To find a point that is a solution to the inequalities \(y > 3x + 1\) and \(y \leq 2x + 5\), we need to identify points where the y-value satisfies both conditions.
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Graph the inequalities:
- The inequality \(y > 3x + 1\) represents the region above the line \(y = 3x + 1\).
- The inequality \(y \leq 2x + 5\) represents the region below or on the line \(y = 2x + 5\).
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Find the intersection: To see where these inequalities may intersect, set the equations equal to each other: \[ 3x + 1 = 2x + 5 \] Subtract \(2x\) from both sides: \[ x + 1 = 5 \] Subtract 1 from both sides: \[ x = 4 \] Now substitute \(x = 4\) back into either equation to find \(y\): \[ y = 3(4) + 1 = 12 + 1 = 13 \quad \text{(from the first equation)} \] or \[ y = 2(4) + 5 = 8 + 5 = 13 \quad \text{(from the second equation)} \] So the intersection point is \((4, 13)\).
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Check regions:
- For the inequality \(y > 3x + 1\), points above the line satisfy the inequality.
- For the inequality \(y \leq 2x + 5\), points below or on the line satisfy the inequality.
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Find valid region: The valid region, therefore, is where \(y\) values are greater than \(3x + 1\) but less than or equal to \(2x + 5\).
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Testing points: We can test points on the graph within both inequalities.
If points on the graph are provided (your request didn't specify which specific points were shown), we can substitute the coordinates into both inequalities to find solutions.
However, if you provide specific (x, y) pairs from the graph, I can confirm if any of those points satisfy both conditions.
For example, if looking at a potential point like \((0, 4)\):
- Check: \[ 4 > 3(0) + 1 \Rightarrow 4 > 1 \quad \text{(True)} \] \[ 4 \leq 2(0) + 5 \Rightarrow 4 \leq 5 \quad \text{(True)} \] So \((0, 4)\) would be a valid solution.
If you list specific points, I can assist in determining which one(s) satisfy the simultaneous inequalities.