OP Malhotra Class 11 Maths Solutions Chapter 11 Inequalities Exercise 11 (B)

Question 1. \( x < 2 \) and \( y > -1 \)
Answer:The given inequalities are: 1. \( x < 2 \) 2. \( y > -1 \) For the region \( x < 2 \): The line \( x = 2 \) is parallel to the y-axis and passes through the point \( (2, 0) \). When we test the origin \( (0, 0) \), it satisfies the inequality \( 0 < 2 \). This means the region containing the origin gives the solution set for \( x < 2 \). For the region \( y > -1 \): The line \( y = -1 \) is parallel to the x-axis and passes through the point \( (0, -1) \). When we test the origin \( (0, 0) \), it satisfies the inequality \( 0 > -1 \). This means the region containing the origin gives the solution set for \( y > -1 \). To find the combined solution, we shade the common region where both inequalities are true. All points within this common shaded area represent the solution set for the given system of inequalities.
In simple words: First, draw the line \( x = 2 \) (vertical) and the line \( y = -1 \) (horizontal) using dashed lines because the inequalities do not include "equal to". Then, shade the area to the left of \( x = 2 \) and above \( y = -1 \). The overlapping shaded part is the solution.

🎯 Exam Tip: Remember to use a dashed line for strict inequalities (< or >) and a solid line for non-strict inequalities (≤ or ≥). Always test the origin (0,0) to easily determine which side of the line to shade, unless the line passes through the origin.

Question 2. \( y > x \) and \( y \le 3 \)
Answer:The given inequalities are: 1. \( y > x \) 2. \( y \le 3 \) For the region \( y > x \): The line \( y = x \) passes through the origin \( (0, 0) \) and makes an angle of \( 45^\circ \) with the positive x-axis. Since the line passes through \( (0,0) \), we cannot use it as a test point. Let's pick another point, for example, \( (0, 1) \). For \( (0,1) \), \( 1 > 0 \) which satisfies \( y > x \). So, the region containing \( (0, 1) \) (the upper half of the plane divided by \( y=x \)) is the solution set for \( y > x \). The line itself should be dashed because the inequality is strict. For the region \( y \le 3 \): The line \( y = 3 \) is parallel to the x-axis and passes through \( (0, 3) \). When we test the origin \( (0, 0) \), it satisfies the inequality \( 0 \le 3 \). This means the region containing the origin (below the line \( y = 3 \)) is the solution set for \( y \le 3 \). The line itself should be solid because the inequality includes "equal to". To find the combined solution, we shade the common region where both inequalities are true. All points in this common shaded portion give the solution set for the given system of inequalities. The lines \( y=x \) and \( y=3 \) intersect at \( (3,3) \).
In simple words: First, draw a dashed line for \( y = x \) (it slopes upwards through the middle). Then, draw a solid horizontal line for \( y = 3 \). Shade the area above the \( y = x \) line and below the \( y = 3 \) line. That overlapping shaded part is your answer.

🎯 Exam Tip: When a boundary line passes through the origin (like \( y=x \) or \( x=0 \)), you cannot use \( (0,0) \) as a test point. Choose any other point not on the line, like \( (1,0) \) or \( (0,1) \), to determine the correct shading region.

Question 3. \( x \le -y \) and \( 2y < x \)
Answer:The given inequalities are: 1. \( x \le -y \) (which can be rewritten as \( y \le -x \)) 2. \( 2y < x \) (which can be rewritten as \( y < \frac{x}{2} \)) For the region \( x \le -y \): The line \( x = -y \) (or \( y = -x \)) passes through \( (0, 0) \) and makes an angle of \( 135^\circ \) with the positive x-axis. Since the line passes through \( (0,0) \), we test another point, e.g., \( (1, 0) \). For \( (1, 0) \), \( 1 \le -0 \) is \( 1 \le 0 \), which is false. So, the region not containing \( (1,0) \) is the solution set for \( x \le -y \). This is the region above and to the left of the line \( y=-x \). The line should be solid. For the region \( 2y < x \): The line \( 2y = x \) (or \( y = \frac{x}{2} \)) passes through \( (0, 0) \) and \( (2, 1) \). Testing the origin \( (0, 0) \) gives \( 2(0) < 0 \), which is \( 0 < 0 \), false. So, the region not containing the origin gives the solution set for \( 2y < x \). This means the area below the line \( y = \frac{x}{2} \). The line itself should be dashed. The solution set consists of all points in the common shaded region on the coordinate plane where both these conditions are met.
In simple words: Draw a solid line for \( y = -x \) (it slopes down from left to right). Then, draw a dashed line for \( y = x/2 \) (it slopes up but is flatter than \( y=x \)). The solution is the area where \( y \) is smaller than \( -x \) (below \( y=-x \)) and also smaller than \( x/2 \) (below \( y=x/2 \)). Find the part where both shaded areas overlap.

