Math Problem Statement

Take S to be the feasible set for the system of inequalities: y ≥ (10/9)x + (105/9) y ≤ 21x + 131 y < (-11/8)x + (153/8) Write S as a union of two sets, where x is a free variable and y is dependent on x. In what intervals does x vary in the two sets?

Solution

Let's break down the system of inequalities step by step and figure out how to represent SS as a union of two sets based on the variable xx.

The system is:

y109x+1059y \geq \frac{10}{9}x + \frac{105}{9} y21x+131y \leq 21x + 131 y<118x+1538y < -\frac{11}{8}x + \frac{153}{8}

Step 1: Finding the intersection points

To determine in what intervals xx varies, we need to find the points where the boundary lines intersect. This gives us the critical values for xx.

Intersection of y=109x+1059y = \frac{10}{9}x + \frac{105}{9} and y=21x+131y = 21x + 131

Equating both expressions for yy:

109x+1059=21x+131\frac{10}{9}x + \frac{105}{9} = 21x + 131

Multiply everything by 9 to clear the denominators:

10x+105=189x+117910x + 105 = 189x + 1179

Simplifying:

10x189x=117910510x - 189x = 1179 - 105 179x=1074-179x = 1074 x=1074179=6x = \frac{1074}{179} = 6

So, the first intersection occurs at x=6x = 6.

Intersection of y=21x+131y = 21x + 131 and y=118x+1538y = -\frac{11}{8}x + \frac{153}{8}

Equating both expressions for yy:

21x+131=118x+153821x + 131 = -\frac{11}{8}x + \frac{153}{8}

Multiply through by 8 to eliminate the fractions:

168x+1048=11x+153168x + 1048 = -11x + 153

Simplifying:

168x+11x=1531048168x + 11x = 153 - 1048 179x=895179x = -895 x=895179=5x = \frac{-895}{179} = -5

So, the second intersection occurs at x=5x = -5.

Step 2: Determining the intervals

Now that we have the critical points x=5x = -5 and x=6x = 6, we can describe the intervals for xx:

  1. For x(,5]x \in (-\infty, -5], SS will be the region defined by the inequalities involving the lines y109x+1059y \geq \frac{10}{9}x + \frac{105}{9} and y<118x+1538y < -\frac{11}{8}x + \frac{153}{8}.
  2. For x[5,6]x \in [-5, 6], SS will be the region bounded by y109x+1059y \geq \frac{10}{9}x + \frac{105}{9} and y21x+131y \leq 21x + 131.
  3. For x[6,)x \in [6, \infty), SS will be the region between y21x+131y \leq 21x + 131 and y<118x+1538y < -\frac{11}{8}x + \frac{153}{8}.

Thus, the feasible set SS is a union of these intervals.

Final Answer:

The intervals where xx varies are:

  • (,5](-\infty, -5]
  • [5,6][-5, 6]
  • [6,)[6, \infty)

Would you like more detailed explanations or have any other questions?

Five Related Questions:

  1. How do you graphically represent the feasible region for this system of inequalities?
  2. What happens to the system if we change the inequality symbols in one of the equations?
  3. How would the solution change if we added another inequality constraint to the system?
  4. What is the geometric meaning of the intersection points we found?
  5. Can this system be solved using matrix methods or linear programming?

Tip:

Always check the slope and intercept of each line to understand how the inequalities divide the plane. This helps visualize the solution.

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Math Problem Analysis

Mathematical Concepts

Inequalities
Linear systems
Intersection of lines

Formulas

y = (10/9)x + (105/9)
y = 21x + 131
y = (-11/8)x + (153/8)

Theorems

Intersection of lines
System of inequalities

Suitable Grade Level

Grades 10-12