To solve the compound inequality \(-2 \leq 2x + \frac{1}{3} - 4 < 1\), we can break it into two parts:
- \(-2 \leq 2x + \frac{1}{3} - 4\)
- \(2x + \frac{1}{3} - 4 < 1\)
Step 1: Solve the first part:
\[ -2 \leq 2x + \frac{1}{3} - 4 \]
First, simplify \(\frac{1}{3} - 4\):
\[ \frac{1}{3} - 4 = \frac{1}{3} - \frac{12}{3} = \frac{1 - 12}{3} = \frac{-11}{3} \]
So the first inequality becomes:
\[ -2 \leq 2x - \frac{11}{3} \]
Add \(\frac{11}{3}\) to both sides:
\[ -2 + \frac{11}{3} \geq 2x \]
Convert \(-2\) to a fraction:
\[ -\frac{6}{3} + \frac{11}{3} = \frac{5}{3} \]
This gives us:
\[ \frac{5}{3} \geq 2x \quad \text{or} \quad 2x \leq \frac{5}{3} \]
Now, divide both sides by 2:
\[ x \leq \frac{5}{6} \]
Step 2: Solve the second part:
\[ 2x + \frac{1}{3} - 4 < 1 \]
Using the same simplification as before:
\[ 2x - \frac{11}{3} < 1 \]
Add \(\frac{11}{3}\) to both sides:
\[ 2x < 1 + \frac{11}{3} \]
Convert \(1\) to a fraction:
\[ 1 = \frac{3}{3} \]
So, this becomes:
\[ 2x < \frac{3}{3} + \frac{11}{3} = \frac{14}{3} \]
Now, divide both sides by 2:
\[ x < \frac{14}{6} = \frac{7}{3} \]
Step 3: Combine the results:
From the first part, we have:
\[ x \leq \frac{5}{6} \]
From the second part, we have:
\[ x < \frac{7}{3} \]
Since \(\frac{5}{6} \approx 0.833\) is less than \(\frac{7}{3} \approx 2.333\), the solution set for the compound inequality can be written as:
\[ x \leq \frac{5}{6} \]
Thus, the correct answer for the set that expresses the solution to the compound inequality is:
\[ { x : -\frac{3}{2} \leq x < 1 } \]
So the correct option is:
\({ x : -\frac{3}{2} \leq x < 1 }\)
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