To find a point that satisfies both inequalities \(y > 4x - 2\) and \(y \geq -\frac{1}{3}x + 2\), we can analyze both inequalities graphically.
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Line for \(y = 4x - 2\): This line has a slope of 4 and a y-intercept of -2. The area above this line (not including the line itself) is where \(y > 4x - 2\).
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Line for \(y = -\frac{1}{3}x + 2\): This line has a slope of \(-\frac{1}{3}\) and a y-intercept of 2. The area above this line (including the line itself) is where \(y \geq -\frac{1}{3}x + 2\).
Now we look for points that lie in the region satisfying both conditions.
One possible way to find a solution is to pick a specific value for \(x\) and calculate the corresponding \(y\) values:
For \(x = 0\):
- From \(y > 4(0) - 2\): \(y > -2\)
- From \(y \geq -\frac{1}{3}(0) + 2\): \(y \geq 2\)
Here, any \(y\) value that is greater than -2 and greater than or equal to 2 will satisfy both inequalities. So, \(y\) can be any value greater than or equal to 2 (like \(y=3\)).
Thus, one solution that meets both conditions is the point \((0, 3)\).
Answer: 0, 3