Question 4

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

(1 point)
Responses

{x
:52
≤ x
< 7
}
{x :5 halves ≤ x < 7}

{x
:52
≤ x
< 1
}
{x :5 halves ≤ x < 1}

{x
:3
≤ x
< 7
}
{x :3 ≤ x < 7}

{x
:−32
≤ x
< 1
}

1 answer

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

Step 1: Solve the left part of the compound inequality

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

  1. Combine like terms: \[ -2 \leq 2x - \frac{11}{3} \]

  2. Add \(\frac{11}{3}\) to both sides: \[ \frac{-6}{3} + \frac{11}{3} \leq 2x \] \[ \frac{5}{3} \leq 2x \]

  3. Divide both sides by 2: \[ \frac{5}{6} \leq x \]

Step 2: Solve the right part of the compound inequality

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

  1. Combine like terms: \[ 2x - \frac{11}{3} < 1 \]

  2. Add \(\frac{11}{3}\) to both sides: \[ 2x < 1 + \frac{11}{3} \] \[ 2x < \frac{3}{3} + \frac{11}{3} \] \[ 2x < \frac{14}{3} \]

  3. Divide both sides by 2: \[ x < \frac{7}{3} \]

Step 3: Combine the results

Now we have the results:

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

Final Step: Convert to set notation

The solution in set notation is:

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

To express \(\frac{5}{6}\) in decimal form (which is 0.833...) and \(\frac{7}{3}\) in decimal form (which is approximately 2.333...), we can see that none of the original answer choices have the notation in this form. The closest aligned standardized choice in term of representation appears to be:

{ x : 5 halves ≤ x < 7 }

Since \( 5/2 = 2.5\) is greater than \( 7/3 \). Therefore, you should check if all options correspond accurately. None of them matched directly. Can you please confirm the original options, or give clarity if you meant to consider fractions directly without converting them?