If we put \( x = 0, y = 0 \) into \( x + y \le 9 \), we get \( 0 + 0 = 0 < 9 \), which is true. This means the origin lies in this region. Therefore, the region below line AB and including all points on it forms the region for \( x + y \le 9 \).
(ii) The line \( y = x \) passes through the origin \( (0, 0) \) and \( (2, 2) \). This line is represented by CD.
If we put a test point, for example, \( (0, 1) \), into \( y - x > 0 \), we get \( 1 - 0 > 0 \), which is true. So, \( (0, 1) \) lies in this region. The inequality \( y > x \) is represented by the region above line CD. Since it's a strict inequality, the line itself is not included.
(iii) The inequality \( x \ge 0 \) represents the region on the right side of the y-axis and includes all points on the y-axis.
The region bounded by triangle PQR (as seen in the graph) is the solution for the given inequalities: \( x + y \le 9 \), \( y > x \), and \( x \ge 0 \).
In simple words: Draw lines for \( x+y=9 \), \( y=x \), and \( x=0 \) (the y-axis). For \( x+y \le 9 \), shade towards the origin. For \( y > x \), shade above the line \( y=x \). For \( x \ge 0 \), shade to the right of the y-axis. The region where all three shaded areas overlap is the answer.

Exam Tip: When an inequality like \( y > x \) is given, the origin cannot be used as a test point as it lies on the line \( y=x \). In such cases, choose a point not on the line, for example \( (0,1) \) or \( (1,0) \), to determine the correct shading side.

Question 9. \( 5x + 4y \le 20, x \ge 1, y \ge 2 \)
Answer: The given inequalities are \( 5x + 4y \le 20 \), \( x \ge 1 \), and \( y \ge 2 \).
(i) The line \( 5x + 4y = 20 \) passes through points \( (4, 0) \) and \( (0, 5) \). This line is represented by AB.
If we put \( x = 0, y = 0 \) into \( 5x + 4y \le 20 \), we get \( 0 + 0 = 0 < 20 \), which is true. This means the origin lies in this region. Therefore, the region below the line \( 5x + 4y = 20 \) and including all points on it belongs to \( 5x + 4y \le 20 \).
(ii) The line \( y = 2 \) is parallel to the x-axis, at a distance of 2 units from the origin. This line is represented by EF.
If we put \( y = 0 \), we get \( 0 \ge 2 \), which is not true. This shows that the origin does not lie in this region. The region above line EF and including the points on it represents the inequality \( y \ge 2 \).
(iii) The line \( x = 1 \) is parallel to the y-axis, at a distance of 1 unit from the origin.
If we put \( x = 0 \) into \( x - 1 \ge 0 \), we get \( -1 \ge 0 \), which is false. This shows that the origin does not lie in this region. The region on the right side of \( x = 1 \) and including all points on it belongs to \( x \ge 1 \).
The double shaded region bounded by triangle PQR is the solution for the given inequalities.
In simple words: Draw lines for \( 5x+4y=20 \), \( x=1 \), and \( y=2 \). For \( 5x+4y \le 20 \), shade towards the origin. For \( x \ge 1 \), shade to the right. For \( y \ge 2 \), shade upwards. The area where all three shaded parts overlap is the solution.

Exam Tip: When dealing with boundary lines like \( x=a \) or \( y=b \), remember that \( x \ge a \) means the region to the right, and \( y \ge b \) means the region above. Always ensure your shaded region correctly reflects these basic principles.

