To solve this problem, we need to work through the concept of direct variation. When [tex]\(y\)[/tex] varies directly as [tex]\(x\)[/tex], we have the following:
[tex]\[ y = kx \][/tex]
where [tex]\(k\)[/tex] is the constant of variation.
### Step 1: Find the constant of variation [tex]\(k\)[/tex]
We're given that [tex]\( y = 48 \)[/tex] when [tex]\( x = 6 \)[/tex]. Plugging these values into the equation, we get:
[tex]\[ 48 = k \cdot 6 \][/tex]
To find [tex]\(k\)[/tex], solve for [tex]\(k\)[/tex]:
[tex]\[ k = \frac{48}{6} \][/tex]
So, the constant of variation [tex]\(k\)[/tex] is 8.
### Step 2: Use the constant to find [tex]\(y\)[/tex] when [tex]\( x = 2 \)[/tex]
Now, we know that the relationship between [tex]\(y\)[/tex] and [tex]\(x\)[/tex] is given by:
[tex]\[ y = 8x \][/tex]
We need to find the value of [tex]\( y \)[/tex] when [tex]\( x = 2 \)[/tex]. Plug [tex]\( x = 2 \)[/tex] into the equation:
[tex]\[ y = 8 \cdot 2 \][/tex]
[tex]\[ y = 16 \][/tex]
### Step 3: Identify the correct expression
Now that we know [tex]\( k = 8 \)[/tex], we can match this with the given expressions. The correct expression for finding [tex]\( y \)[/tex] when [tex]\( x = 2 \)[/tex] is:
[tex]\[ y = \frac{48}{6}(2) \][/tex]
Hence, the answer is:
[tex]\[ \boxed{y = \frac{48}{6}(2)} \][/tex]
