\(|\frac{y}{2} - \frac{1}{6}| < \frac{2}{3}\)

Case 1: \(\frac{y}{2} - \frac{1}{6} <\frac{2}{3}\)
\(\frac{y}{2} < \frac{2}{3} + \frac{1}{6}\)
\(\frac{y}{2} < \frac{4+ 1}{6}\)
\(\frac{y}{2} < \frac{5}{6}\)
\(y<\frac{5}{3}\)

Case 2: \(- (\frac{y}{2} - \frac{1}{6}) < \frac{2}{3}\)
\(-\frac{y}{2} + \frac{1}{6} < \frac{2}{3}\)
\(-\frac{y}{2} < \frac{2}{3} - \frac{1}{6}\)
\(-\frac{y}{2} < \frac{4 - 1}{6}\)
\(-\frac{y}{2} < \frac{3}{6}\)
\(-\frac{y}{2} < \frac{1}{2}\)
\(-y < 1\)
\(y > -1\)

Value of \(y ==> -1 < y < \frac{5}{3}\)

Values which come between this range are \(-\frac{5}{6}, 0, \frac{5}{6}, \frac{4}{3}.\)

Therefore y cannot be \(-\frac{4}{3}\). Answer A...

For this absolute value problem, we have two cases to consider: when (y/2 - ⅙) is positive and when (y/2 - ⅙) is negative. Let’s start with the positive case:

Case 1: (y/2 - ⅙) is positive

y/2 - 1/6 < 2/3

Multiplying by 6, we have:

3y - 1 < 4

3y < 5

y < 5/3

Case 2: (y/2 - ⅙) is negative

-(y/2 - ⅙) < 2/3

-y/2 + 1/6 < ⅔

Multiplying by 6, we have:

-3y + 1 < 4

-3y < 3

y > -1

Thus, -1 < y < 5/3. Thus, y cannot be -4/3.

Answer: A
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