The question involves a direct variation between x and y,therefore
[tex]\begin{gathered} y\text{ }\alpha\text{ x} \\ y=kx\ldots\ldots\ldots\ldots\text{.}(1) \end{gathered}[/tex]Given that,
[tex]\begin{gathered} y=8 \\ \text{when } \\ x=9 \end{gathered}[/tex]Substitute the values of x and y in equation (1)
[tex]\begin{gathered} y=kx \\ 8=k\times9 \\ 8=9k \\ \frac{9k}{9}=\frac{8}{9} \\ k=\frac{8}{9} \end{gathered}[/tex]Substituting the value of k=8/9 in equation (1), we will have the equation of variation to be
[tex]\begin{gathered} y=kx,\text{ becomes} \\ y=\frac{8}{9}x \end{gathered}[/tex]Hence,
The equation of the variation is y= 8x/9
Next, we will calculate the value of y when x=72 using the equation of variation above
[tex]\begin{gathered} \text{equation of variation is} \\ y=\frac{8}{9}x \\ \text{when x=72,we will have y to be} \\ y=\frac{8}{9}\times72 \\ y=\frac{576}{9} \\ y=64 \end{gathered}[/tex]Hence,
The value of y when x=72 is y=64