🎯 Exam Tip: When rearranging inequalities like \( x \le -y \) to \( y \le -x \), be careful with the inequality sign if you multiply or divide by a negative number. Also, always use a solid line for \( \le \) or \( \ge \) and a dashed line for \( < \) or \( > \).

Question 4. \( y > 2-x \) and \( y > x-2 \)
Answer:The given inequalities are: 1. \( y > 2-x \) (or \( x+y > 2 \)) 2. \( y > x-2 \) (or \( x-y < 2 \)) For the region \( y > 2-x \): The line \( y = 2-x \) (or \( x+y = 2 \)) meets the coordinate axes at \( A(2, 0) \) and \( B(0, 2) \). Testing the origin \( (0, 0) \) gives \( 0 > 2-0 \), which is \( 0 > 2 \), false. So, the region not containing the origin gives the solution set for \( y > 2-x \). This means the area above the line \( x+y=2 \). The line should be dashed. For the region \( y > x-2 \): The line \( y = x-2 \) (or \( x-y = 2 \)) meets the coordinate axes at \( C(2, 0) \) and \( D(0, -2) \). Testing the origin \( (0, 0) \) gives \( 0 > 0-2 \), which is \( 0 > -2 \), true. So, the region containing the origin gives the solution set for \( y > x-2 \). This means the area above the line \( x-y=2 \). The line should be dashed. Both lines, \( y = 2-x \) and \( y = x-2 \), intersect at the point \( P(2, 0) \). The solution set includes all points in the common shaded region, which is where both conditions are satisfied.
In simple words: Draw two dashed lines: one for \( y = 2-x \) and another for \( y = x-2 \). These lines will cross each other. Find the area that is above both of these lines. This V-shaped area, opening upwards, is the solution.

🎯 Exam Tip: When both inequalities are strict (> or <), the intersection points of the boundary lines are not included in the solution set. If the lines are parallel and the regions point away from each other, there might be no common solution, and you should state "No Solution".

Question 5. \( -2y > x - 8 \) and \( 4y > x - 12 \)
Answer:The given inequalities are: 1. \( -2y > x - 8 \) (or \( 2y < -x + 8 \)) 2. \( 4y > x - 12 \) (or \( 4y - x > -12 \)) For the region \( -2y > x - 8 \): The boundary line is \( -2y = x - 8 \) (or \( x + 2y = 8 \)). This line meets the coordinate axes at \( A(8, 0) \) and \( B(0, 4) \). Testing the origin \( (0, 0) \) gives \( -2(0) > 0 - 8 \), which is \( 0 > -8 \), true. So, the region containing the origin gives the solution set for \( -2y > x - 8 \). The line should be dashed. For the region \( 4y > x - 12 \): The boundary line is \( 4y = x - 12 \) (or \( x - 4y = 12 \)). This line meets the coordinate axes at \( C(12, 0) \) and \( D(0, -3) \). Testing the origin \( (0, 0) \) gives \( 4(0) > 0 - 12 \), which is \( 0 > -12 \), true. So, the region containing the origin gives the solution set for \( 4y > x - 12 \). The line should be dashed. The two lines, \( x + 2y = 8 \) and \( x - 4y = 12 \), intersect at point \( P(\frac{28}{3}, -\frac{2}{3}) \). All points in the common shaded region represent the solution set for the given system of inequalities.
In simple words: Draw two dashed lines for the given equations. For the first line, \( x+2y=8 \), and for the second line, \( x-4y=12 \). Since the origin \( (0,0) \) satisfies both inequalities, shade the region that contains the origin and is between these two lines. This overlapping area is the solution.

🎯 Exam Tip: When both inequalities point towards the origin, the solution region will be bounded by the lines and contain the origin. Always label your axes and intercepts clearly.

Question 6. \( 3x - 4y < 6 \) and \( 2x - y < -1 \)
Answer:The given inequalities are: 1. \( 3x - 4y < 6 \) 2. \( 2x - y < -1 \) For the region \( 3x - 4y < 6 \): The boundary line is \( 3x - 4y = 6 \). This line meets the coordinate axes at \( A(2, 0) \) and \( B(0, -\frac{3}{2}) \). Testing the origin \( (0, 0) \) gives \( 3(0) - 4(0) < 6 \), which is \( 0 < 6 \), true. So, the region containing the origin gives the solution set for \( 3x - 4y < 6 \). The line should be dashed. For the region \( 2x - y < -1 \): The boundary line is \( 2x - y = -1 \). This line meets the coordinate axes at \( C(-\frac{1}{2}, 0) \) and \( D(0, 1) \). Testing the origin \( (0, 0) \) gives \( 2(0) - 0 < -1 \), which is \( 0 < -1 \), false. So, the region not containing the origin gives the solution set for \( 2x - y < -1 \). This means the area above the line \( 2x-y = -1 \). The line should be dashed. The two lines, \( 3x - 4y = 6 \) and \( 2x - y = -1 \), intersect at the point \( P(-2, -3) \). All points in the common shaded region represent the solution set for the given system of inequalities.
In simple words: Draw two dashed lines for \( 3x - 4y = 6 \) and \( 2x - y = -1 \). For the first inequality, shade the side of the line that includes the origin. For the second inequality, shade the side that does NOT include the origin. The area where these two shaded regions overlap is the solution.

