To determine which point is a solution to the simultaneous inequalities \( y > 3x + 1 \) and \( y \leq 2x + 5 \), we will check each point against both inequalities.
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Check the point (-5, 5):
- For the first inequality \( y > 3x + 1 \): \[ 5 > 3(-5) + 1 \Rightarrow 5 > -15 + 1 \Rightarrow 5 > -14 \quad \text{(True)} \]
- For the second inequality \( y \leq 2x + 5 \): \[ 5 \leq 2(-5) + 5 \Rightarrow 5 \leq -10 + 5 \Rightarrow 5 \leq -5 \quad \text{(False)} \]
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Check the point (-1, 2):
- For the first inequality \( y > 3x + 1 \): \[ 2 > 3(-1) + 1 \Rightarrow 2 > -3 + 1 \Rightarrow 2 > -2 \quad \text{(True)} \]
- For the second inequality \( y \leq 2x + 5 \): \[ 2 \leq 2(-1) + 5 \Rightarrow 2 \leq -2 + 5 \Rightarrow 2 \leq 3 \quad \text{(True)} \]
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Check the point (9, 5):
- For the first inequality \( y > 3x + 1 \): \[ 5 > 3(9) + 1 \Rightarrow 5 > 27 + 1 \Rightarrow 5 > 28 \quad \text{(False)} \]
- For the second inequality \( y \leq 2x + 5 \): \[ 5 \leq 2(9) + 5 \Rightarrow 5 \leq 18 + 5 \Rightarrow 5 \leq 23 \quad \text{(True)} \]
The only point that satisfies both inequalities is (-1, 2).
Thus, the answer is: \[ ( -1 , 2 ) \]
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