Question

Which set expresses the solution to the compound inequality −2≤2x+1/3 −4<1
?(1 point)
Responses

{x:3≤x<7}
left brace x colon 3 less-than-or-equal-to x less than 7 right brace

{x:−3/2≤x<1}
left brace x colon Start Fraction negative 3 over 2 End Fraction less-than-or-equal-to x less than 1 right brace

{x:5/2≤x<7}
left brace x colon Start Fraction 5 over 2 End Fraction less-than-or-equal-to x less than 7 right brace

{x:5/2≤x<1}

Answer 1

To solve the compound inequality \(-2 \leq 2x + \frac{1}{3} - 4 < 1\), we need to break it down into two parts:

1. \(-2 \leq 2x + \frac{1}{3} - 4\)
2. \(2x + \frac{1}{3} - 4 < 1\)

### Step 1: Solve the first inequality

\[
-2 \leq 2x + \frac{1}{3} - 4
\]

First, simplify the right side:

\[
-2 \leq 2x - \frac{11}{3}
\]

Next, add \(\frac{11}{3}\) to both sides:

\[
-2 + \frac{11}{3} \leq 2x
\]

Converting \(-2\) to a fraction gives \(-\frac{6}{3}\):

\[
-\frac{6}{3} + \frac{11}{3} \leq 2x
\]

\[
\frac{5}{3} \leq 2x
\]

Now, divide both sides by 2:

\[
\frac{5}{6} \leq x
\]

### Step 2: Solve the second inequality

\[
2x + \frac{1}{3} - 4 < 1
\]

Again, simplifying the left side:

\[
2x - \frac{11}{3} < 1
\]

Now add \(\frac{11}{3}\) to both sides:

\[
2x < 1 + \frac{11}{3}
\]

Convert \(1\) to a fraction:

\[
1 = \frac{3}{3}
\]

Thus,

\[
2x < \frac{3}{3} + \frac{11}{3} = \frac{14}{3}
\]

Divide both sides by 2:

\[
x < \frac{7}{3}
\]

### Combine the Inequalities

Putting the two results together, we have:

\[
\frac{5}{6} \leq x < \frac{7}{3}
\]

### Final Expression in Set Notation

This can be expressed in set notation as:

\[
\{ x : \frac{5}{6} \leq x < \frac{7}{3} \}
\]

However, none of the options explicitly match this range. Let's check the options:

1. \(\{x:3 \leq x < 7\}\)
2. \(\{x: -\frac{3}{2} \leq x < 1\}\)
3. \(\{x: \frac{5}{2} \leq x < 7\}\)
4. \(\{x: \frac{5}{2} \leq x < 1\}\)

None of these options include the correct interval between \(\frac{5}{6}\) and \(\frac{7}{3}\).

Thus, it looks like all given options are incorrect concerning the provided compound inequality.