RS Aggarwal Solutions for Class 11 Chapter 07 Linear Inequations (In two variables)

Question 1. Draw the graph of the solution set of x + y ≥ 4
Answer: Transform the inequality into the equation x + y = 4, then rearrange to get y = 4 - x. To find points on this line, substitute x = 0 to obtain y = 4, and substitute y = 0 to obtain x = 4. Mark the points (0, 4) and (4, 0) on the coordinate plane and draw a solid line connecting them. Test the origin (0, 0) by plugging it into the original inequality: 0 + 0 ≥ 4, which is false. Since this statement is false, the origin is not included in the solution. Shade the region above and to the right of the line, which represents all points satisfying the inequality, and include the line itself since the inequality uses ≥.
In simple words: Change the inequality to an equation and find two points to draw the line. Check if (0,0) works in the original inequality - it doesn't here. Shade the side that doesn't include the origin, and include the line in your solution.

Exam Tip: Always test the origin (0, 0) to determine which side of the line to shade. For ≥ and ≤ inequalities, draw a solid line; for > and <, use a dashed line.

Question 2. Draw the graph of the solution set of x - y ≤ 3
Answer: Convert the inequality into an equation: x - y = 3. Rewrite this in slope-intercept form as y = x - 3. To find coordinates, let x = 0 to get y = -3, and let y = 0 to get x = 3. Plot the points (0, -3) and (3, 0) on the graph and draw a solid line through them. Verify using the test point (0, 0): substituting into x - y ≤ 3 gives 0 - 0 ≤ 3, which simplifies to 0 ≤ 3, a true statement. Because the test point satisfies the inequality, shade the region that contains the origin, including all points on the boundary line.
In simple words: Turn the inequality into a line equation and plot two points. Test point (0, 0) - if it makes the inequality true, shade the region with the origin. Keep the line solid when using ≤ or ≥.

Exam Tip: When the test point satisfies the inequality, shade toward it; when it doesn't, shade away from it. The solid line indicates that boundary points are part of the solution set.

Question 3. Draw the graph of the solution set of y - 2 ≤ 3x
Answer: First, convert to an equation: y - 2 = 3x. Rearrange to slope-intercept form: y = 3x + 2. Find two key points by setting x = 0 to get y = 2, and setting y = 0 to get x = -2/3. Plot (0, 2) and (-2/3, 0) and draw a solid line through these points. Test the origin by substituting x = 0 and y = 0 into the original inequality: 0 - 2 ≤ 0 becomes -2 ≤ 0, which is true. Since the origin satisfies the condition, shade the region containing the origin, with the line included in the solution set.
In simple words: Rewrite as y = 3x + 2 and find two points on the line. Check if (0, 0) works - it does, so shade the region containing the origin with the line included.

Exam Tip: Pay careful attention to rearranging the inequality into standard form before finding intercepts. Always mark the two intercept points clearly for accuracy.

Question 4. Draw the graph of the solution set of x ≥ y - 2
Answer: Change the inequality to the equation x = y - 2 and rearrange to y = x + 2. Determine two coordinates by letting x = 0 to obtain y = 2, and letting y = 0 to obtain x = -2. Mark (0, 2) and (-2, 0) on the plane and connect with a solid line. Verify using (0, 0): substitute into x ≥ y - 2 to get 0 ≥ 0 - 2, which gives 0 ≥ -2, a true statement. The origin is part of the solution, so shade the region that contains it, keeping the boundary line as part of the shaded area.
In simple words: Convert to y = x + 2 and plot the two intercepts. Test (0, 0) - it satisfies the inequality, so shade the side with the origin and include the line.

Exam Tip: When rearranging inequalities to standard form, be consistent with your algebraic steps. Include the line for ≥ and ≤ symbols only.

Question 5. Draw the graph of the solution set of 3x + 2y > 6
Answer: Convert to the equation 3x + 2y = 6 and solve for y to get y = (6 - 3x)/2. When x = 0, y = 3; when y = 0, x = 2. Plot (0, 3) and (2, 0) and draw a dashed line between them (because the inequality is strict). Test the origin: 3(0) + 2(0) > 6 becomes 0 > 6, which is false. Since the origin does not satisfy the inequality, shade the region that does not contain the origin, and exclude the boundary line from the solution.
In simple words: Create the line from the equation and find intercepts. Test (0, 0) - it fails, so shade away from the origin. Use a dashed line since the inequality is strictly greater than.

