Which point would not be a solution to the system of linear inequalities shown below?
y, is less than, 2, x, plus, 2, y, is less than or equal to, minus, start fraction, 5, divided by, 2, end fraction, x, minus, 5
y<2x+2y≤−
2
5
​
x−5

To determine which point would not be a solution to the given system of linear inequalities, we need to verify the points against both inequalities:

1. **\( y < 2x + 2 \)**
2. **\( y \leq -\frac{5}{2}x - 5 \)**

Let's check each point by substituting its coordinates into both inequalities. We'll use a few test points:

- **(0, 0)**
- **(1, 0)**
- **(-1, -6)**
- **(3, 4)**

### Point (0, 0)

1. **For \( y < 2x + 2 \):**
   \( 0 < 2(0) + 2 \)
   \( 0 < 2 \) (True)

2. **For \( y \leq -\frac{5}{2}x - 5 \):**
   \( 0 \leq -\frac{5}{2}(0) - 5 \)
   \( 0 \leq -5 \) (False)

**Point (0, 0) does not satisfy the second inequality, so it is not a solution to the system.**

### Point (1, 0)

1. **For \( y < 2x + 2 \):**
   \( 0 < 2(1) + 2 \)
   \( 0 < 4 \) (True)

2. **For \( y \leq -\frac{5}{2}x - 5 \):**
   \( 0 \leq -\frac{5}{2}(1) - 5 \)
   \( 0 \leq -\frac{15}{2} \) (False)

**Point (1, 0) does not satisfy the second inequality, so it is not a solution to the system.**

### Point (-1, -6)

1. **For \( y < 2x + 2 \):**
   \( -6 < 2(-1) + 2 \)
   \( -6 < 0 \) (True)

2. **For \( y \leq -\frac{5}{2}x - 5 \):**
   \( -6 \leq -\frac{5}{2}(-1) - 5 \)
   \( -6 \leq \frac{5}{2} - 5 \)
   \( -6 \leq -\frac{5}{2} \) (True)

**Point (-1, -6) satisfies both inequalities.**

### Point (3, 4)

1. **For \( y < 2x + 2 \):**
   \( 4 < 2(3) + 2 \)
   \( 4 < 8 \) (True)

2. **For \( y \leq -\frac{5}{2}x - 5 \):**
   \( 4 \leq -\frac{5}{2}(3) - 5 \)
   \( 4 \leq -\frac{15}{2} - 5 \)
   \( 4 \leq -\frac{25}{2} \) (False)

**Point (3, 4) does not satisfy the second inequality, so it is not a solution to the system.**

Based on the above checks, the points that are **not solutions** to the system of inequalities are:

- **(0, 0)**
- **(1, 0)**
- **(3, 4)**

Let me know if you need further details or have any questions.

**Relative questions:**

1. How can we graphically represent the solution set for a system of linear inequalities?
2. What are the steps to solve a system of linear inequalities graphically?
3. How do we interpret the intersection of inequalities in a system?
4. Can a point satisfy only one inequality in a system of inequalities?
5. How do the coefficients in a linear inequality affect the boundary line on a graph?
6. What changes in the solution set if we switch the inequality symbols?
7. How can we verify a solution point algebraically?
8. What is the significance of testing corner points in linear programming?

**Tip:** Always test potential solution points with all inequalities in the system to confirm they satisfy all conditions.

Learn how to identify points that do not satisfy a system of linear inequalities through detailed algebraic verification. This problem covers key concepts in linear inequalities and is suitable for students in Grades 7-9.

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