To find a point on the graph that satisfies the simultaneous inequalities \( y \leq 14x + 2 \) and \( y > 12x + 3 \), we can test a few values of \( x \) in the inequalities.
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Choose a value for \( x \): Let's try \( x = 0 \).
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For \( y \leq 14(0) + 2 \): \[ y \leq 2 \]
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For \( y > 12(0) + 3 \): \[ y > 3 \]
So there is no \( y \) value that can satisfy both conditions simultaneously with \( x = 0 \).
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Choose another value for \( x \): Let’s try \( x = 1 \).
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For \( y \leq 14(1) + 2 \): \[ y \leq 16 \]
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For \( y > 12(1) + 3 \): \[ y > 15 \]
To satisfy both inequalities, \( y \) must be greater than 15 but less than or equal to 16. An example point would be \( (1, 15.5) \).
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Verifying the solution:
- Is \( 15.5 \leq 16 \)? Yes.
- Is \( 15.5 > 15 \)? Yes.
Thus, the point \( (1, 15.5) \) is indeed a solution to the simultaneous inequalities.
The final answer is:
1, 15.5
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