If [tex]\( y \)[/tex] varies directly as [tex]\( x \)[/tex], it means there is a constant [tex]\( k \)[/tex] such that [tex]\( y = kx \)[/tex].
Given the information:
- [tex]\( y = 48 \)[/tex] when [tex]\( x = 6 \)[/tex], we can find [tex]\( k \)[/tex] as follows:
[tex]\[ 48 = k \cdot 6 \][/tex]
Solving for [tex]\( k \)[/tex]:
[tex]\[ k = \frac{48}{6} \][/tex]
[tex]\[ k = 8 \][/tex]
Now that we have the constant of variation ([tex]\( k = 8 \)[/tex]), we need to find the value of [tex]\( y \)[/tex] when [tex]\( x = 2 \)[/tex].
Using the direct variation equation [tex]\( y = kx \)[/tex]:
[tex]\[ y = 8 \cdot 2 \][/tex]
[tex]\[ y = 16 \][/tex]
To match this result with the expressions given in the problem:
[tex]\[ y = \frac{48}{6} \cdot 2 \][/tex]
Therefore, the correct expression to use is:
[tex]\[ y = \frac{48}{6} (2) \][/tex]
The other expressions are incorrect because:
- [tex]\( y = \frac{6}{48} (2) \)[/tex] does not use the correct ratio.
- [tex]\( y = \frac{(48)(6)}{2} \)[/tex] incorrectly multiplies and then divides by [tex]\( 2 \)[/tex].
- [tex]\( y = \frac{2}{(48)(6)} \)[/tex] incorrectly places [tex]\( 48 \)[/tex] and [tex]\( 6 \)[/tex] in the denominator.
So the expression that can be used to find the value of [tex]\( y \)[/tex] when [tex]\( x \)[/tex] is 2, is:
[tex]\[ y = \frac{48}{6} (2) \][/tex]
The value of [tex]\( y \)[/tex] when [tex]\( x = 2 \)[/tex] is [tex]\( 16 \)[/tex], and the corresponding expression is [tex]\( y = \frac{48}{6} (2) \)[/tex].
