Which set expresses the solution to the compound inequality —2 ≤

Question

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!