Question 10. \( 3x + 4y < 60, x + 3y \le 30, x \ge 0, y \ge 0 \)
Answer: The given inequalities are \( 3x + 4y < 60 \), \( x + 3y \le 30 \), \( x \ge 0 \), and \( y \ge 0 \).
First, we draw the graphs of the lines \( 3x + 4y = 60 \), \( x + 3y = 30 \), \( x = 0 \) (y-axis), and \( y = 0 \) (x-axis).
The line \( 3x + 4y = 60 \) passes through point A \( (20, 0) \) and point B \( (0, 15) \). Since \( 0 < 60 \) (by substituting \( x=0, y=0 \)), the origin lies in this region. Thus, \( 3x + 4y < 60 \) represents the region below this line, but *not* including the line itself (because of the strict inequality \( < \)).
Furthermore, the line \( x + 3y = 30 \) passes through \( (0, 10) \) and \( (30, 0) \). This line is represented by CD. Since \( 0 < 30 \) (by substituting \( x=0, y=0 \)), the origin lies in this region. This inequality represents the region below the line \( x + 3y = 30 \) and includes the line itself.
Inequality \( x \ge 0 \) represents the region on the right side of the y-axis and the y-axis itself.
Inequality \( y \ge 0 \) represents the region above the x-axis and the x-axis itself.
The multishaded region in the figure is the solution area.
In simple words: Draw the lines for \( 3x+4y=60 \), \( x+3y=30 \), \( x=0 \), and \( y=0 \). For \( 3x+4y < 60 \), shade below the line (dashed line). For \( x+3y \le 30 \), shade below the line (solid line). For \( x \ge 0 \) and \( y \ge 0 \), shade in the first quadrant. The common area of these shaded regions is the answer.

Exam Tip: Remember to use a dashed line for strict inequalities \( (<, >) \) and a solid line for non-strict inequalities \( (\le, \ge) \). The regions \( x \ge 0 \) and \( y \ge 0 \) always restrict the solution to the first quadrant of the coordinate plane.

Question 11. \( 2x + y \ge 4, x + y \le 3, 2x - 3y \le 6 \)
Answer: First, we draw the graphs of the equations \( 2x + y = 4 \), \( x + y = 3 \) and \( 2x - 3y = 6 \).
For the line \( 2x + y = 4 \), it passes through points A \( (2, 0) \) and B \( (0, 4) \). The graph of this line is AB. Since \( 0 \not\ge 4 \) (when testing the origin), the origin does not lie in the region for \( 2x + y \ge 4 \). This inequality represents the region above this line and includes all points on it.
Next, for the line \( x + y = 3 \), it is represented by CD, where C is \( (3, 0) \) and D is \( (0, 3) \). As \( 0 < 3 \) (when testing the origin), the origin lies in the region for \( x + y \le 3 \). This inequality represents the region below line CD and includes all points on it.
Further, for the line \( 2x - 3y = 6 \), it is represented by EC, where E is \( (0, -2) \) and C is \( (3, 0) \). As \( 0 < 6 \) (when testing the origin), the origin lies in it. Therefore, \( 2x - 3y \le 6 \) represents the region above line EC and includes all points on it. The triple shaded triangular region in the figure is the solution area.
In simple words: Draw the three lines \( 2x+y=4 \), \( x+y=3 \), and \( 2x-3y=6 \). For \( 2x+y \ge 4 \), shade away from the origin. For \( x+y \le 3 \), shade towards the origin. For \( 2x-3y \le 6 \), shade towards the origin. The area where all three shaded regions overlap forms the triangular solution.

Exam Tip: Be careful with inequalities involving negative coefficients, such as \( -3y \). When dividing by a negative number to isolate \( y \), always remember to reverse the inequality sign. For instance, \( -3y \le 6 \) becomes \( y \ge -2 \).