Exam Tip: Use a dashed line for strict inequalities (> or <) because the boundary is not included. For ≥ or ≤, use a solid line.

Question 6. Draw the graph of the solution set of 3x + 5y < 15
Answer: Change to the equation 3x + 5y = 15 and express as y = (15 - 3x)/5. Setting x = 0 gives y = 3, and setting y = 0 gives x = 5. Mark (0, 3) and (5, 0) and connect with a dashed line (since the inequality is strict). Check point (0, 0): 3(0) + 5(0) < 15 becomes 0 < 15, which is true. The origin satisfies the inequality, so shade the region containing it, excluding the boundary line from the solution set.
In simple words: Find the line 3x + 5y = 15 using intercepts. Test (0, 0) - it works, so shade toward the origin with a dashed line since < is strictly less than.

Exam Tip: For < inequalities, always use a dashed boundary line. The direction of shading depends entirely on whether the test point satisfies the inequality.

Question 7. Solve graphically the system of inequalities: x ≥ 2 and y ≥ 3
Answer: Treat each inequality as an equation: x = 2 is a vertical line and y = 3 is a horizontal line. Draw both lines as solid lines on the coordinate plane. For x ≥ 2, test (0, 0): 0 ≥ 2 is false, so shade the region to the right of x = 2 (away from the origin). For y ≥ 3, test (0, 0): 0 ≥ 3 is false, so shade the region above y = 3 (away from the origin). The solution region is where both shaded areas overlap - this is the rectangular region where x ≥ 2 and y ≥ 3 simultaneously, including the boundary lines.
In simple words: Draw vertical line x = 2 and horizontal line y = 3. Test (0, 0) for each - both fail, so shade right of x = 2 and above y = 3. The overlap is your solution.

Exam Tip: For a system of inequalities, always find the common intersection region where all conditions are satisfied simultaneously. The solution is bounded by the boundary lines.

Question 8. Solve graphically the system of inequalities: 3x + 2y ≤ 12, x ≤ 1, y ≥ 2
Answer: Write each as an equation: 3x + 2y = 12 (find intercepts: (4, 0) and (0, 6)), x = 1 (vertical line), and y = 2 (horizontal line). Draw all three lines with solid boundaries. Test (0, 0) in 3x + 2y ≤ 12: 0 ≤ 12 is true, so shade toward the origin for this inequality. Test (0, 0) in x ≤ 1: 0 ≤ 1 is true, so shade toward the origin. Test (0, 0) in y ≥ 2: 0 ≥ 2 is false, so shade away from the origin (above the line). The solution is the triangular region bounded by all three lines where all three conditions hold simultaneously.
In simple words: Plot three lines from the three inequalities. Shade regions based on test point (0, 0) for each. Find where all three shaded regions meet - that's your solution.

Exam Tip: In systems with three or more inequalities, carefully shade each region and identify the final common area. Mark key vertices of the solution region if needed.

Question 9. Solve graphically the system of inequalities: x + y ≤ 6, x + y ≥ 4
Answer: Treat the inequalities as equations: x + y = 6 and x + y = 4. For x + y = 6, plot (0, 6) and (6, 0); for x + y = 4, plot (0, 4) and (4, 0). Draw both as solid lines. Test (0, 0) in x + y ≤ 6: 0 ≤ 6 is true, so shade toward the origin (below the line). Test (0, 0) in x + y ≥ 4: 0 ≥ 4 is false, so shade away from the origin (above the line). The solution is the strip-shaped region between the two parallel lines, including both boundary lines.
In simple words: Draw two parallel lines from x + y = 6 and x + y = 4. Shade between them based on the test point (0, 0). The region between and including both lines is the solution.

Exam Tip: When two inequalities involve the same expression, the solution often forms a band or strip between two parallel lines. Always verify the boundaries are included (solid) or excluded (dashed).

Question 10. Solve graphically the system of inequalities: 2x + y ≥ 6, 3x + 4y ≤ 12
Answer: Convert to equations: 2x + y = 6 (intercepts: (3, 0) and (0, 6)) and 3x + 4y = 12 (intercepts: (4, 0) and (0, 3)). Draw both lines solid. Test (0, 0) in 2x + y ≥ 6: 0 ≥ 6 is false, so shade away from the origin (above and to the right of the line). Test (0, 0) in 3x + 4y ≤ 12: 0 ≤ 12 is true, so shade toward the origin (below and to the left of the line). The solution is the quadrilateral region where these two shaded areas overlap, bounded by both solid lines.
In simple words: Plot both lines and test (0, 0) for each inequality. For the first, shade away; for the second, shade toward. Where the two shaded regions meet is your solution.

