Solve this?

Question

If y varies directly as x and inversely as z^2, and y = 3 when x = 2 and z = 4, what is the calue of x when y = 9 and z = 4?

Answer 1

let y = k [ x / (z^2) ] , where k is a constant.

when y = 3, x = 2 & z = 4:
3 = k [ 2 / (4^2) ]
k = 24

so, y = (24)[ x / (z^2) ]

when y = 9 & z = 4,
9 = (24)[ x / (4^2) ] = (24)[ x / 16] = (3/2) x

hence, x = 6.

Answer 2

y = k x / z²
k = y z² / x = 3 x 16 / 2 = 24
x = y z² / k
x = (9 x 16) / 24
x = 9 x 2 / 3
x = 6