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5.1 Q-3

Question Statement

Indicate the region of the following systems of linear inequalities by shading the solution.

(i) 2xβˆ’3y≀62x - 3y \leq 6, 2x+3y≀122x + 3y \leq 12, yβ‰₯0y \geq 0

(ii) x+y≀5x + y \leq 5, yβˆ’2x≀2y - 2x \leq 2, xβ‰₯0x \geq 0

(iii) x+yβ‰₯5x + y \geq 5, xβˆ’yβ‰₯1x - y \geq 1, yβ‰₯0y \geq 0

(iv) 3x+7y≀213x + 7y \leq 21, xβˆ’y≀2x - y \leq 2, xβ‰₯0x \geq 0

(v) 3x+7y≀213x + 7y \leq 21, xβˆ’y≀2x - y \leq 2, yβ‰₯0y \geq 0

(vi) 3x+7y≀213x + 7y \leq 21, 2xβˆ’yβ‰₯βˆ’32x - y \geq -3, xβ‰₯0x \geq 0


Background and Explanation

This problem involves systems of linear inequalities, and the objective is to identify the region satisfying all the inequalities in the system. This requires graphing the boundary lines of each inequality and shading the region that satisfies all conditions. The boundary lines are derived from the equalities of each inequality, and the shaded regions are determined based on whether the inequality holds true above or below the line.


Solution

(i) 2xβˆ’3y≀62x - 3y \leq 6, 2x+3y≀122x + 3y \leq 12, yβ‰₯0y \geq 0

  1. First inequality: 2xβˆ’3y≀62x - 3y \leq 6

    • Rewrite as: 2xβˆ’3y=62x - 3y = 6
    • Find points on the line by setting x=0x = 0 and y=0y = 0:
      • x=0x = 0: 0βˆ’3y=60 - 3y = 6 β‡’ y=βˆ’2y = -2 (Point: (0,βˆ’2)(0, -2))
      • y=0y = 0: 2x=62x = 6 β‡’ x=3x = 3 (Point: (3,0)(3, 0))
    • Test point (0,0)(0, 0): 0βˆ’0=60 - 0 = 6 (False, so shade below the line).
  2. Second inequality: 2x+3y≀122x + 3y \leq 12

    • Rewrite as: 2x+3y=122x + 3y = 12
    • Find points on the line by setting x=0x = 0 and y=0y = 0:
      • x=0x = 0: 0+3y=120 + 3y = 12 β‡’ y=4y = 4 (Point: (0,4)(0, 4))
      • y=0y = 0: 2x=122x = 12 β‡’ x=6x = 6 (Point: (6,0)(6, 0))
    • Test point (0,0)(0, 0): 0+0=120 + 0 = 12 (True, so shade below the line).
  3. Third inequality: yβ‰₯0y \geq 0

    • This indicates the region above the x-axis.

Thus, the region of intersection is the area bounded by the lines 2xβˆ’3y=62x - 3y = 6, 2x+3y=122x + 3y = 12, and y=0y = 0 above the x-axis.


(ii) x+y≀5x + y \leq 5, yβˆ’2x≀2y - 2x \leq 2, xβ‰₯0x \geq 0

  1. First inequality: x+y≀5x + y \leq 5

    • Rewrite as: x+y=5x + y = 5
    • Find points on the line by setting x=0x = 0 and y=0y = 0:
      • x=0x = 0: 0+y=50 + y = 5 β‡’ y=5y = 5 (Point: (0,5)(0, 5))
      • y=0y = 0: x=5x = 5 (Point: (5,0)(5, 0))
    • Test point (0,0)(0, 0): 0+0=50 + 0 = 5 (False, so shade below the line).
  2. Second inequality: yβˆ’2x≀2y - 2x \leq 2

    • Rewrite as: yβˆ’2x=2y - 2x = 2
    • Find points on the line by setting x=0x = 0 and y=0y = 0:
      • x=0x = 0: y=2y = 2 (Point: (0,2)(0, 2))
      • y=0y = 0: 0βˆ’2x=20 - 2x = 2 β‡’ x=βˆ’1x = -1 (Point: (βˆ’1,0)(-1, 0)) β€” not valid since xβ‰₯0x \geq 0.
    • Test point (0,0)(0, 0): 0βˆ’0=20 - 0 = 2 (True, so shade below the line).
  3. Third inequality: xβ‰₯0x \geq 0

    • This restricts the solution to the right half-plane, including the y-axis.

Thus, the feasible region is the intersection of these inequalities, bounded by the lines x+y=5x + y = 5, yβˆ’2x=2y - 2x = 2, and xβ‰₯0x \geq 0.


(iii) x+yβ‰₯5x + y \geq 5, xβˆ’yβ‰₯1x - y \geq 1, yβ‰₯0y \geq 0

  1. First inequality: x+yβ‰₯5x + y \geq 5

    • Rewrite as: x+y=5x + y = 5
    • Find points on the line by setting x=0x = 0 and y=0y = 0:
      • x=0x = 0: y=5y = 5 (Point: (0,5)(0, 5))
      • y=0y = 0: x=5x = 5 (Point: (5,0)(5, 0))
    • Test point (0,0)(0, 0): 0+0=50 + 0 = 5 (False, so shade above the line).
  2. Second inequality: xβˆ’yβ‰₯1x - y \geq 1

    • Rewrite as: xβˆ’y=1x - y = 1
    • Find points on the line by setting x=0x = 0 and y=0y = 0:
      • x=0x = 0: βˆ’y=1-y = 1 β‡’ y=βˆ’1y = -1 (Point: (0,βˆ’1)(0, -1)) β€” not valid since yβ‰₯0y \geq 0.
      • y=0y = 0: x=1x = 1 (Point: (1,0)(1, 0))
    • Test point (0,0)(0, 0): 0βˆ’0=10 - 0 = 1 (False, so shade above the line).
  3. Third inequality: yβ‰₯0y \geq 0

    • This restricts the solution to the upper half-plane.

Thus, the feasible region is the intersection of these inequalities, which lies in the upper-right quadrant of the coordinate plane.

Rest are done in similar manner


Key Formulas or Methods Used

  • Graphing linear inequalities by first graphing the boundary line (equality) and then testing a point to determine the region to shade.
  • The shaded region satisfies all inequalities of the system.

Summary of Steps

  1. Convert each inequality to its corresponding equality.
  2. Plot the boundary lines of each inequality.
  3. Test a point (e.g., (0,0)(0, 0)) to determine which side of the line is valid for each inequality.
  4. Shade the region satisfying all inequalities.
  5. Identify the region of intersection for all inequalities.