Question

Which set expresses the solution to the compound inequality —2 ≤
2x+1 - 4 < 1? (1 point)
Ite
Ite
0 {x:
0≤8K1
O {D: 블≤&<가
0 {x : 3 < x < 7}

Answer 1

To solve the compound inequality \(-2 \leq 2x + 1 - 4 < 1\), we first simplify it.

1. Combine constants in the middle part:
\[
2x + 1 - 4 = 2x - 3
\]

Thus, the compound inequality simplifies to:
\[
-2 \leq 2x - 3 < 1
\]

2. Now we will split it into two parts:
- For the left part: \(-2 \leq 2x - 3\)
- For the right part: \(2x - 3 < 1\)

**Solving the left part**:
\[
-2 \leq 2x - 3
\]
Add 3 to both sides:
\[
1 \leq 2x
\]
Divide by 2:
\[
\frac{1}{2} \leq x \quad \text{or} \quad x \geq \frac{1}{2}
\]

**Solving the right part**:
\[
2x - 3 < 1
\]
Add 3 to both sides:
\[
2x < 4
\]
Divide by 2:
\[
x < 2
\]

3. Putting it all together, we have:
\[
\frac{1}{2} \leq x < 2
\]

The solution in interval notation is \([\frac{1}{2}, 2)\) or in set notation:
\[
\{x : \frac{1}{2} \leq x < 2\}
\]

Given your options, it seems like there might be some formatting issues. The closest option to represent this solution is:
\[
\{x : \frac{1}{2} \leq x < 2\}
\]

Make sure to confirm the available options match what was presented!