# If Y varies directly as x and inversely as the square of z. y=20 when x=50 and z =5. How do you find y when x=3 and z=6?

May 25, 2016

$y = \frac{c}{z} ^ 2 \text{ "->" "y=500/6^2" "->" } y = 13 \frac{8}{9} \to \frac{125}{9}$

$y = k x \text{ "->" " y=2/5xx3" "->" } y = \frac{6}{5}$

#### Explanation:

As both $x$ and $z$ are related to $y$ we can express them as follows:

Given:$\text{ "color(red)( y)" "color(blue)(alpha)" "color(red)( x)" "color(blue)(alpha)" } \textcolor{red}{\frac{1}{z} ^ 2}$

Where $\alpha$ means proportional to.

Let $k \text{ and } c$ be constants of variation. Then we have:

$y = k x = \frac{c}{z} ^ 2$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Determine the values of "k" and } c}$

Base condition is that at y=20 ; x=50 ; z=5

$\textcolor{b r o w n}{\text{To determine } k}$
$\implies y = k x \text{ "->" } 20 = k \left(50\right)$

Divide both sides by 50

$\frac{20}{50} = k \times \frac{50}{50} \text{ }$

but $\frac{50}{50} = 1$

color(brown)(k=20/50=2/5

'...........................................................
$\textcolor{b r o w n}{\text{To determine } c}$

$\implies y = \frac{c}{z} ^ 2 \text{ "->" } 20 = \frac{c}{{5}^{2}}$

Multiply both sides by 25

$\textcolor{b r o w n}{c = 20 \times 25 = 500}$

'.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Determine the value of "y" at } x = 3}$

$y = k x \text{ "->" } y = \frac{2}{5} \times 3$

$\textcolor{b l u e}{y = \frac{6}{5}}$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Determine the value of "y" at } z = 6}$

$y = \frac{c}{z} ^ 2 \text{ "->" } y = \frac{500}{6} ^ 2$

$y = 13 \frac{8}{9} \to \frac{125}{9}$