To determine which points are solutions to the simultaneous inequalities \( y > 4x - 2 \) and \( y \geq -\frac{1}{3}x + 2 \), we can evaluate each point one at a time.
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Point (-1, 5):
- For \( y > 4x - 2 \): \[ 5 > 4(-1) - 2 \Rightarrow 5 > -4 - 2 \Rightarrow 5 > -6 \quad \text{(True)} \]
- For \( y \geq -\frac{1}{3}x + 2 \): \[ 5 \geq -\frac{1}{3}(-1) + 2 \Rightarrow 5 \geq \frac{1}{3} + 2 \Rightarrow 5 \geq \frac{7}{3} \quad \text{(True)} \]
- Result: True for both inequalities.
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Point (0.923, 1.6):
- For \( y > 4x - 2 \): \[ 1.6 > 4(0.923) - 2 \Rightarrow 1.6 > 3.692 - 2 \Rightarrow 1.6 > 1.692 \quad \text{(False)} \]
- Result: False for the first inequality.
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Point (5, 2):
- For \( y > 4x - 2 \): \[ 2 > 4(5) - 2 \Rightarrow 2 > 20 - 2 \Rightarrow 2 > 18 \quad \text{(False)} \]
- Result: False for the first inequality.
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Point (-4, -2):
- For \( y > 4x - 2 \): \[ -2 > 4(-4) - 2 \Rightarrow -2 > -16 - 2 \Rightarrow -2 > -18 \quad \text{(True)} \]
- For \( y \geq -\frac{1}{3}x + 2 \): \[ -2 \geq -\frac{1}{3}(-4) + 2 \Rightarrow -2 \geq \frac{4}{3} + 2 \Rightarrow -2 \geq \frac{4}{3} + \frac{6}{3} \Rightarrow -2 \geq \frac{10}{3} \quad \text{(False)} \]
- Result: False for the second inequality.
The only point that satisfies both inequalities is (-1, 5).
Thus, the solution is: \[ \boxed{-1 , 5} \]