Answer :
Let's determine which of the given points [tex]\((4, 1)\)[/tex] and [tex]\((0, -3)\)[/tex] satisfy the system of linear inequalities:
[tex]\[ \begin{array}{l} y \leq x+1 \\ y < -\frac{x}{2} - 1 \end{array} \][/tex]
### Checking Point (4, 1):
1. For the inequality [tex]\(y \leq x + 1\)[/tex]:
[tex]\[ y = 1 \quad \text{and} \quad x = 4 \][/tex]
Substitute the values into the inequality:
[tex]\[ 1 \leq 4 + 1 \][/tex]
Simplify:
[tex]\[ 1 \leq 5 \quad \text{(This is True)} \][/tex]
2. For the inequality [tex]\(y < -\frac{x}{2} - 1\)[/tex]:
[tex]\[ y = 1 \quad \text{and} \quad x = 4 \][/tex]
Substitute the values into the inequality:
[tex]\[ 1 < -\frac{4}{2} - 1 \][/tex]
Simplify:
[tex]\[ 1 < -2 - 1 \][/tex]
Simplify further:
[tex]\[ 1 < -3 \quad \text{(This is False)} \][/tex]
Since the first inequality is satisfied but the second one is not, the point [tex]\((4, 1)\)[/tex] does not satisfy both inequalities.
### Checking Point (0, -3):
1. For the inequality [tex]\(y \leq x + 1\)[/tex]:
[tex]\[ y = -3 \quad \text{and} \quad x = 0 \][/tex]
Substitute the values into the inequality:
[tex]\[ -3 \leq 0 + 1 \][/tex]
Simplify:
[tex]\[ -3 \leq 1 \quad \text{(This is True)} \][/tex]
2. For the inequality [tex]\(y < -\frac{x}{2} - 1\)[/tex]:
[tex]\[ y = -3 \quad \text{and} \quad x = 0 \][/tex]
Substitute the values into the inequality:
[tex]\[ -3 < -\frac{0}{2} - 1 \][/tex]
Simplify:
[tex]\[ -3 < -1 \quad \text{(This is True)} \][/tex]
Since both inequalities are satisfied, the point [tex]\((0, -3)\)[/tex] does satisfy both inequalities.
### Conclusion:
The point [tex]\((0, -3)\)[/tex] is a solution to the system of linear inequalities, whereas the point [tex]\((4, 1)\)[/tex] is not.
[tex]\[ \begin{array}{l} y \leq x+1 \\ y < -\frac{x}{2} - 1 \end{array} \][/tex]
### Checking Point (4, 1):
1. For the inequality [tex]\(y \leq x + 1\)[/tex]:
[tex]\[ y = 1 \quad \text{and} \quad x = 4 \][/tex]
Substitute the values into the inequality:
[tex]\[ 1 \leq 4 + 1 \][/tex]
Simplify:
[tex]\[ 1 \leq 5 \quad \text{(This is True)} \][/tex]
2. For the inequality [tex]\(y < -\frac{x}{2} - 1\)[/tex]:
[tex]\[ y = 1 \quad \text{and} \quad x = 4 \][/tex]
Substitute the values into the inequality:
[tex]\[ 1 < -\frac{4}{2} - 1 \][/tex]
Simplify:
[tex]\[ 1 < -2 - 1 \][/tex]
Simplify further:
[tex]\[ 1 < -3 \quad \text{(This is False)} \][/tex]
Since the first inequality is satisfied but the second one is not, the point [tex]\((4, 1)\)[/tex] does not satisfy both inequalities.
### Checking Point (0, -3):
1. For the inequality [tex]\(y \leq x + 1\)[/tex]:
[tex]\[ y = -3 \quad \text{and} \quad x = 0 \][/tex]
Substitute the values into the inequality:
[tex]\[ -3 \leq 0 + 1 \][/tex]
Simplify:
[tex]\[ -3 \leq 1 \quad \text{(This is True)} \][/tex]
2. For the inequality [tex]\(y < -\frac{x}{2} - 1\)[/tex]:
[tex]\[ y = -3 \quad \text{and} \quad x = 0 \][/tex]
Substitute the values into the inequality:
[tex]\[ -3 < -\frac{0}{2} - 1 \][/tex]
Simplify:
[tex]\[ -3 < -1 \quad \text{(This is True)} \][/tex]
Since both inequalities are satisfied, the point [tex]\((0, -3)\)[/tex] does satisfy both inequalities.
### Conclusion:
The point [tex]\((0, -3)\)[/tex] is a solution to the system of linear inequalities, whereas the point [tex]\((4, 1)\)[/tex] is not.