Question

If y varies inversely as the cube of x and directly as the square of z and y = -6 when x=3 and z =9, how do you find y when x =6 and z= -4?

Answer 1

If y varies inversely as the cube of x and directly as the square of z
then we can write:
`color(white)("XXXXX")` `(y*x^3)/(z^2) = c` for some constant `c`

If
`color(white)("XXXXX")` `(x,y,z) = (3,-6,9)`
is a solution for this equation, then

`color(white)("XXXXX")` `((-6)*3^3)/(9^2) = c`
`color(white)("XXXXX")` `=162/81 = c`
`color(white)("XXXXX")` `c= 2`

When `(x,z) =(6,-4)`
`color(white)("XXXXX")` `(y*x^3)/(z^2) = 2`
becomes
`color(white)("XXXXX")` `(y*216)/16 = 2`

`color(white)("XXXXX")` `y = (16*2)/(216)`

`y = 4/27`