Math Problem Statement
Locate all corners of the system of inequalities based on the provided graph.
Solution
To solve this system of inequalities and locate the corners of the feasible region, we need to solve the system of boundary equations, which are derived from the inequalities provided in the image. The inequalities are:
-5x + 2y &\leq -55, \\ -2x + y &\leq -13, \\ 7x + 3y &\leq 91, \\ x &\geq 0, \\ y &\geq 0. \end{aligned}$$ We will solve the boundary equations (treating them as equalities) to find the intersection points, or corners of the feasible region. ### Step 1: Solve the system of equations The boundary lines corresponding to the inequalities are: 1. $$-5x + 2y = -55$$ 2. $$-2x + y = -13$$ 3. $$7x + 3y = 91$$ 4. $$x = 0$$ (y-axis) 5. $$y = 0$$ (x-axis) ### Step 2: Find intersections #### Intersection of $$-5x + 2y = -55$$ and $$-2x + y = -13$$ Solve the second equation for $$y$$: $$y = 2x - 13$$ Substitute this into the first equation: $$-5x + 2(2x - 13) = -55$$ $$-5x + 4x - 26 = -55$$ $$-x - 26 = -55$$ $$-x = -29 \quad \Rightarrow \quad x = 29$$ Substitute $$x = 29$$ into $$y = 2x - 13$$: $$y = 2(29) - 13 = 58 - 13 = 45$$ Thus, the intersection point is $$(29, 45)$$. #### Intersection of $$-5x + 2y = -55$$ and $$7x + 3y = 91$$ Multiply the first equation by 3 and the second by 2 to eliminate $$y$$: $$3(-5x + 2y) = 3(-55) \quad \Rightarrow \quad -15x + 6y = -165$$ $$2(7x + 3y) = 2(91) \quad \Rightarrow \quad 14x + 6y = 182$$ Now, subtract the first from the second: $$14x + 6y - (-15x + 6y) = 182 - (-165)$$ $$14x + 15x = 182 + 165$$ $$29x = 347 \quad \Rightarrow \quad x = \frac{347}{29} = 11.97$$ Substitute $$x = 11.97$$ into $$-5x + 2y = -55$$: $$-5(11.97) + 2y = -55$$ $$-59.85 + 2y = -55$$ $$2y = 4.85 \quad \Rightarrow \quad y = 2.425$$ Thus, the intersection point is approximately $$(11.97, 2.43)$$. #### Intersection of $$-2x + y = -13$$ and $$7x + 3y = 91$$ Solve $$-2x + y = -13$$ for $$y$$: $$y = 2x - 13$$ Substitute into $$7x + 3y = 91$$: $$7x + 3(2x - 13) = 91$$ $$7x + 6x - 39 = 91$$ $$13x = 130 \quad \Rightarrow \quad x = 10$$ Substitute $$x = 10$$ into $$y = 2x - 13$$: $$y = 2(10) - 13 = 20 - 13 = 7$$ Thus, the intersection point is $$(10, 7)$$. ### Step 3: Verify intersection with axes - Intersection with $$x = 0$$: Set $$x = 0$$ in each equation to find $$y$$. - Intersection with $$y = 0$$: Set $$y = 0$$ in each equation to find $$x$$. ### Final Corners The vertices or corners of the feasible region are approximately: 1. $$(0, 0)$$ — from the origin. 2. $$(29, 45)$$. 3. $$(11.97, 2.43)$$. 4. $$(10, 7)$$. These vertices define the region that satisfies the system of inequalities. --- Do you have any questions, or would you like more details on any step? Here are five related questions to expand on this topic: 1. How do we determine the feasible region from these inequalities? 2. What happens when there are no feasible solutions in such systems? 3. Can this system be solved graphically without solving the equations manually? 4. How do we check if a point lies within the feasible region? 5. How are systems of inequalities used in optimization problems? **Tip:** Always substitute the solution points back into the original inequalities to verify they satisfy all conditions.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Systems of Linear Inequalities
Graphing
Algebra
Formulas
-5x + 2y = -55
-2x + y = -13
7x + 3y = 91
Theorems
System of Linear Equations
Intersection of Lines
Suitable Grade Level
Grades 9-12
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