If x varies inversely as y and \(x = \frac{2}{3}\) when y = 9, find the value of y when \(x=\frac{3}{4}\)

A.

\(\frac{1}{18}\)

B.

\(\frac{8}{81}\)

C.

\(\frac{9}{2}\)

D.

8

Correct answer is D

\(x \propto \frac{1}{y}\)

\(x = \frac{k}{y}\)

\(\frac{2}{3} = \frac{k}{9}\)

\(3k = 18 \implies k = 6\)

\(x = \frac{6}{y}\)

When y = \(\frac{3}{4}\),

x = \(\frac{6}{\frac{3}{4}}\)

= \(\frac{6 \times 4}{3}\)

= 8