Math Problem Statement

Question

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Show Examples Which point would not be a solution to the system of linear inequalities shown below? y, is less than or equal to, start fraction, 5, divided by, 4, end fraction, x, minus, 3, y, is less than, minus, 3, x, plus, 5 y≤ 4 5 ​ x−3y<−3x+5 Answer Multiple Choice Answers left parenthesis, minus, 4, comma, minus, 10, right parenthesis(−4,−10) left parenthesis, 0, comma, minus, 10, right parenthesis(0,−10) left parenthesis, minus, 8, comma, 0, right parenthesis(−8,0) left parenthesis, 4, comma, minus, 8, right parenthesis(4,−8)

Solution

To determine which point is not a solution to the system of inequalities, we need to plug the coordinates of each point into the system and check if they satisfy both inequalities. The system given is:

  1. y≤54x−3y \leq \frac{5}{4}x - 3
  2. y<−3x+5y < -3x + 5

Step 1: Test each point

Point 1: (-4, -10)

  1. Plugging into y≤54x−3y \leq \frac{5}{4}x - 3: −10≤54(−4)−3⇒−10≤−5−3⇒−10≤−8(True)-10 \leq \frac{5}{4}(-4) - 3 \quad \Rightarrow \quad -10 \leq -5 - 3 \quad \Rightarrow \quad -10 \leq -8 \quad \text{(True)}
  2. Plugging into y<−3x+5y < -3x + 5: −10<−3(−4)+5⇒−10<12+5⇒−10<17(True)-10 < -3(-4) + 5 \quad \Rightarrow \quad -10 < 12 + 5 \quad \Rightarrow \quad -10 < 17 \quad \text{(True)}

Point 2: (0, -10)

  1. Plugging into y≤54x−3y \leq \frac{5}{4}x - 3: −10≤54(0)−3⇒−10≤−3(True)-10 \leq \frac{5}{4}(0) - 3 \quad \Rightarrow \quad -10 \leq -3 \quad \text{(True)}
  2. Plugging into y<−3x+5y < -3x + 5: −10<−3(0)+5⇒−10<5(True)-10 < -3(0) + 5 \quad \Rightarrow \quad -10 < 5 \quad \text{(True)}

Point 3: (-8, 0)

  1. Plugging into y≤54x−3y \leq \frac{5}{4}x - 3: 0≤54(−8)−3⇒0≤−10−3⇒0≤−13(False)0 \leq \frac{5}{4}(-8) - 3 \quad \Rightarrow \quad 0 \leq -10 - 3 \quad \Rightarrow \quad 0 \leq -13 \quad \text{(False)} This fails the first inequality, so (-8, 0) is not a solution.

Point 4: (4, -8)

  1. Plugging into y≤54x−3y \leq \frac{5}{4}x - 3: −8≤54(4)−3⇒−8≤5−3⇒−8≤2(True)-8 \leq \frac{5}{4}(4) - 3 \quad \Rightarrow \quad -8 \leq 5 - 3 \quad \Rightarrow \quad -8 \leq 2 \quad \text{(True)}
  2. Plugging into y<−3x+5y < -3x + 5: −8<−3(4)+5⇒−8<−12+5⇒−8<−7(True)-8 < -3(4) + 5 \quad \Rightarrow \quad -8 < -12 + 5 \quad \Rightarrow \quad -8 < -7 \quad \text{(True)}

Step 2: Conclusion

The point that does not satisfy both inequalities is (-8, 0).

Final Answer:

The point that would not be a solution is (-8, 0).

Do you need further details on this? Let me know if you want to discuss the solution step by step. Here are some related questions for you:

  1. What is the method for graphing linear inequalities?
  2. How can you determine whether a point is a solution to a system of inequalities by graphical methods?
  3. What is the difference between a linear inequality and a linear equation?
  4. How do you handle a system of two linear inequalities that have an overlap or no solution?
  5. How do you interpret the meaning of "y < -3x + 5" graphically?

Tip: When working with linear inequalities, always check both inequalities for the given point to ensure it satisfies all the conditions.

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Math Problem Analysis

Mathematical Concepts

Linear Inequalities
Systems of Inequalities
Graphing Inequalities

Formulas

y ≤ 5/4x - 3
y < -3x + 5

Theorems

Solving Systems of Inequalities
Substitution Method for Systems of Inequalities

Suitable Grade Level

Grades 9-10

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