To solve the given problem, we need to use the concept of inverse variation. When two quantities vary inversely, their product remains constant. In mathematical terms, if [tex]\( y \)[/tex] varies inversely with [tex]\( x \)[/tex], then:
[tex]\[ y \cdot x = k \][/tex]
where [tex]\( k \)[/tex] is a constant.
Given:
[tex]\[ y_1 = 3 \][/tex]
[tex]\[ x_1 = 10 \][/tex]
First, let's determine the constant [tex]\( k \)[/tex]:
[tex]\[ k = y_1 \cdot x_1 \][/tex]
[tex]\[ k = 3 \cdot 10 \][/tex]
[tex]\[ k = 30 \][/tex]
Now, we need to find [tex]\( y_2 \)[/tex] when [tex]\( x_2 = 6 \)[/tex]. According to the inverse variation relationship:
[tex]\[ y_2 \cdot x_2 = k \][/tex]
Substitute [tex]\( k \)[/tex] and [tex]\( x_2 \)[/tex] into the equation:
[tex]\[ y_2 \cdot 6 = 30 \][/tex]
To solve for [tex]\( y_2 \)[/tex], divide both sides of the equation by 6:
[tex]\[ y_2 = \frac{30}{6} \][/tex]
[tex]\[ y_2 = 5 \][/tex]
Thus, the value of [tex]\( y \)[/tex] when [tex]\( x \)[/tex] is 6 is [tex]\( 5 \)[/tex].
