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?

May 31, 2015

If y varies inversely as the cube of x and directly as the square of z
then we can write:
$\textcolor{w h i t e}{\text{XXXXX}}$$\frac{y \cdot {x}^{3}}{{z}^{2}} = c$ for some constant $c$

If
$\textcolor{w h i t e}{\text{XXXXX}}$$\left(x , y , z\right) = \left(3 , - 6 , 9\right)$
is a solution for this equation, then

$\textcolor{w h i t e}{\text{XXXXX}}$$\frac{\left(- 6\right) \cdot {3}^{3}}{{9}^{2}} = c$
$\textcolor{w h i t e}{\text{XXXXX}}$$= \frac{162}{81} = c$
$\textcolor{w h i t e}{\text{XXXXX}}$$c = 2$

When $\left(x , z\right) = \left(6 , - 4\right)$
$\textcolor{w h i t e}{\text{XXXXX}}$$\frac{y \cdot {x}^{3}}{{z}^{2}} = 2$
becomes
$\textcolor{w h i t e}{\text{XXXXX}}$$\frac{y \cdot 216}{16} = 2$

$\textcolor{w h i t e}{\text{XXXXX}}$$y = \frac{16 \cdot 2}{216}$

$y = \frac{4}{27}$