# Suppose that y varies directly with x and inversely with z^2, & x=48 when y=8 and z=3. How do you find x when y=12 & z=2?

$x = 32$

#### Explanation:

Equation can be built
$y = k \cdot \frac{x}{z} ^ 2$

we will find $k$

$8 = k \cdot \frac{48}{3} ^ 2 \implies k = \frac{9 \cdot 8}{48} = \frac{9}{6} = \frac{3}{2}$

now solve for 2nd part

$12 = \frac{3}{2} \cdot \frac{x}{2} ^ 2 \implies 12 = \frac{3 x}{8}$
$4 = \frac{x}{8}$

$x = 32$