To find a point that satisfies both inequalities \( y > 4x - 2 \) and \( y \geq -\frac{1}{3}x + 2 \), we first need to analyze what each inequality represents graphically.
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For the inequality \( y > 4x - 2 \):
- The line \( y = 4x - 2 \) has a slope of 4 and a y-intercept at -2.
- The region above this line (not including the line itself) represents all the points that satisfy this inequality.
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For the inequality \( y \geq -\frac{1}{3}x + 2 \):
- The line \( y = -\frac{1}{3}x + 2 \) has a slope of -1/3 and a y-intercept at 2.
- The region on or above this line represents all points that satisfy this inequality.
Next, we need to find a point (x, y) that is in the region defined by both inequalities.
Let's evaluate some points.
Test Point (0, 3):
- For \( y > 4x - 2 \): \[ 3 > 4(0) - 2 \quad \Rightarrow \quad 3 > -2 \quad \text{(True)} \]
- For \( y \geq -\frac{1}{3}x + 2 \): \[ 3 \geq -\frac{1}{3}(0) + 2 \quad \Rightarrow \quad 3 \geq 2 \quad \text{(True)} \]
Thus, the point (0, 3) satisfies both inequalities.
Conclusion
A possible solution point (x, y) that satisfies both inequalities is: \[ \boxed{0 ; 3} \]
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