To determine the solutions to the simultaneous inequalities \(y > 12x + 5\) and \(y < -2x + 1\), we can analyze the given options by substituting the \(x\) and \(y\) values from each point into both inequalities.
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For point A \((1, 5.5)\):
- Check \(y > 12x + 5\): \[ 5.5 > 12(1) + 5 \] \[ 5.5 > 12 + 5 \] \[ 5.5 > 17 \] (False)
- Since the first inequality is false, A is not a solution.
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For point B \((-5, 5)\):
- Check \(y > 12x + 5\): \[ 5 > 12(-5) + 5 \] \[ 5 > -60 + 5 \] \[ 5 > -55 \] (True)
- Check \(y < -2x + 1\): \[ 5 < -2(-5) + 1 \] \[ 5 < 10 + 1 \] \[ 5 < 11 \] (True)
- Since both inequalities are true, B is a solution.
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For point C \((-1.6, 4.2)\):
- Check \(y > 12x + 5\): \[ 4.2 > 12(-1.6) + 5 \] \[ 4.2 > -19.2 + 5 \] \[ 4.2 > -14.2 \] (True)
- Check \(y < -2x + 1\): \[ 4.2 < -2(-1.6) + 1 \] \[ 4.2 < 3.2 + 1 \] \[ 4.2 < 4.2 \] (False)
- Since the second inequality is false, C is not a solution.
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For point D \((4, 8)\):
- Check \(y > 12x + 5\): \[ 8 > 12(4) + 5 \] \[ 8 > 48 + 5 \] \[ 8 > 53 \] (False)
- Since the first inequality is false, D is not a solution.
The only point that satisfies both inequalities is: B) (-5, 5).
Now, for the second part regarding the system of inequalities:
Since the two inequalities represent two lines that have a certain area between them, we can assume:
- The inequality \(y > 12x + 5\) represents the area above the line \(y = 12x + 5\).
- The inequality \(y < -2x + 1\) represents the area below the line \(y = -2x + 1\).
Given that the two areas (for the systems of inequalities) are distinct, and provided that one side limits the solutions of the other, it’s correct to conclude:
A) The system has infinite solutions.
So, in summary:
- Point showing a solution: B) (−5, 5).
- True statement about the system of inequalities: A) The system has infinite solutions.