To solve the problem where [tex]\( y \)[/tex] varies directly as [tex]\( x \)[/tex] and [tex]\( z \)[/tex], we start by writing the relationship as:
[tex]\[ y = k \cdot x \cdot z \][/tex]
where [tex]\( k \)[/tex] is a constant.
Step-by-Step Solution:
1. Finding the constant [tex]\( k \)[/tex]:
We're given that [tex]\( y = \frac{8}{3} \)[/tex] when [tex]\( x = 1 \)[/tex] and [tex]\( z = 4 \)[/tex].
Substitute these values into the equation to find [tex]\( k \)[/tex]:
[tex]\[
\frac{8}{3} = k \cdot 1 \cdot 4
\][/tex]
Simplify this to:
[tex]\[
k = \frac{\frac{8}{3}}{4} = \frac{8}{3} \cdot \frac{1}{4} = \frac{8}{12} = \frac{2}{3}
\][/tex]
2. Using the constant [tex]\( k \)[/tex] to find [tex]\( y \)[/tex] when [tex]\( x = 6 \)[/tex] and [tex]\( z = 3 \)[/tex]:
Now, substitute [tex]\( x = 6 \)[/tex] and [tex]\( z = 3 \)[/tex], along with the constant [tex]\( k = \frac{2}{3} \)[/tex], back into the equation:
[tex]\[
y = \frac{2}{3} \cdot 6 \cdot 3
\][/tex]
Simplify the expression:
[tex]\[
y = \frac{2}{3} \cdot 18 = 2 \cdot 6 = 12
\][/tex]
So, when [tex]\( x = 6 \)[/tex] and [tex]\( z = 3 \)[/tex], we find that [tex]\( y = 12 \)[/tex].
Therefore, the correct answer is:
d. [tex]\(\ y = \frac{2}{3} x z ; y(6, 3) = 12 \)[/tex]