🎯 Exam Tip: Always clearly indicate test points and whether they satisfy the inequality. Pay close attention to the origin test; if the origin fails, shade the opposite side of the line.

Question 7. \( x < y < x + 3 \)
Answer:The inequality \( x < y < x + 3 \) represents two separate inequalities: 1. \( x < y \) (or \( y > x \)) 2. \( y < x + 3 \) For the region \( y > x \): The line \( y = x \) passes through \( (0, 0) \) and makes an angle of \( 45^\circ \) with the positive x-axis. Testing \( (0, 1) \) gives \( 1 > 0 \), which is true. So, the region containing \( (0, 1) \) (above \( y=x \)) gives the solution set. The line should be dashed. For the region \( y < x + 3 \): The line \( y = x + 3 \) meets the coordinate axes at \( A(-3, 0) \) and \( B(0, 3) \). Testing the origin \( (0, 0) \) gives \( 0 < 0 + 3 \), which is \( 0 < 3 \), true. So, the region containing the origin (below \( y = x+3 \)) gives the solution set. The line should be dashed. Notice that the lines \( y=x \) and \( y=x+3 \) are parallel to each other. All points in the common shaded region between these two parallel lines give the solution set for the system of inequalities.
In simple words: Draw two dashed lines: one for \( y = x \) and one for \( y = x + 3 \). These lines will be parallel. The solution is the strip of area that lies between these two parallel lines.

🎯 Exam Tip: When dealing with combined inequalities like \( A < B < C \), always split them into two separate inequalities: \( A < B \) and \( B < C \). If the boundary lines are parallel, the solution is usually the region between them.

Question 8. \( 2x + 1 \ge y \ge 2x - 3 \)
Answer:The inequality \( 2x + 1 \ge y \ge 2x - 3 \) represents two separate inequalities: 1. \( 2x + 1 \ge y \) (or \( y \le 2x + 1 \)) 2. \( y \ge 2x - 3 \) For the region \( y \le 2x + 1 \): The boundary line is \( y = 2x + 1 \). This line intersects the coordinate axes at \( A(-\frac{1}{2}, 0) \) and \( B(0, 1) \). Testing the origin \( (0, 0) \) gives \( 0 \le 2(0) + 1 \), which is \( 0 \le 1 \), true. So, the region containing the origin gives the solution set. The line should be solid. For the region \( y \ge 2x - 3 \): The boundary line is \( y = 2x - 3 \). This line meets the coordinate axes at \( C(\frac{3}{2}, 0) \) and \( D(0, -3) \). Testing the origin \( (0, 0) \) gives \( 0 \ge 2(0) - 3 \), which is \( 0 \ge -3 \), true. So, the region containing the origin gives the solution set. The line should be solid. The lines \( y = 2x + 1 \) and \( y = 2x - 3 \) are parallel to each other because they have the same slope \( (m=2) \). All points in the common shaded region between these two parallel lines represent the solution set for the system of inequalities.
In simple words: Draw two solid lines: one for \( y = 2x + 1 \) and another for \( y = 2x - 3 \). These lines will be parallel. The solution is the strip of area that lies exactly between these two parallel lines.

🎯 Exam Tip: When inequalities include "equal to" (≤ or ≥), the boundary lines are part of the solution set and should be drawn as solid lines. Parallel lines usually mean the solution is a band or strip between them.

Question 9. \( 2 < x < 4 \) and \( 2y > x-6 \)
Answer:The given inequalities are: 1. \( 2 < x < 4 \) (which means \( x > 2 \) and \( x < 4 \)) 2. \( 2y > x - 6 \) For the region \( 2 < x < 4 \): The line \( x = 2 \) is parallel to the y-axis and passes through \( A(2, 0) \). The line \( x = 4 \) is also parallel to the y-axis and passes through \( B(4, 0) \). The region \( 2 < x < 4 \) represents the strip between these two dashed parallel lines. For the region \( 2y > x - 6 \): The boundary line is \( 2y = x - 6 \). This line meets the coordinate axes at \( C(6, 0) \) and \( D(0, -3) \). Testing the origin \( (0, 0) \) gives \( 2(0) > 0 - 6 \), which is \( 0 > -6 \), true. So, the region containing the origin gives the solution set for \( 2y > x - 6 \). This means the area above the line \( 2y = x-6 \). The line should be dashed. The line \( 2y = x - 6 \) meets \( x = 2 \) at \( E(2, -2) \) and \( x = 4 \) at \( F(4, -1) \). All points in the common shaded region, which is the intersection of the strip \( 2 < x < 4 \) and the region \( 2y > x - 6 \), represent the solution set.
In simple words: First, draw two dashed vertical lines: one at \( x = 2 \) and another at \( x = 4 \). The area between these lines is one part of the solution. Then, draw a dashed line for \( 2y = x - 6 \). Since the origin \( (0,0) \) satisfies this, shade the area above this line. The final solution is the area where the vertical strip and the region above \( 2y = x - 6 \) overlap.

