Certainly! Let's go through the problem step by step.
1. Understanding inverse variation: When [tex]\( y \)[/tex] varies inversely as [tex]\( x \)[/tex], it means that the product of [tex]\( y \)[/tex] and [tex]\( x \)[/tex] is a constant. Mathematically, we can express this as:
[tex]\[
y \cdot x = k
\][/tex]
where [tex]\( k \)[/tex] is the constant.
2. Finding the constant [tex]\( k \)[/tex]: Given in the problem, [tex]\( y = 12 \)[/tex] when [tex]\( x = 5 \)[/tex]. We can use these values to find the constant [tex]\( k \)[/tex]:
[tex]\[
12 \cdot 5 = k
\][/tex]
[tex]\[
k = 60
\][/tex]
3. Using the constant [tex]\( k \)[/tex] to find [tex]\( y \)[/tex] when [tex]\( x = 2 \)[/tex]:
We now use the constant [tex]\( k \)[/tex] to find the new value of [tex]\( y \)[/tex] when [tex]\( x = 2 \)[/tex]. We use the inverse variation equation again:
[tex]\[
y \cdot 2 = 60
\][/tex]
4. Solving for [tex]\( y \)[/tex]:
[tex]\[
y = \frac{60}{2}
\][/tex]
[tex]\[
y = 30
\][/tex]
Thus, when [tex]\( x = 2 \)[/tex], the value of [tex]\( y \)[/tex] is [tex]\( 30 \)[/tex].
