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.