🎯 Exam Tip: Always tackle compound inequalities like \( 2 < x < 4 \) by breaking them into simpler inequalities (\( x > 2 \) and \( x < 4 \)). This helps visualize the bounded region more easily.

Question 10. \( y \le x \), \( x + y < 4 \), and \( x - 2y < 1 \)
Answer:The given inequalities are: 1. \( y \le x \) 2. \( x + y < 4 \) 3. \( x - 2y < 1 \) For the region \( y \le x \): The boundary line \( y = x \) passes through the origin \( (0, 0) \) and makes an angle of \( 45^\circ \) with the x-axis. Testing \( (1, 0) \) gives \( 0 \le 1 \), which is true. So, the region containing \( (1, 0) \) (below \( y=x \)) gives the solution set. The line should be solid. For the region \( x + y < 4 \): The boundary line \( x + y = 4 \) meets the coordinate axes at \( A(4, 0) \) and \( B(0, 4) \). Testing the origin \( (0, 0) \) gives \( 0 + 0 < 4 \), which is \( 0 < 4 \), true. So, the region containing the origin (below \( x+y=4 \)) gives the solution set. The line should be dashed. For the region \( x - 2y < 1 \): The boundary line \( x - 2y = 1 \) meets the coordinate axes at \( C(1, 0) \) and \( D(0, -\frac{1}{2}) \). Testing the origin \( (0, 0) \) gives \( 0 - 2(0) < 1 \), which is \( 0 < 1 \), true. So, the region containing the origin (above \( x-2y=1 \)) gives the solution set. The line should be dashed. The lines intersect at: * \( y = x \) and \( x + y = 4 \) intersect at \( P(2, 2) \). * \( y = x \) and \( x - 2y = 1 \) intersect at \( Q(-1, -1) \). * \( x + y = 4 \) and \( x - 2y = 1 \) intersect at \( R(3, 1) \). All points in the common shaded region, which is bounded by these three lines, represent the solution set for the given system of inequalities.
In simple words: Draw three lines: a solid line for \( y=x \), a dashed line for \( x+y=4 \), and a dashed line for \( x-2y=1 \). Shade the area below \( y=x \), below \( x+y=4 \), and above \( x-2y=1 \). The common region where all three shaded areas overlap is the solution, forming a triangle.

🎯 Exam Tip: For systems with three or more inequalities, identifying the intersection points (vertices) of the boundary lines is crucial. The solution set will typically be a polygon formed by these lines. Always pay attention to whether lines should be solid or dashed.

Question 11. \( 2x + y < 2 \), \( x - y > -2 \), and \( x + y > -2 \)
Answer:The given inequalities are: 1. \( 2x + y < 2 \) 2. \( x - y > -2 \) 3. \( x + y > -2 \) For the region \( 2x + y < 2 \): The boundary line is \( 2x + y = 2 \). This line meets the coordinate axes at \( A(1, 0) \) and \( B(0, 2) \). Testing the origin \( (0, 0) \) gives \( 2(0) + 0 < 2 \), which is \( 0 < 2 \), true. So, the region containing the origin gives the solution set. The line should be dashed. For the region \( x - y > -2 \): The boundary line is \( x - y = -2 \). This line meets the coordinate axes at \( D(-2, 0) \) and \( E(0, 2) \). Testing the origin \( (0, 0) \) gives \( 0 - 0 > -2 \), which is \( 0 > -2 \), true. So, the region containing the origin gives the solution set. The line should be dashed. For the region \( x + y > -2 \): The boundary line is \( x + y = -2 \). This line meets the coordinate axes at \( E(-2, 0) \) and \( F(0, -2) \). Testing the origin \( (0, 0) \) gives \( 0 + 0 > -2 \), which is \( 0 > -2 \), true. So, the region containing the origin gives the solution set. The line should be dashed. The lines intersect at: * \( 2x + y = 2 \) and \( x - y = -2 \) intersect at \( P(0, 2) \). * \( 2x + y = 2 \) and \( x + y = -2 \) intersect at \( Q(4, -6) \). * \( x - y = -2 \) and \( x + y = -2 \) intersect at \( R(-2, 0) \). All points in the common shaded region represent the solution set for the given system of inequalities.
In simple words: Draw three dashed lines for the given equations. Since the origin \( (0,0) \) satisfies all three inequalities, shade the region that contains the origin and is enclosed by these three lines. This will form a triangular area.

