To determine which points are part of the solution set for the inequality \( y \leq 25x + 1 \), we need to evaluate each point by plugging in its coordinates into the inequality.
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For the point \((10, 5.5)\): \[ y = 5.5 \quad \text{and} \quad 25x + 1 = 25(10) + 1 = 250 + 1 = 251 \] \[ 5.5 \leq 251 \quad \text{(True)} \]
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For the point \((10, 6)\): \[ y = 6 \quad \text{and} \quad 25x + 1 = 25(10) + 1 = 250 + 1 = 251 \] \[ 6 \leq 251 \quad \text{(True)} \]
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For the point \((10, 4)\): \[ y = 4 \quad \text{and} \quad 25x + 1 = 25(10) + 1 = 250 + 1 = 251 \] \[ 4 \leq 251 \quad \text{(True)} \]
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For the point \((5, 3)\): \[ y = 3 \quad \text{and} \quad 25x + 1 = 25(5) + 1 = 125 + 1 = 126 \] \[ 3 \leq 126 \quad \text{(True)} \]
All four points satisfy the inequality \( y \leq 25x + 1 \):
- \((10, 5.5)\)
- \((10, 6)\)
- \((10, 4)\)
- \((5, 3)\)
Thus, all points are part of the solution set for the inequality.