To find the points that satisfy the simultaneous inequalities \( y > \frac{1}{2}x + 5 \) and \( y < -2x + 1 \), we will evaluate each point one by one.
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For point a (1, 5.5):
- Check \( y > \frac{1}{2}x + 5 \): \[ 5.5 > \frac{1}{2}(1) + 5 \Rightarrow 5.5 > 0.5 + 5 \Rightarrow 5.5 > 5.5 \text{ (False)} \]
- So, point a does not satisfy the inequalities.
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For point b (-5, 5):
- Check \( y > \frac{1}{2}x + 5 \): \[ 5 > \frac{1}{2}(-5) + 5 \Rightarrow 5 > -2.5 + 5 \Rightarrow 5 > 2.5 \text{ (True)} \]
- Check \( y < -2x + 1 \): \[ 5 < -2(-5) + 1 \Rightarrow 5 < 10 + 1 \Rightarrow 5 < 11 \text{ (True)} \]
- Point b satisfies both inequalities.
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For point c (4, 8):
- Check \( y > \frac{1}{2}x + 5 \): \[ 8 > \frac{1}{2}(4) + 5 \Rightarrow 8 > 2 + 5 \Rightarrow 8 > 7 \text{ (True)} \]
- Check \( y < -2x + 1 \): \[ 8 < -2(4) + 1 \Rightarrow 8 < -8 + 1 \Rightarrow 8 < -7 \text{ (False)} \]
- Point c does not satisfy both inequalities.
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For point d (-1.6, 4.2):
- Check \( y > \frac{1}{2}x + 5 \): \[ 4.2 > \frac{1}{2}(-1.6) + 5 \Rightarrow 4.2 > -0.8 + 5 \Rightarrow 4.2 > 4.2 \text{ (False)} \]
- So, point d does not satisfy the inequalities.
In conclusion, the only point that shows a solution to the simultaneous inequalities is: b (-5, 5).
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