🎯 Exam Tip: When all inequalities require the region containing the origin, the solution will be the polygon formed by the intersection of the "origin side" of all lines. Make sure your lines are dashed for strict inequalities.

Question 12. \( x + y \ge 3 \), \( x - y \le 3 \), and \( x + 5y \ge 15 \)
Answer:The given inequalities are: 1. \( x + y \ge 3 \) 2. \( x - y \le 3 \) 3. \( x + 5y \ge 15 \) For the region \( x + y \ge 3 \): The boundary line is \( x + y = 3 \). This line meets the coordinate axes at \( A(3, 0) \) and \( B(0, 3) \). Testing the origin \( (0, 0) \) gives \( 0 + 0 \ge 3 \), which is \( 0 \ge 3 \), false. So, the region not containing the origin gives the solution set (above the line). The line should be solid. For the region \( x - y \le 3 \): The boundary line is \( x - y = 3 \). This line meets the coordinate axes at \( C(3, 0) \) and \( D(0, -3) \). Testing the origin \( (0, 0) \) gives \( 0 - 0 \le 3 \), which is \( 0 \le 3 \), true. So, the region containing the origin gives the solution set (above the line). The line should be solid. For the region \( x + 5y \ge 15 \): The boundary line is \( x + 5y = 15 \). This line meets the coordinate axes at \( E(15, 0) \) and \( F(0, 3) \). Testing the origin \( (0, 0) \) gives \( 0 + 5(0) \ge 15 \), which is \( 0 \ge 15 \), false. So, the region not containing the origin gives the solution set (above the line). The line should be solid. All points in the common shaded region represent the solution set for the given system of inequalities. The final region is unbounded.
In simple words: Draw three solid lines for the given equations. For \( x+y=3 \) and \( x+5y=15 \), shade the side that does NOT include the origin. For \( x-y=3 \), shade the side that DOES include the origin. The common region where all three shaded areas overlap is the solution, and it will stretch out infinitely in one direction.

🎯 Exam Tip: When a region is unbounded, it means the solution set extends infinitely in at least one direction. Always test the origin for each inequality to determine which side of the line to shade. Double-check your arithmetic for intercepts, especially when dealing with fractions.

Question 13. \( y - 2 < 0 \), \( x + 3 > 0 \), \( 2y + x < 2 \), and \( 3y + 3 < 2x \)
Answer:The given inequalities are: 1. \( y - 2 < 0 \) (or \( y < 2 \)) 2. \( x + 3 > 0 \) (or \( x > -3 \)) 3. \( 2y + x < 2 \) 4. \( 3y + 3 < 2x \) (or \( 3y - 2x < -3 \)) For the region \( y < 2 \): The boundary line \( y = 2 \) passes through \( (0, 2) \) and is parallel to the x-axis. Testing \( (0, 0) \) gives \( 0 < 2 \), which is true. So, the region containing the origin (below \( y=2 \)) gives the solution set. The line should be dashed. For the region \( x > -3 \): The boundary line \( x = -3 \) passes through \( (-3, 0) \) and is parallel to the y-axis. Testing \( (0, 0) \) gives \( 0 > -3 \), which is true. So, the region containing the origin (to the right of \( x=-3 \)) gives the solution set. The line should be dashed. For the region \( 2y + x < 2 \): The boundary line \( 2y + x = 2 \) meets the coordinate axes at \( A(2, 0) \) and \( B(0, 1) \). Testing \( (0, 0) \) gives \( 2(0) + 0 < 2 \), which is \( 0 < 2 \), true. So, the region containing the origin (below \( 2y+x=2 \)) gives the solution set. The line should be dashed. For the region \( 3y + 3 < 2x \): The boundary line \( 3y + 3 = 2x \) meets the coordinate axes at \( C(\frac{3}{2}, 0) \) and \( D(0, -1) \). Testing \( (0, 0) \) gives \( 3(0) + 3 < 2(0) \), which is \( 3 < 0 \), false. So, the region not containing the origin gives the solution set (below the line \( y = \frac{2x-3}{3} \)). The line should be dashed. All points in the common shaded region represent the solution set for the given system of inequalities.
In simple words: Draw four dashed lines for the given equations. For \( y=2 \), shade below it. For \( x=-3 \), shade to its right. For \( 2y+x=2 \), shade below it. For \( 3y+3=2x \), shade the side that does NOT include the origin (this means above it). The overlapping area of all these shaded regions is the solution.