Question 12. \( x - 2y \le 3, 3x + 4y \ge 12, x \ge 0, y \ge 1 \)
Answer: The given inequalities are \( x - 2y \le 3 \), \( 3x + 4y \ge 12 \), \( x \ge 0 \), and \( y \ge 1 \).
(i) The line \( x - 2y = 3 \) passes through \( (3, 0) \) and \( (0, \frac{3}{2}) \). This line is represented by AB. Since \( 0 < 3 \) (when testing the origin), the origin lies in the region for \( x - 2y \le 3 \). Thus, the region above this line and including its points represents \( x - 2y \le 3 \).
(ii) The line \( 3x + 4y = 12 \) passes through \( (4, 0) \) and \( (0, 3) \). This line is represented by CD. Since \( 0 \not\ge 12 \) (when testing the origin), the origin does not lie in the region for \( 3x + 4y \ge 12 \). Therefore, the region above line CD and including points on CD represents \( 3x + 4y \ge 12 \).
(iii) The inequality \( x \ge 0 \) represents the region on the right side of the y-axis and includes all the points on it.
(iv) The line \( y = 1 \) is parallel to the x-axis at a distance of 1 unit from it. For \( y \ge 1 \), if we test the origin \( y=0 \), we get \( 0 \ge 1 \), which is false. So, the origin does not lie in this region. Therefore, \( y \ge 1 \) is the region above the line \( y = 1 \) and includes all points lying on it.
The common shaded region YC PQT (from the figure) represents the solution to the given inequalities.
In simple words: Draw lines for \( x-2y=3 \), \( 3x+4y=12 \), \( x=0 \) (y-axis), and \( y=1 \). For \( x-2y \le 3 \), shade above the line. For \( 3x+4y \ge 12 \), shade above the line. For \( x \ge 0 \), shade right of the y-axis. For \( y \ge 1 \), shade above the line \( y=1 \). The intersection of all these shaded areas is the answer.

Exam Tip: For inequalities like \( x \ge 0 \) and \( y \ge 0 \), remember these typically mean the solution must be in the first quadrant (or a part of it). Always sketch these boundary lines first.

Question 13. \( 4x + 3y \le 60, y \le 2x, x \ge 3, y \ge 0 \)
Answer: The given inequalities are \( 4x + 3y \le 60 \), \( y \le 2x \), \( x \ge 3 \), and \( y \ge 0 \).
(i) The line \( 4x + 3y = 60 \) passes through points \( (15, 0) \) and \( (0, 20) \). This line is represented by AB.
If we put \( x = 0, y = 0 \) into \( 4x + 3y \le 60 \), we get \( 0 + 0 = 0 \le 60 \), which is true. Therefore, the origin lies in this region. Thus, the region below line AB and including the points lying on AB represents the inequality \( 4x + 3y \le 60 \).
(ii) The line \( y = 2x \) passes through \( (0, 0) \) and \( (5, 10) \). This line is represented by CD.
If we put a test point, for example, \( (0, 5) \), into \( y - 2x \le 0 \), we get \( 5 - 2(0) \le 0 \), which simplifies to \( 5 \le 0 \), which is not true. This shows that \( (0, 5) \) does not lie in this region. The region lying above the line CD and including the points on CD represents \( y \ge 2x \). (Note: There's an inconsistency in the source description here. For \( y \le 2x \), we would typically shade *below* the line \( y=2x \). Following the source's explicit description, we represent it as "above line CD" as stated).
(iii) The inequality \( x \ge 3 \) is the region lying on the right side of the line \( x = 3 \) and includes all points on it.
(iv) The inequality \( y \ge 0 \) represents the region above the x-axis and includes all points on the x-axis.
The solution is the common shaded region satisfying all these inequalities.
In simple words: Draw lines for \( 4x+3y=60 \), \( y=2x \), \( x=3 \), and \( y=0 \). For \( 4x+3y \le 60 \), shade towards the origin. For \( y \le 2x \), shade below the line (based on standard interpretation, though source states above). For \( x \ge 3 \), shade right. For \( y \ge 0 \), shade up. The common overlapping area is the solution.

Exam Tip: Be careful with inequalities of the form \( y \le mx \) or \( y \ge mx \). If \( (0,0) \) is on the line, choose a point like \( (1,0) \) or \( (0,1) \) to determine the correct side to shade. For \( y \le 2x \), points like \( (1,0) \) give \( 0 \le 2 \), so the region below \( y=2x \) contains \( (1,0) \).

