To solve the compound inequality \(-2 \leq \frac{2x + 1}{3} - 4 < 1\), we will break it down into two parts.
First, let's address the left side of the compound inequality:
- \(-2 \leq \frac{2x + 1}{3} - 4\)
Add 4 to both sides:
\[ -2 + 4 \leq \frac{2x + 1}{3} \]
This simplifies to:
\[ 2 \leq \frac{2x + 1}{3} \]
Next, multiply both sides by 3:
\[ 6 \leq 2x + 1 \]
Subtract 1 from both sides:
\[ 5 \leq 2x \]
Now divide by 2:
\[ \frac{5}{2} \leq x \]
This can also be expressed as:
\[ x \geq \frac{5}{2} \]
Now let's address the right side of the compound inequality:
- \(\frac{2x + 1}{3} - 4 < 1\)
Add 4 to both sides:
\[ \frac{2x + 1}{3} < 5 \]
Multiply both sides by 3:
\[ 2x + 1 < 15 \]
Subtract 1 from both sides:
\[ 2x < 14 \]
Now divide by 2:
\[ x < 7 \]
Now we have the solutions from both parts:
\[ \frac{5}{2} \leq x < 7 \]
In set notation, this is expressed as:
\[ { x : \frac{5}{2} \leq x < 7 } \]
Now we need to find the correct answer from the given options. The answer is closest to:
B) \({ x : \frac{5}{2} \leq x < 1 }\)
However, notice that it looks like there's a typo in the answer options because \(< 1\) is not correct based on our derived inequalities. The correct condition should state \(< 7\), which is not present.
Therefore, the answer depends on clarifying your available options, but based on the original compilation provided, the correct solution is:
\({ x : \frac{5}{2} \leq x < 7 }. \)
If we go strictly by the available options, there's no completely correct one. If we were to pick from these options strictly on the available constructions, and given that B is repeated, this might be a copy error.
Thus, I can only say none of the provided options strictly represents the solution. Please check the options again, as the correct derived solution doesn't match the available selections.
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