🎯 Exam Tip: With multiple inequalities, it's helpful to shade each region lightly with different patterns or colors, then identify the area where all patterns overlap. Always be careful about how the "not containing origin" rule applies for inequalities.

Question 14. \( 2y - 6 \le x \), \( 2y + 3x \ge -6 \), \( 5y + 15 \ge 2x \), and \( 3y + 5x \le 15 \)
Answer:The given inequalities are: 1. \( 2y - 6 \le x \) (or \( x - 2y \ge -6 \)) 2. \( 2y + 3x \ge -6 \) 3. \( 5y + 15 \ge 2x \) (or \( 2x - 5y \le 15 \)) 4. \( 3y + 5x \le 15 \) For the region \( 2y - 6 \le x \): The boundary line is \( x - 2y = -6 \). This line meets the coordinate axes at \( A(-6, 0) \) and \( B(0, 3) \). Testing the origin \( (0, 0) \) gives \( 0 - 2(0) \ge -6 \), which is \( 0 \ge -6 \), true. So, the region containing the origin gives the solution set. The line should be solid. For the region \( 2y + 3x \ge -6 \): The boundary line is \( 2y + 3x = -6 \). This line meets the coordinate axes at \( C(-2, 0) \) and \( D(0, -3) \). Testing the origin \( (0, 0) \) gives \( 2(0) + 3(0) \ge -6 \), which is \( 0 \ge -6 \), true. So, the region containing the origin gives the solution set. The line should be solid. For the region \( 5y + 15 \ge 2x \): The boundary line is \( 2x - 5y = 15 \). This line meets the coordinate axes at \( E(\frac{15}{2}, 0) \) (or \( (7.5, 0) \)) and \( F(0, -3) \). Testing the origin \( (0, 0) \) gives \( 5(0) + 15 \ge 2(0) \), which is \( 15 \ge 0 \), true. So, the region containing the origin gives the solution set. The line should be solid. For the region \( 3y + 5x \le 15 \): The boundary line is \( 3y + 5x = 15 \). This line meets the coordinate axes at \( G(3, 0) \) and \( M(0, 5) \). Testing the origin \( (0, 0) \) gives \( 3(0) + 5(0) \le 15 \), which is \( 0 \le 15 \), true. So, the region containing the origin gives the solution set. The line should be solid. All points in the common shaded region, which contains the origin and is bounded by these four lines, represent the solution set for the given system of inequalities.
In simple words: Draw four solid lines for the given equations. Since the origin \( (0,0) \) satisfies all four inequalities, the solution is the region that contains the origin and is enclosed by these four lines. This will form a polygon.

🎯 Exam Tip: When all inequalities hold true for the origin, the feasible region is a polygon enclosing the origin. For questions with many inequalities, organize your work by finding intercepts for each line and testing the origin systematically.

Question 15. (i) Find the linear constraints for which the shaded area in the figure given below is the solution set.
Answer:The shaded area in the first figure is bounded by the following lines: 1. \( -7x + 4y = 14 \) 2. \( 3x + 4y = 18 \) 3. \( 2x + 3y = 3 \) 4. \( x - 6y = 3 \) Let's determine the inequality for each line by testing the origin \( (0, 0) \): For the line \( -7x + 4y = 14 \): The shaded area and the origin \( (0, 0) \) are on the same side of this line. Testing \( (0,0) \): \( -7(0) + 4(0) = 0 \). Since \( 0 < 14 \), the corresponding constraint is \( -7x + 4y \le 14 \). For the line \( 3x + 4y = 18 \): The shaded area and the origin \( (0, 0) \) are on the same side of this line. Testing \( (0,0) \): \( 3(0) + 4(0) = 0 \). Since \( 0 < 18 \), the corresponding constraint is \( 3x + 4y \le 18 \). For the line \( 2x + 3y = 3 \): The shaded area and the origin \( (0, 0) \) are on the opposite side of this line. Testing \( (0,0) \): \( 2(0) + 3(0) = 0 \). Since \( 0 < 3 \), and the shaded region is opposite to the origin, the corresponding constraint is \( 2x + 3y \ge 3 \). For the line \( x - 6y = 3 \): The origin \( (0, 0) \) and the shaded area are on the same side of this line. Testing \( (0,0) \): \( 0 - 6(0) = 0 \). Since \( 0 < 3 \), the corresponding constraint is \( x - 6y \le 3 \). Additionally, the shaded area lies in the first quadrant of the XOY plane, which implies: \( x \ge 0 \) \( y \ge 0 \) Thus, the required system of constraints is: \( -7x + 4y \le 14 \) \( 3x + 4y \le 18 \) \( 2x + 3y \ge 3 \) \( x - 6y \le 3 \) \( x \ge 0 \) \( y \ge 0 \)
In simple words: Look at each line in the graph. For each line, check if the shaded area is on the same side as the origin \( (0,0) \) or the opposite side. If it's the same side, use \( \le \) or \( \ge \) that keeps \( (0,0) \) true. If it's the opposite side, use \( \le \) or \( \ge \) that makes \( (0,0) \) false. Also, since the shaded part is only in the top-right quarter of the graph, add \( x \ge 0 \) and \( y \ge 0 \).

