To solve this problem, we need to recall the concept of inverse variation. When we say that [tex]\( y \)[/tex] varies inversely as [tex]\( 3x - 2 \)[/tex], we mean that:
[tex]\[ y \cdot (3x - 2) = k \][/tex]
where [tex]\( k \)[/tex] is a constant.
Given in the problem:
[tex]\[ y = 4 \text{ when } x = 2 \][/tex]
Plug these values into the inverse variation equation to find [tex]\( k \)[/tex]:
[tex]\[ 4 \cdot (3 \cdot 2 - 2) = k \][/tex]
[tex]\[ 4 \cdot (6 - 2) = k \][/tex]
[tex]\[ 4 \cdot 4 = k \][/tex]
[tex]\[ k = 16 \][/tex]
Now we need to find the value of [tex]\( x \)[/tex] when [tex]\( y = -4 \)[/tex]. Use the constant [tex]\( k \)[/tex] that we just found:
[tex]\[ -4 \cdot (3x - 2) = 16 \][/tex]
Solve for [tex]\( x \)[/tex]:
[tex]\[ -4 \cdot (3x - 2) = 16 \][/tex]
[tex]\[ 3x - 2 = \frac{16}{-4} \][/tex]
[tex]\[ 3x - 2 = -4 \][/tex]
Next, isolate [tex]\( x \)[/tex]:
[tex]\[ 3x = -4 + 2 \][/tex]
[tex]\[ 3x = -2 \][/tex]
[tex]\[ x = \frac{-2}{3} \][/tex]
Thus, the value of [tex]\( x \)[/tex] when [tex]\( y = -4 \)[/tex] is:
[tex]\[ x = -\frac{2}{3} \][/tex]
So, the correct answer is:
A. [tex]\(-\frac{2}{3}\)[/tex]
