To determine which inequality Willow solved, we can examine the provided options and see which one would yield a solution that can be graphically represented.
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\(x + 212 < 212\)
- This simplifies to \(x < 0\).
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\(x + 2 \frac{1}{2} < 2 \frac{1}{2}\)
- This simplifies to \(x + 2.5 < 2.5\), which gives \(x < 0\).
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\(x - 2 \frac{1}{3} < 2 \frac{1}{6}\)
- This simplifies to \(x - \frac{7}{3} < \frac{13}{6}\).
- Adding \(\frac{7}{3}\) to both sides converts \(\frac{7}{3}\) into a common denominator of \(6\), making \(x < \frac{13}{6} + \frac{14}{6}\).
- Hence, \(x < \frac{27}{6}\), or \(x < 4.5\).
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\(x - 1.5 \leq 3\)
- This simplifies to \(x \leq 4.5\).
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\(x + 1.25 \leq 5.75\)
- This simplifies to \(x \leq 4.5\).
From the solutions obtained, we have:
- For options 1 and 2, the solution is \(x < 0\).
- For options 3 and 4, and 5, the solution \(x \leq 4.5\).
If Willow graphed a solution that includes a boundary or mark indicating possible equality up to \(4.5\), then she solved either \(x - 1.5 \leq 3\) or \(x + 1.25 \leq 5.75\).
If her graph indicated an open interval or a more exclusive range, it could imply either \(x + 212 < 212\) or \(x + 2 \frac{1}{2} < 2 \frac{1}{2}\).
Without the actual graph, it's hard to conclude definitively. However, the inequalities that suggest a specific endpoint with inclusivity (noted by ≤) are likely the most relevant if the graph shows that kind of interval.
Given the nature of typical inequalities and their graphing interpretations, the most likely candidate that Willow solved would be \(x - 1.5 \leq 3\) (also reflects \(x \leq 4.5\)).
So the final identified inequality would be:
\(x - 1.5 \leq 3\)
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