🎯 Exam Tip: Always include the non-negativity constraints \( x \ge 0 \) and \( y \ge 0 \) if the shaded region is confined to the first quadrant, as this is a common condition in linear programming problems.

(ii) Here the shaded area bounded by the lines \( 3x + y - 6 = 0 \), \( 4x + 9y - 36 = 0 \), \( x + 3y - 6 = 0 \), and \( 4x - 3y - 12 = 0 \).
Answer:The shaded area in the second figure is bounded by the following lines: 1. \( 3x + y - 6 = 0 \) (or \( 3x + y = 6 \)) 2. \( 4x + 9y - 36 = 0 \) (or \( 4x + 9y = 36 \)) 3. \( x + 3y - 6 = 0 \) (or \( x + 3y = 6 \)) 4. \( 4x - 3y - 12 = 0 \) (or \( 4x - 3y = 12 \)) Let's determine the inequality for each line by testing the origin \( (0, 0) \): For the line \( 3x + y = 6 \): The shaded area and the origin \( (0, 0) \) are on the opposite side of this line. Testing \( (0,0) \): \( 3(0) + 0 = 0 \). Since \( 0 < 6 \), and the shaded region is opposite to the origin, the corresponding constraint is \( 3x + y \ge 6 \). For the line \( 4x + 9y = 36 \): The shaded area and the origin \( (0, 0) \) are on the same side of this line. Testing \( (0,0) \): \( 4(0) + 9(0) = 0 \). Since \( 0 < 36 \), the corresponding constraint is \( 4x + 9y \le 36 \). For the line \( x + 3y = 6 \): The shaded area and the origin \( (0, 0) \) are on the opposite side of this line. Testing \( (0,0) \): \( 0 + 3(0) = 0 \). Since \( 0 < 6 \), and the shaded region is opposite to the origin, the corresponding constraint is \( x + 3y \ge 6 \). For the line \( 4x - 3y = 12 \): The shaded area and the origin \( (0, 0) \) are on the same side of this line. Testing \( (0,0) \): \( 4(0) - 3(0) = 0 \). Since \( 0 < 12 \), the corresponding constraint is \( 4x - 3y \le 12 \). Additionally, the shaded area lies in the first quadrant of the XOY plane, which implies: \( x \ge 0 \) \( y \ge 0 \) Thus, the required system of constraints is: \( 3x + y \ge 6 \) \( 4x + 9y \le 36 \) \( x + 3y \ge 6 \) \( 4x - 3y \le 12 \) \( x \ge 0 \) \( y \ge 0 \)
In simple words: For each of the four lines shown, check whether the shaded area is on the same side as the origin \( (0,0) \) or not. This tells you if you need \( \le \) or \( \ge \). Also, because the shaded area is only in the top-right quarter of the graph, remember to add \( x \ge 0 \) and \( y \ge 0 \) as conditions.

🎯 Exam Tip: When given a shaded region and asked to find the inequalities, first identify all boundary lines. Then, for each line, use a test point (like the origin) to correctly determine the direction of the inequality sign. Don't forget the \( x \ge 0, y \ge 0 \) constraints for regions in the first quadrant.

Question 16. \( x + y \geq 3 \), \( 7x + 6y \leq 42 \), \( x \leq 5 \), \( y \leq 4 \), \( x, y > 0 \).
Answer: The given system of inequalities defines a region. Let's look at each one:
1. For \( x + y \geq 3 \): The line \( x + y = 3 \) passes through \( A(3, 0) \) and \( B(0, 3) \). Since \( (0, 0) \) does not satisfy \( x + y \geq 3 \), the solution region is the area not containing the origin. This line helps form one boundary of the feasible region.
2. For \( 7x + 6y \leq 42 \): The line \( 7x + 6y = 42 \) meets the axes at \( C(6, 0) \) and \( (0, 7) \). Since \( (0, 0) \) satisfies this inequality, the solution region includes the origin. This line cuts off a part of the plane.
3. For \( x \leq 5 \): The line \( x = 5 \) is parallel to the y-axis and passes through \( (5, 0) \). The region satisfying this inequality is to the left of or on this line, as \( (0, 0) \) satisfies it.
4. For \( y \leq 4 \): The line \( y = 4 \) is parallel to the x-axis and passes through \( (0, 4) \). The region satisfying this inequality is below or on this line, as \( (0, 0) \) satisfies it.
5. For \( x > 0, y > 0 \): These mean the solution must be in the first quadrant of the coordinate plane.
The common shaded region that satisfies all these inequalities together is the solution set. This region represents all possible points that fit all the conditions at once.In simple words: We need to find the area on a graph where all these math rules are true at the same time. We draw each line, then check which side of the line is correct using the point \( (0, 0) \) if possible. The area where all the shaded parts overlap is our final answer.