Question 14. \( 3x + 2y \le 150, x + 4y \le 80, x \le 15, y \ge 0 \)
Answer: The given inequalities are \( 3x + 2y \le 150 \), \( x + 4y \le 80 \), \( x \le 15 \), and \( y \ge 0 \).
(i) The line \( 3x + 2y = 150 \) passes through the points \( (50, 0) \) and \( (0, 75) \). This line is represented by AB.
If we put \( x = 0, y = 0 \) into \( 3x + 2y \le 150 \), we get \( 0 \le 150 \), which is true. This shows that the origin lies in this region. Therefore, the region below line AB and including the points lying on AB represents the inequality \( 3x + 2y \le 150 \).
(ii) The line \( x + 4y = 80 \) passes through the points \( (80, 0) \) and \( (0, 20) \). This line is represented by CD.
Since \( 0 < 80 \) (when testing the origin), the origin lies in this region. So, the region below line CD and including the points on CD represents the inequality \( x + 4y \le 80 \).
(iii) The inequality \( x \le 15 \) is the region lying on the left side of the line EF (where the equation is \( x = 15 \)) and includes all the points on EF.
(iv) The inequality \( y \ge 0 \) represents the region above the x-axis and includes all points on the x-axis.
Thus, the region X'PQR, which is a double shaded region in the figure, is the solution to the given inequalities. It should be noted that any point lying in this region also satisfies the inequality \( 3x + 2y \le 150 \).
In simple words: Draw lines for \( 3x+2y=150 \), \( x+4y=80 \), \( x=15 \), and \( y=0 \). For \( 3x+2y \le 150 \), shade towards the origin. For \( x+4y \le 80 \), shade towards the origin. For \( x \le 15 \), shade to the left. For \( y \ge 0 \), shade upwards. The overlapping area of all these shadings is the solution.

Exam Tip: For problems with many inequalities, systematically graph each boundary line and determine its valid region. The overall feasible region for all inequalities will be the polygon formed by the intersection of all individual regions.

Question 15. \( x + 2y < 10, x + y \ge 1, x - y \le 0, x \ge 0, y \ge 0 \)
Answer: The given inequalities are \( x + 2y < 10 \), \( x + y \ge 1 \), \( x - y \le 0 \), \( x \ge 0 \), and \( y \ge 0 \).
(i) The line \( x + 2y = 10 \) passes through \( (10, 0) \) and \( (0, 5) \). This line is represented by AB. Since \( 0 < 10 \) (when testing the origin), the origin lies in this region. Therefore, the region below line AB and including the points on it represents \( x + 2y < 10 \) (the line itself is not included for strict inequality).
(ii) The line \( x + y = 1 \) passes through \( (1, 0) \) and \( (0, 1) \). This line is represented by CD. Since \( 0 \not\ge 1 \) (when testing the origin), the origin does not lie in this region. Therefore, the region above line CD and including the points on CD represents \( x + y \ge 1 \).
(iii) The line \( x - y = 0 \) passes through \( (0, 0) \) and \( (1, 1) \). This line is represented by EF. Also, the point \( (0, 1) \) lies in \( x - y \le 0 \) (since \( 0 \le 1 \)). The region above line EF and including the points on it represents \( x - y \le 0 \).
(iv) The inequality \( x \ge 0 \) represents the region lying on the right side of the y-axis and includes all points on the y-axis.
(v) The inequality \( y \ge 0 \) represents the region lying above the x-axis and includes all points on the x-axis.
The region PQRS (from the figure) represents the given inequalities, including \( y \ge 0 \), since every point in region PQRS satisfies the condition \( y \ge 0 \).
In simple words: Draw lines for \( x+2y=10 \), \( x+y=1 \), \( x-y=0 \), \( x=0 \), and \( y=0 \). Shade for each: \( x+2y < 10 \) (below, dashed), \( x+y \ge 1 \) (above, solid), \( x-y \le 0 \) (above, solid), \( x \ge 0 \) (right of y-axis), \( y \ge 0 \) (above x-axis). The common area of these shaded regions is the solution.

Exam Tip: When \( x \ge 0 \) and \( y \ge 0 \) are included, the solution region is always constrained to the first quadrant. Remember to use dashed lines for strict inequalities and solid lines for non-strict ones.