Question

Which point on the graph is a solution to the simultaneous inequalities y≤ 1/3x +2 and y< 2/3x +4
A) -4, -2
B) -10, -2
C) -6,0

Answer 1

To determine which point is a solution to the inequalities \( y \leq \frac{1}{3}x + 2 \) and \( y < \frac{2}{3}x + 4 \), we need to substitute each of the provided points into both inequalities.

### Inequalities:
1. \( y \leq \frac{1}{3}x + 2 \)
2. \( y < \frac{2}{3}x + 4 \)

### Point A: (-4, -2)
- For \( y \leq \frac{1}{3}x + 2 \):
\[
-2 \leq \frac{1}{3}(-4) + 2 \\
-2 \leq -\frac{4}{3} + 2 \\
-2 \leq \frac{2}{3} \quad \text{(True)}
\]

- For \( y < \frac{2}{3}x + 4 \):
\[
-2 < \frac{2}{3}(-4) + 4 \\
-2 < -\frac{8}{3} + 4 \\
-2 < \frac{4}{3} \quad \text{(True)}
\]

### Point B: (-10, -2)
- For \( y \leq \frac{1}{3}x + 2 \):
\[
-2 \leq \frac{1}{3}(-10) + 2 \\
-2 \leq -\frac{10}{3} + 2 \\
-2 \leq -\frac{10}{3} + \frac{6}{3} \\
-2 \leq -\frac{4}{3} \quad \text{(False)}
\]

### Point C: (-6, 0)
- For \( y \leq \frac{1}{3}x + 2 \):
\[
0 \leq \frac{1}{3}(-6) + 2 \\
0 \leq -2 + 2 \\
0 \leq 0 \quad \text{(True)}
\]

- For \( y < \frac{2}{3}x + 4 \):
\[
0 < \frac{2}{3}(-6) + 4 \\
0 < -4 + 4 \\
0 < 0 \quad \text{(False)}
\]

### Conclusion
The only point that satisfies both inequalities is **Point A: (-4, -2)**.

Thus, the correct answer is **A) (-4, -2)**.