Answer:
[tex]x[/tex] would be [tex](1/3)[/tex] when [tex]y = 3[/tex].
Step-by-step explanation:
The question states that [tex]x[/tex] varies "inversely" as [tex]y^2[/tex]. In other words, there is a non-zero number [tex]a[/tex] (a constant) such that:
[tex]\displaystyle x = \frac{a}{y^2}[/tex].
The challenging part is to find the value of [tex]a[/tex] in this equation.
Given that [tex]x = 3[/tex] when [tex]y = -1[/tex], replace the [tex]x[/tex] in this equation with [tex]3[/tex] and [tex]y[/tex] with [tex](-1)[/tex]. This equation should still be valid:
[tex]\displaystyle 3 = \frac{a}{(-1)^{2}}[/tex].
Solve for [tex]a[/tex]:
[tex]a = 3 \times (-1)^2 = 3[/tex].
Hence, the relation between [tex]x[/tex] and [tex]y[/tex] becomes:
[tex]\displaystyle x = \frac{3}{y^2}[/tex].
Find the value of [tex]x[/tex] when [tex]y = 3[/tex] by replacing the [tex]y[/tex] in this equation with [tex]3[/tex].
[tex]\displaystyle x = \frac{3}{3^2} = \frac{1}{3}[/tex].
In other words, the value of [tex]x[/tex] would be [tex]\displaystyle \frac{1}{3}[/tex] when [tex]y = 3[/tex].
