Question 246151
if y varies directly as x and inversely as z^2, and y=3 when x=2 and z=4 what is the value of x when y=9 and z=4?

Restated the above says y = k*(x/(z^2)) where k is a constant of proportionality yet to be determined.

As stated above y=3 when x=2 and z=4 so substituting in our formula:

3 = k*(2/(4^2))
3 = k*(2/16)= k/8
k=24

So our equation becomes y=24*(x/(z^2))

To solve the problem we need to substitute y = 9 and z = 4 and then solve for x in the equation above.