Linear Inequalities Ex 6.3 Class 11 NCERT Solutions

Understanding Systems of Linear Inequalities

A 'system' of linear inequalities is simply a collection of two or more linear inequalities involving the same variables (usually x and y). The 'solution' to a single inequality is a half-plane on the graph. The 'solution' to a system of inequalities is the region where all these half-planes overlap. This overlapping area is called the feasible region. Every single point (x, y) within this feasible region will satisfy every single inequality in the system. If a point is in the shaded region for one inequality but not another, it is not part of the solution to the system. Exercise 6.3 is all about the graphical method to find this common region. The process involves drawing the boundary line for each inequality and then shading the correct region, ultimately finding the common ground for all.

Step-by-Step Guide: Solving Systems Graphically

1. Step 1: Convert Inequalities to Equations — For each inequality in the system, replace the inequality sign (<, >, ≤, ≥) with an equals sign (=) to get the equation of the boundary line.
2. Step 2: Plot the Boundary Lines — Draw each line on the Cartesian plane. Use a solid line if the inequality is ≤ or ≥ (inclusive). Use a dashed line if the inequality is < or > (exclusive).
3. Step 3: Perform the Test Point Check — For each inequality, choose a test point that is not on the line. The origin (0,0) is the easiest choice. Substitute the coordinates of the test point into the original inequality.
4. Step 4: Shade the Individual Solution Regions — If the test point makes the inequality true, shade the entire half-plane containing that point. If it's false, shade the other side. It's helpful to use different shading styles or light colours for each region.
5. Step 5: Identify the Common Feasible Region — The final solution is the region on the graph where all the shaded areas overlap. This common area is the 'feasible region'. Clearly mark or darken this final region.

Worked Examples: Solving Systems of Inequalities

• Example 1: Solve the system of inequalities graphically: x + y ≤ 6 and x + y ≥ 4 Step 1: Convert to equations: x + y = 6 (Line 1) and x + y = 4 (Line 2). Step 2: Plot the lines. Both will be solid lines because of the ≤ and ≥ signs. Both lines are parallel. Step 3 & 4 (For x + y ≤ 6): - Test point (0,0): 0 + 0 ≤ 6 which is 0 ≤ 6. This is TRUE. - Shade the region containing the origin (towards the origin) for the line x + y = 6. Step 3 & 4 (For x + y ≥ 4): - Test point (0,0): 0 + 0 ≥ 4 which is 0 ≥ 4. This is FALSE. - Shade the region not containing the origin (away from the origin) for the line x + y = 4. Step 5: The feasible region is the area between the two parallel lines x + y = 4 and x + y = 6, including the lines themselves.
• Example 2: Solve the system graphically: 2x + y ≥ 6, 3x + 4y ≤ 12 Step 1: Equations are 2x + y = 6 and 3x + 4y = 12. Both are solid lines. Step 2: - For 2x+y=6: Points are (3,0) and (0,6). - For 3x+4y=12: Points are (4,0) and (0,3). - Plot these solid lines. Step 3 & 4 (For 2x + y ≥ 6): - Test point (0,0): 2(0) + 0 ≥ 6 → 0 ≥ 6. FALSE. - Shade the side of 2x+y=6 that does not contain the origin. Step 3 & 4 (For 3x + 4y ≤ 12): - Test point (0,0): 3(0) + 4(0) ≤ 12 → 0 ≤ 12. TRUE. - Shade the side of 3x+4y=12 that does contain the origin. Step 5: The feasible region is the overlapping area of the two shadings. It will be a triangular-like region bounded by the two lines and the y-axis.

Exam Tips & Common Mistakes

Pay close attention to these points to avoid losing marks:

• Solid vs. Dashed Lines: This is a very common mistake. Always use a solid line for ≤ and ≥ to show that points on the line are included in the solution. Use a dashed line for < and > to show they are not. An incorrect line type can cost you marks.
• Shading the Correct Region: If the test point (0,0) gives a TRUE result, shade towards the origin. If it's FALSE, shade away from it. Always double-check your substitution.
• When (0,0) Fails: If a line passes through the origin (e.g., y = 2x or x - 3y = 0), you cannot use (0,0) as a test point. Choose any other simple point not on the line, like (1,0) or (0,1).
• Clearly Mark the Final Region: After shading for each inequality, it's crucial to clearly indicate the final, overlapping 'feasible region'. You can do this by using a darker shade or by explicitly labelling it. The examiner needs to see that you've identified the common solution.

Practice Questions with Solutions

• Q: Solve the following system of inequalities graphically: x ≥ 3, y ≥ 2. A: Step 1: The corresponding equations are x = 3 (a vertical line) and y = 2 (a horizontal line). Step 2: Draw both as solid lines, since the inequalities are ≥. Step 3: For x ≥ 3, shading will be to the right of the line x = 3. For y ≥ 2, shading will be above the line y = 2. Final answer: The feasible region is the intersection of these two areas, which is the quadrant-like region starting from the point (3,2) and extending infinitely up and to the right.
• Q: Solve the system graphically: x + 2y ≤ 8, 2x + y ≤ 8, x ≥ 0, y ≥ 0. A: Step 1: Equations are x + 2y = 8, 2x + y = 8, x=0 (y-axis), and y=0 (x-axis). All lines are solid. Step 2: Plot the lines. x + 2y = 8 passes through (8,0) and (0,4). 2x + y = 8 passes through (4,0) and (0,8). Step 3: The inequalities x ≥ 0 and y ≥ 0 restrict the solution to the first quadrant. Step 4: For x + 2y ≤ 8, test (0,0): 0 ≤ 8 is TRUE. Shade towards the origin. For 2x + y ≤ 8, test (0,0): 0 ≤ 8 is TRUE. Shade towards the origin. Final answer: The feasible region is a quadrilateral in the first quadrant with vertices at (0,0), (4,0), the intersection point of the two lines, and (0,4).
• Q: Solve the system graphically: x + y < 5, x - y > 3. A: Step 1: The equations are x + y = 5 and x - y = 3. Both lines must be dashed because of the < and > signs. Step 2: Plot the dashed lines. x+y=5 passes through (5,0) and (0,5). x-y=3 passes through (3,0) and (0,-3). Step 3: For x + y < 5, test (0,0): 0 < 5 is TRUE. Shade the region containing the origin. Step 4: For x - y > 3, test (0,0): 0 > 3 is FALSE. Shade the region not containing the origin. Final answer: The feasible region is the overlapping area, which is an infinite region bounded by the two dashed lines.
• Q: Find the solution region for the inequalities: 8x + 3y ≤ 100, x ≥ 0, y ≥ 0. A: Step 1: The equations are 8x + 3y = 100, x=0, and y=0. All lines are solid. Step 2: The lines x=0 (y-axis) and y=0 (x-axis) are the boundaries. 8x + 3y = 100 passes through (12.5, 0) and (0, 33.3). Step 3: x ≥ 0 and y ≥ 0 together mean the solution is in the first quadrant. Step 4: For 8x + 3y ≤ 100, test (0,0): 0 ≤ 100 is TRUE. Shade towards the origin. Final answer: The feasible region is the triangle formed by the origin (0,0), the x-intercept (12.5, 0), and the y-intercept (0, 33.3), including the boundary lines.

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