Question 303177
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=6 and z=2 

{{{y = kx/(z^2)}}} where k is constant

when y = 3 , x = 2 ,z =4 

{{{3=2k/(4^2)}}}

{{{3*16 = 2k }}}
{{{ k = 48/2 }}}
{{{ k = 24 }}}

Therefore, 

{{{y = 24x/(z^2) }}}

when y=6 ,z= 2 

{{{ 6 =24x/(2^2) }}}

{{{ 24x = 6*(4)}}}

x = 24/24
x = 1