The problem requires solving an absolute value equation and then evaluating an algebraic expression.

Solving the Absolute Value Equation \( |9y-6|=3 \)

An absolute value equation of the form \( |ax-b|=c \) has two possible cases:

• Case 1: \( 9y-6 = 3 \)
• Case 2: \( 9y-6 = -3 \)

Case 1 Calculation

Solve for \( y \) when \( 9y-6 = 3 \):

$9y = 3 + 6$

$9y = 9$

$y = \frac{9}{9}$

$y = 1$

Case 2 Calculation

Solve for \( y \) when \( 9y-6 = -3 \):

$9y = -3 + 6$

$9y = 3$

$y = \frac{3}{9}$

$y = \frac{1}{3}$

The possible values for \( y \) are \( 1 \) and \( \frac{1}{3} \).

Evaluating the Expression \( y^2 - 4y/3 \)

Now, substitute each value of \( y \) into the expression \( y^2 - \frac{4y}{3} \).

Evaluation for \( y = 1 \)

Substitute \( y = 1 \):

$ (1)^2 - \frac{4(1)}{3} = 1 - \frac{4}{3} $

$ = \frac{3}{3} - \frac{4}{3} = -\frac{1}{3} $

Evaluation for \( y = 1/3 \)

Substitute \( y = \frac{1}{3} \):

$ \left(\frac{1}{3}\right)^2 - \frac{4\left(\frac{1}{3}\right)}{3} = \frac{1}{9} - \frac{\frac{4}{3}}{3} $

$ = \frac{1}{9} - \frac{4}{9} = -\frac{3}{9} $

$ = -\frac{1}{3} $

Conclusion

For both possible values of \( y \), the expression \( y^2 - \frac{4y}{3} \) evaluates to \( -\frac{1}{3} \).