If [tex]$y$[/tex] varies directly as [tex]$x$[/tex] and [tex]$z$[/tex], and [tex]$y=\frac{8}{3}$[/tex] when [tex]$x=1$[/tex] and [tex]$z=4$[/tex], find [tex]$y$[/tex] when [tex]$x=6$[/tex] and [tex]$z=3$[/tex].

a. [tex]$y=\frac{3}{4} x z ; y(6,3)=\frac{27}{2}$[/tex]

b. [tex]$y=\frac{3}{2} x z ; y(6,3)=27$[/tex]

c. [tex]$y=\frac{4}{3} x z ; y(6,3)=24$[/tex]

d. [tex]$y=\frac{2}{3} x z ; y(6,3)=12$[/tex]



Answer :

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]