Exam Tip: When inequalities have opposite directions (≥ and ≤), the solution region is typically a bounded polygon. Identify and label the vertices of this region for clarity.

Question 11. Solve graphically the system of inequalities: x + y ≤ 9, y > x, x ≥ 0
Answer: Convert to equations: x + y = 9 (intercepts: (0, 9) and (9, 0)), y = x (passes through origin with slope 1), and x = 0 (the y-axis). Draw x + y = 9 and x = 0 as solid lines; draw y = x as a dashed line (since the inequality is strict >). Test (1, 0) in x + y ≤ 9: 1 + 0 ≤ 9 is true, shade below the line. Test (1, 0) in y > x: 0 > 1 is false, shade above the line y = x. Test any point with x < 0 in x ≥ 0: false, so keep x ≥ 0 (right of y-axis). The solution is the triangular region bounded by x = 0, y = x (dashed), and x + y = 9 (solid), where x ≥ 0 and the region lies above y = x and below x + y = 9.
In simple words: Draw three boundary lines. Use a dashed line for y = x (strict inequality) and solid lines for the others. The solution is the region in the first quadrant, above the line y = x, and below x + y = 9.

Exam Tip: Strict inequalities (> or <) are shown with dashed boundaries, meaning those points are not included. Always identify the three vertices of the solution triangle for reference.

Question 11. Solve the system of linear inequations: x + y ≤ 9, y > x, x ≥ 0
Answer: First, convert the inequations into their equation forms: x + y = 9, y = x, and x = 0. Plot these three lines on a coordinate plane.

For the line x + y = 9, identify two points: when x = 0, y = 9, giving (0, 9); when x = 9, y = 0, giving (9, 0). Draw a line through these points.

For the line y = x, this passes through the origin with slope 1, so points like (0, 0), (1, 1), (2, 2) lie on it.

For the line x = 0, this is the y-axis itself.

Now test the origin (0, 0) in each inequality:
- For x + y ≤ 9: substituting gives 0 ≤ 9, which holds true. Shade the side containing the origin.
- For y > x: substituting gives 0 > 0, which is false. Shade the region above the line y = x (away from the origin).
- For x ≥ 0: substituting gives 0 ≥ 0, which is true. Shade to the right of the y-axis.

The solution region is where all three shaded areas overlap - a triangular region bounded by these three lines, including boundary points on x + y = 9 and x = 0, but excluding the line y = x.
In simple words: Graph the three boundary lines, then test the origin to decide which side of each line to shade. The answer is the region where all three shaded areas meet - a triangle in the upper right, not including the diagonal line y = x.

Exam Tip: Always test a point (usually the origin) in each inequality to determine which region to shade; the solution is the intersection of all individual shaded regions, and remember to check whether boundary lines are included (solid) or excluded (dashed).

Question 12. Solve the system of linear inequations: 2x - y > 1, x - 2y < -1
Answer: Start by expressing the inequations as equations: 2x - y = 1 and x - 2y = -1. These represent the boundary lines.

For the line 2x - y = 1, find two points by assigning values. When x = 0, then -y = 1, so y = -1, giving (0, -1). When y = 0, then 2x = 1, so x = 0.5, giving (0.5, 0). Plot both points and draw the line through them.

For the line x - 2y = -1, when x = 0, then -2y = -1, so y = 0.5, giving (0, 0.5). When y = 0, then x = -1, giving (-1, 0). Plot these points and draw the line through them.

Test the origin (0, 0) in both inequalities:
- For 2x - y > 1: substituting gives 0 > 1, which is false. Shade the region not containing the origin (below and to the right of the line).
- For x - 2y < -1: substituting gives 0 < -1, which is false. Shade the region not containing the origin (above and to the left of the line).

The solution is the region where both shaded areas meet, excluding all boundary points since both inequalities are strict (> and <).
In simple words: Plot the two boundary lines using points from each equation. Test the origin in both inequations to find which regions to shade. The final answer is where both shaded regions overlap, but do not include the boundary lines themselves.

Exam Tip: When both inequalities are strict (> or <), the boundary lines are dashed and never included in the solution set; always verify the shading direction by testing a convenient point like the origin.