To find a point that satisfies both inequalities \( y > 4x - 2 \) and \( y \geq -13x + 2 \), we need to analyze the two equations that represent the boundaries of the inequalities.
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First Inequality: \( y > 4x - 2 \)
- This line has a slope of 4 and a y-intercept of -2.
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Second Inequality: \( y \geq -13x + 2 \)
- This line has a slope of -13 and a y-intercept of 2.
The solution to the system of inequalities will be the region that is above the line \( y = 4x - 2 \) and above (or on) the line \( y = -13x + 2 \).
To find potential points that lie in the solution area, we can substitute some values of \( x \) into both equations and check for their corresponding values of \( y \). Let's check a couple of points:
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For \( x = 0 \):
- From \( y > 4(0) - 2 \): \( y > -2 \)
- From \( y \geq -13(0) + 2 \): \( y \geq 2 \)
- The point (0, 2) satisfies the second inequality but not the first, as \( 2 > -2 \).
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For \( x = 1 \):
- From \( y > 4(1) - 2 \): \( y > 2 \)
- From \( y \geq -13(1) + 2 \): \( y \geq -11 \)
- The point (1, 3) satisfies both inequalities as \( 3 > 2 \) and \( 3 \geq -11 \).
Thus, the point (1, 3) is a solution for the simultaneous inequalities \( y > 4x - 2 \) and \( y \geq -13x + 2 \).
Final result:
(1, 3)
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