To determine which point is a solution to the simultaneous inequalities \( y > 3x + 1 \) and \( y \leq 2x + 5 \), we can analyze the inequalities.
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Inequality \( y > 3x + 1 \):
- This inequality represents the region above the line \( y = 3x + 1 \).
-
Inequality \( y \leq 2x + 5 \):
- This inequality represents the region below or on the line \( y = 2x + 5 \).
A solution to the simultaneous inequalities must be in the region where both conditions are satisfied:
- The point must be above the line \( y = 3x + 1 \).
- The point must be on or below the line \( y = 2x + 5 \).
Without a specific graph or points provided, you would need to test any given points (x, y) against both inequalities to find a point that fulfills both conditions.
For example:
- Suppose you have the point (1, 5).
- Check \( y > 3x + 1 \): \( 5 > 3(1) + 1 \) → \( 5 > 4 \) (True)
- Check \( y \leq 2x + 5 \): \( 5 \leq 2(1) + 5 \) → \( 5 \leq 7 \) (True)
Since both conditions are satisfied, (1, 5) is a solution to the simultaneous inequalities.
If you have specific points from a graph you would like to check, please provide them, and I can help verify which one is a solution!
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