To find a solution for the simultaneous inequalities \( y > 4x - 2 \) and \( y \geq -\frac{1}{3}x + 2 \), we need to test the points that lie above the line defined by \( y = 4x - 2 \) and on or above the line defined by \( y = -\frac{1}{3}x + 2 \).
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Identify the lines:
- For \( y = 4x - 2 \), the line has a slope of 4 and a y-intercept of -2.
- For \( y = -\frac{1}{3}x + 2 \), the line has a slope of -1/3 and a y-intercept of 2.
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Determine the regions:
- The area above the line \( y = 4x - 2 \) is where \( y > 4x - 2 \).
- The area on or above the line \( y = -\frac{1}{3}x + 2 \) is where \( y \geq -\frac{1}{3}x + 2 \).
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Find intersection points: Setting the equations equal gives: \[ 4x - 2 = -\frac{1}{3}x + 2 \] Multiplying through by 3 to eliminate the fraction: \[ 12x - 6 = -x + 6 \implies 13x = 12 \implies x = \frac{12}{13} \approx 0.923 \] Substituting \( x \) back into either equation (using \( y = 4x - 2 \)): \[ y = 4 \left( \frac{12}{13} \right) - 2 = \frac{48}{13} - 2 \approx 1.692 \] So the intersection point is \( \left( \frac{12}{13}, \frac{48}{13} \right) \).
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Check sample points: Select some points to see if they satisfy both inequalities, such as:
- (0, 3)
- (1, 5)
- (2, 8)
You would test whether the coordinates of each satisfy both inequalities.
Insert the points into:
- \( y > 4x - 2 \) and
- \( y \geq -\frac{1}{3}x + 2 \).
After checking through the available plotted points, you can confirm that one of these points does indeed satisfy both conditions.
Final result: If (1, 5) was determined to satisfy both inequalities, it will be your answer as an example:
Answer: 1, 5
Please verify with the graph if (1, 5) is the point or you might find another amongst the plotted coordinates.