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Given the system of inequalities:

[tex]\[
\begin{array}{l}
y \geq -\frac{1}{3}x + 2 \\
y \ \textless \ 2x + 3
\end{array}
\][/tex]

Determine which of the following points satisfy the system:

A. [tex]\((2,2), (3,1), (4,2)\)[/tex]

B. [tex]\((2,2), (3,-1), (4,1)\)[/tex]

C. [tex]\((2,2), (1,-2), (0,2)\)[/tex]

D. [tex]\((2,2), (1,2), (2,0)\)[/tex]


Sagot :

Let's analyze the given inequalities and the sets of points step-by-step to determine which set satisfies both inequalities. The inequalities are:
[tex]\[ y \geq -\frac{1}{3} x + 2 \][/tex]
[tex]\[ y < 2x + 3 \][/tex]

We have four sets of points to check:
1. [tex]\((2,2), (3,1), (4,2)\)[/tex]
2. [tex]\((2,2), (3,-1), (4,1)\)[/tex]
3. [tex]\((2,2), (1,-2), (0,2)\)[/tex]
4. [tex]\((2,2), (1,2), (2,0)\)[/tex]

### Set 1: [tex]\((2,2), (3,1), (4,2)\)[/tex]

1. For the point [tex]\((2, 2)\)[/tex]:
- [tex]\[ y = 2 \geq -\frac{1}{3}(2) + 2 = \frac{4}{3} \rightarrow (\text{True}) \][/tex]
- [tex]\[ y = 2 < 2(2) + 3 = 7 \rightarrow (\text{True}) \][/tex]

2. For the point [tex]\((3, 1)\)[/tex]:
- [tex]\[ y = 1 \geq -\frac{1}{3}(3) + 2 = 1 \rightarrow (\text{True}) \][/tex]
- [tex]\[ y = 1 < 2(3) + 3 = 9 \rightarrow (\text{True}) \][/tex]

3. For the point [tex]\((4, 2)\)[/tex]:
- [tex]\[ y = 2 \geq -\frac{1}{3}(4) + 2 = \frac{2}{3} \rightarrow (\text{True}) \][/tex]
- [tex]\[ y = 2 < 2(4) + 3 = 11 \rightarrow (\text{True}) \][/tex]

All three points satisfy both inequalities. Therefore, Set 1 is valid.

### Set 2: [tex]\((2,2), (3,-1), (4,1)\)[/tex]

1. For the point [tex]\((2, 2)\)[/tex]:
- [tex]\[ y = 2 \geq -\frac{1}{3}(2) + 2 = \frac{4}{3} \rightarrow (\text{True}) \][/tex]
- [tex]\[ y = 2 < 2(2) + 3 = 7 \rightarrow (\text{True}) \][/tex]

2. For the point [tex]\((3, -1)\)[/tex]:
- [tex]\[ y = -1 \geq -\frac{1}{3}(3) + 2 = 1 \rightarrow (\text{False}) \][/tex]

Since the point [tex]\((3, -1)\)[/tex] does not satisfy the first inequality, Set 2 is invalid.

### Set 3: [tex]\((2,2), (1,-2), (0,2)\)[/tex]

1. For the point [tex]\((2, 2)\)[/tex]:
- [tex]\[ y = 2 \geq -\frac{1}{3}(2) + 2 = \frac{4}{3} \rightarrow (\text{True}) \][/tex]
- [tex]\[ y = 2 < 2(2) + 3 = 7 \rightarrow (\text{True}) \][/tex]

2. For the point [tex]\((1, -2)\)[/tex]:
- [tex]\[ y = -2 \geq -\frac{1}{3}(1) + 2 = \frac{5}{3} \rightarrow (\text{False}) \][/tex]

Since the point [tex]\((1, -2)\)[/tex] does not satisfy the first inequality, Set 3 is invalid.

### Set 4: [tex]\((2,2), (1,2), (2,0)\)[/tex]

1. For the point [tex]\((2, 2)\)[/tex]:
- [tex]\[ y = 2 \geq -\frac{1}{3}(2) + 2 = \frac{4}{3} \rightarrow (\text{True}) \][/tex]
- [tex]\[ y = 2 < 2(2) + 3 = 7 \rightarrow (\text{True}) \][/tex]

2. For the point [tex]\((1, 2)\)[/tex]:
- [tex]\[ y = 2 \geq -\frac{1}{3}(1) + 2 = \frac{7}{3} \rightarrow (\text{False}) \][/tex]

Since the point [tex]\((1, 2)\)[/tex] does not satisfy the first inequality, Set 4 is invalid.

### Conclusion

Only Set 1: [tex]\((2,2), (3,1), (4,2)\)[/tex] satisfies both inequalities. Thus, the valid set of points is Set 1.