To determine which points are part of the solution set for the inequality \( y \leq \frac{2}{5}x + 1 \), we will substitute each point into the inequality and check if the inequality holds.
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For the point \( (10, 6) \): \[ y = 6 \quad \text{and} \quad \frac{2}{5}(10) + 1 = 4 + 1 = 5 \] Check: \( 6 \leq 5 \) (False)
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For the point \( (10, 4) \): \[ y = 4 \quad \text{and} \quad \frac{2}{5}(10) + 1 = 5 \] Check: \( 4 \leq 5 \) (True)
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For the point \( (-10, 6) \): \[ y = 6 \quad \text{and} \quad \frac{2}{5}(-10) + 1 = -4 + 1 = -3 \] Check: \( 6 \leq -3 \) (False)
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For the point \( (10, 5.5) \): \[ y = 5.5 \quad \text{and} \quad \frac{2}{5}(10) + 1 = 5 \] Check: \( 5.5 \leq 5 \) (False)
The only point that satisfies the inequality \( y \leq \frac{2}{5}x + 1 \) is: (10, 4).