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Linear Inequalities - Graphical Solution of Linear Inequalities in Two Variables

Grade 11CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A linear inequality in two variables defines a half-plane in the Cartesian coordinate system. The boundary is the line ax+by=cax + by = c. If the inequality is strict (<< or >>), the line is drawn dashed; if it is non-strict (≤\le or ≥\ge), the line is solid.

Illustration showing dashed line for strict inequality and solid line for non-strict inequality.
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To determine which side of the boundary line to shade, we use a 'test point' not on the line, usually (0,0)(0,0). If the coordinates satisfy the inequality, the half-plane containing (0,0)(0,0) is the solution region; otherwise, the opposite half-plane is chosen.

Graph showing the test point method for x + y >= 4.
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Horizontal inequalities y≥ky \ge k or y≤ky \le k represent regions above or below a horizontal line, while vertical inequalities x≥hx \ge h or x≤hx \le h represent regions to the right or left of a vertical line.

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The solution to a system of linear inequalities is the intersection (overlapping area) of all the individual solution regions. This common region is called the feasible region.

📐Formulae

General form of linear inequalities: ax+by<cax + by < c, ax+by>cax + by > c, ax+by≤cax + by \le c, ax+by≥cax + by \ge c

Boundary line equation: ax+by=cax + by = c

Slope-intercept form for plotting: y=mx+cy = mx + c, where m=−abm = -\frac{a}{b}

Intercept form for plotting: xxintercept+yyintercept=1\frac{x}{x_{intercept}} + \frac{y}{y_{intercept}} = 1

Horizontal line inequality: y≤ky \le k (region below the line y=ky=k) or y≥ky \ge k (region above the line y=ky=k)

Vertical line inequality: x≤hx \le h (region to the left of line x=hx=h) or x≥hx \ge h (region to the right of line x=hx=h)

💡Examples

Problem 1:

Solve the linear inequality 3x+2y>63x + 2y > 6 graphically.

Solution:

Step 1: Convert the inequality into an equation to find the boundary line: 3x+2y=63x + 2y = 6. \nStep 2: Find the intercepts. When x=0,y=3x = 0, y = 3 (point (0,3)(0, 3)). When y=0,x=2y = 0, x = 2 (point (2,0)(2, 0)). \nStep 3: Draw a dashed line passing through (0,3)(0, 3) and (2,0)(2, 0) because the inequality is strict (>>). \nStep 4: Use (0,0)(0,0) as a test point. Substitute into 3x+2y>63x + 2y > 6: 3(0)+2(0)>6⇒0>63(0) + 2(0) > 6 \Rightarrow 0 > 6. This is False. \nStep 5: Since the origin does not satisfy the inequality, shade the half-plane that does not contain the origin.

Explanation:

The solution is the region above the line 3x+2y=63x + 2y = 6, excluding the points on the line itself.

Problem 2:

Find the graphical solution for the system of inequalities: x+y≤5x + y \le 5, 4x+y≥44x + y \ge 4, x≥0x \ge 0, y≥0y \ge 0.

Solution:

Step 1: For x+y=5x + y = 5, intercepts are (5,0)(5,0) and (0,5)(0,5). Draw a solid line. Testing (0,0)(0,0) gives 0≤50 \le 5 (True), so shade towards the origin. \nStep 2: For 4x+y=44x + y = 4, intercepts are (1,0)(1,0) and (0,4)(0,4). Draw a solid line. Testing (0,0)(0,0) gives 0≥40 \ge 4 (False), so shade away from the origin. \nStep 3: x≥0x \ge 0 and y≥0y \ge 0 restrict the solution to the first quadrant. \nStep 4: Identify the common region where all conditions overlap.

Explanation:

The solution is a bounded quadrilateral region in the first quadrant with vertices determined by the intersection of the boundary lines.

Problem 3:

Solve the inequality 2x−y>42x - y > 4 graphically.

Graph of 2x - y > 4 showing a dashed boundary line and shaded region below/to the right.

Solution:

  1. Write the boundary line equation: 2x−y=42x - y = 4.
  2. Find intercepts: When x=0,y=−4x=0, y=-4 (Point A(0,−4)A(0,-4)). When y=0,x=2y=0, x=2 (Point B(2,0)B(2,0)).
  3. Draw a dashed line through AA and BB because the inequality is strict (>>).
  4. Test point (0,0)(0,0): 2(0)−0>4  ⟹  0>42(0) - 0 > 4 \implies 0 > 4 is False.
  5. Since the test point fails, shade the region that does NOT contain (0,0)(0,0).

Explanation:

The boundary line splits the plane. Since (0,0)(0,0) results in a false statement, the solution set is the open half-plane on the side of the line opposite to the origin.

Problem 4:

Solve the system of inequalities graphically: x+2y≤8x + 2y \le 8 and x≥2x \ge 2.

System of inequalities showing the intersection region between a slanted line and a vertical line.

Solution:

  1. For x+2y≤8x + 2y \le 8: Boundary is x+2y=8x + 2y = 8. Intercepts are (8,0)(8,0) and (0,4)(0,4). Test (0,0)(0,0): 0+0≤80+0 \le 8 (True). Shade towards origin.
  2. For x≥2x \ge 2: Boundary is vertical line x=2x = 2. Shade to the right of x=2x = 2.
  3. The solution is the region satisfied by both.

Explanation:

The intersection of the half-plane below x+2y=8x+2y=8 and the half-plane to the right of x=2x=2 forms the solution area.

Graphical Solution of Linear Inequalities in Two Variables Class 11 Notes & Examples