To determine which points on the graph are solutions for the simultaneous inequalities \( y > 4x - 2 \) and \( y \geq -\frac{1}{3}x + 2 \), let's analyze the inequalities one by one.
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Inequality 1: \( y > 4x - 2 \)
- This represents a region above the line \( y = 4x - 2 \). The line itself is not included in the solution because it's a strict inequality (greater than).
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Inequality 2: \( y \geq -\frac{1}{3}x + 2 \)
- This represents a region above or on the line \( y = -\frac{1}{3}x + 2 \). Here, the line is included in the solution since it has a non-strict inequality (greater than or equal to).
To find a point that satisfies both inequalities, we can test points that are in the region described by both inequalities, ideally above both lines:
- Graph the lines:
- Find the intersection points of the lines, if any.
- Check for points above both lines. For example, testing a simple point like \( (0, 3) \):
- For \( y > 4x - 2 \): \( 3 > 4(0) - 2 \) ⇒ \( 3 > -2 \) (True)
- For \( y \geq -\frac{1}{3}x + 2 \): \( 3 \geq -\frac{1}{3}(0) + 2 \) ⇒ \( 3 \geq 2 \) (True)
Both conditions are satisfied, thus \( (0, 3) \) is a solution.
Feel free to test other points or calculate exact boundaries if you need specific coordinates!
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