To determine which ordered pair makes both inequalities true, let's analyze each inequality and the corresponding points:
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First inequality: \( y < 3x - 1 \)
This means we need \( y \) to be less than the value of the expression \( 3x - 1 \). -
Second inequality: \( y > -x + 4 \)
This means we need \( y \) to be greater than the value of the expression \( -x + 4 \).
Now let's evaluate each of the four ordered pairs to see which satisfy both inequalities:
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For (4,0):
- \( y < 3x - 1 \) becomes \( 0 < 3(4) - 1 \) → \( 0 < 12 - 1 \) → \( 0 < 11 \) (True)
- \( y > -x + 4 \) becomes \( 0 > -4 + 4 \) → \( 0 > 0 \) (False)
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For (1,2):
- \( y < 3x - 1 \) becomes \( 2 < 3(1) - 1 \) → \( 2 < 3 - 1 \) → \( 2 < 2 \) (False)
- \( y > -x + 4 \) becomes \( 2 > -1 + 4 \) → \( 2 > 3 \) (False)
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For (0,4):
- \( y < 3x - 1 \) becomes \( 4 < 3(0) - 1 \) → \( 4 < -1 \) (False)
- \( y > -x + 4 \) becomes \( 4 > -0 + 4 \) → \( 4 > 4 \) (False)
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For (2,1):
- \( y < 3x - 1 \) becomes \( 1 < 3(2) - 1 \) → \( 1 < 6 - 1 \) → \( 1 < 5 \) (True)
- \( y > -x + 4 \) becomes \( 1 > -2 + 4 \) → \( 1 > 2 \) (False)
None of the pairs satisfy both inequalities. To find a point that satisfies both conditions, we can try checking points on the graph within the regions defined by the inequalities:
From the graphical representation, we are looking for a point in the intersection of the shaded areas from the two inequalities.
Given that the pairs provided do not fit the criteria for both inequalities, you might want to check writing in specific values close to the boundaries defined by the lines or try free or graphically using points that break boundaries set by the inequalities!
In conclusion, in the context of the pairs given, none of the options (4,0), (1,2), (0,4), and (2,1) satisfy both inequalities.