🎯 Exam Tip: When dealing with multiple inequalities, always check the origin \( (0,0) \) against each inequality to determine which side of the line to shade. This is a quick way to find the correct region if the line does not pass through the origin.

Question 17. Find the region where all the inequations \( x + y \geq 0 \), \( 2x + y \leq 4 \), \( x \geq 0 \) and \( y \leq 2 \) hold good. Find the coordinates of the vertices of the region.
Answer: The given system of inequalities defines a specific region. Let's analyze each inequality:
1. For \( x + y \geq 0 \): The line \( x + y = 0 \) passes through the origin \( (0, 0) \). For points not on this line, choose a test point like \( (1, 0) \). Since \( 1 + 0 \geq 0 \) is true, the region containing \( (1, 0) \) is the solution. This defines the area above or on the line \( y = -x \).
2. For \( 2x + y \leq 4 \): The line \( 2x + y = 4 \) meets the coordinate axes at \( A(2, 0) \) and \( B(0, 4) \). Since \( (0, 0) \) satisfies \( 2(0) + 0 \leq 4 \), the solution region includes the origin. This means the area below or on the line \( 2x + y = 4 \).
3. For \( x \geq 0 \): This represents the right side of the y-axis, including the axis itself.
4. For \( y \leq 2 \): The line \( y = 2 \) is parallel to the x-axis and passes through \( (0, 2) \). Since \( (0, 0) \) satisfies \( 0 \leq 2 \), the solution region is below or on the line \( y = 2 \).
The line \( y = 2 \) intersects with \( x + y = 0 \) at \( Q(-2, 2) \) and with \( 2x + y = 4 \) at \( P(1, 2) \).
The common shaded region that satisfies all these inequalities is OAPQ. The vertices (corner points) of this region are:
- \( O(0, 0) \)
- \( A(2, 0) \)
- \( P(1, 2) \) (intersection of \( y = 2 \) and \( 2x + y = 4 \))
- \( Q(-2, 2) \) (intersection of \( y = 2 \) and \( x + y = 0 \))
These vertices define the boundaries of the feasible solution area.In simple words: We are looking for an area on the graph that follows four specific rules. We draw lines for each rule and decide which side of the line is correct. The space where all the correct sides overlap is our answer. The points where these lines cross over at the corners of this special area are called vertices.

🎯 Exam Tip: When finding vertices of the feasible region, remember to solve the equations of the intersecting lines to get the exact coordinates. This ensures accuracy in defining the boundary points.

Question 18. Find the region where all the inequations \( x + y \leq 6 \), \( x \geq y \), \( x \geq 0 \), \( y \geq 0 \) hold good. Find the ordered pairs of the vertices of the region so formed.
Answer: We need to find the region where all these inequalities are true at the same time:
1. For \( x + y \leq 6 \): The line \( x + y = 6 \) intersects the coordinate axes at \( A(6, 0) \) and \( B(0, 6) \). Since \( (0, 0) \) satisfies \( 0 + 0 \leq 6 \), the solution region includes the origin.
2. For \( x \geq y \): The line \( x = y \) passes through the origin and makes a 45° angle with the positive x-axis. Since \( (0, 0) \) satisfies \( 0 \geq 0 \), this inequality is true at the origin. If you pick a point not on the line, say \( (1, 0) \), it satisfies \( 1 \geq 0 \), so the region below the line \( y=x \) is included.
3. For \( x \geq 0 \) and \( y \geq 0 \): These conditions mean the solution must be located in the first quadrant of the coordinate plane.
The lines \( x = y \) and \( x + y = 6 \) intersect at point \( P(3, 3) \). To find this, substitute \( y=x \) into \( x+y=6 \), which gives \( x+x=6 \), so \( 2x=6 \), and \( x=3 \). Then \( y=3 \).
The common shaded region that satisfies all these inequalities is OAP. The ordered pairs of the vertices of this region are:
- \( O(0, 0) \)
- \( A(6, 0) \) (x-intercept of \( x+y=6 \))
- \( P(3, 3) \) (intersection of \( x=y \) and \( x+y=6 \))
These points form the corners of the solution area.In simple words: We have several mathematical rules, and we need to find the area on a graph where all of them are true together. We draw the boundary lines for each rule. The area that is "correct" for every rule is our solution. The important corner points of this solution area are called vertices.

🎯 Exam Tip: Always clearly label your lines on the graph and show the intersection points to make it easy for the examiner to follow your solution and ensure you get full marks.