If $\displaystyle y$ varies inversely with $\displaystyle x$, $\displaystyle y=\frac{K}{x}$.

1. Using any of the two $\displaystyle x,y$ combinations given, solve for $\displaystyle K$:

Using $\displaystyle (2,18)$:

$\displaystyle 18=\frac{K}{2}$

$\displaystyle K=36$

2. Use your new equation $\displaystyle y=\frac{36}{x}$ and solve when $\displaystyle x=12$:

$\displaystyle y=\frac{36}{12}=3$

An inverse variation is a function in the form: $\displaystyle xy = k$ or $\displaystyle y = \frac{k}{x}$, where $\displaystyle k$ is not equal to 0. 

Substitute each $\displaystyle \left ( x,y \right )$ in $\displaystyle xy = k$.

$\displaystyle 5(4.8) = 24$

$\displaystyle 6.4(3.75) = 24$

$\displaystyle 20(1.2) = 24$

Therefore, the constant of variation, $\displaystyle k$, must equal 24. If $\displaystyle y$ varies inversely as $\displaystyle x$, $\displaystyle 3n$ must equal 24. Solve for $\displaystyle n$.

$\displaystyle 3n = 24$

$\displaystyle n = 8$

If $\displaystyle x=3$ when $\displaystyle y=8$, and the variation is direct, then $\displaystyle xy = 24$. Using this, we know that if $\displaystyle y=6$, $\displaystyle x= \frac{24}{6} = 4$.

The absolute value will always be positive or 0, therefore all values of z will create a true statement as long as $\displaystyle 2-z\neq 0$. Thus all values except for 2 will work.

In order to find excluded values, find the values that make the denominator equal zero.

To do this, you must factor the denominator:

$\displaystyle (2x-1)(x+5)$

Now, set each part equal to zero and solve for $\displaystyle x$.

$\displaystyle 2x-1=0$

$\displaystyle x=\frac{1}{2}$

$\displaystyle x+5=0$

$\displaystyle x=-5$

In order to find the excluded values for x, find the values that make the denominator zero. 

In order to find the zeroes, factor the denominator: 

$\displaystyle (x+3)(x+2)$

Now, set each part equal to zero.

$\displaystyle (x+3)=0$

$\displaystyle (x+2)=0$

$\displaystyle x=-3$

$\displaystyle x=-2$

To find the excluded values, find the values that make the denominator zero. 

$\displaystyle 4-x^{2}=0$

$\displaystyle 4=x^{2}$

$\displaystyle x=2, -2$

To find the excluded values, find the values that make the denominator equal zero. 

To do that, set the denominator equal to zero and solve for $\displaystyle x$. 

Begin by taking each multiple in the denominator and setting it equal to zero. 

$\displaystyle x=0$

$\displaystyle x^{2}-9=0$

$\displaystyle x^{2}=9$

$\displaystyle x=3, -3$

$\displaystyle x=3, -3, 0$

To find the excluded values, find the values that make the denominator equal zero. 

To do that, begin by factoring the denominator.

$\displaystyle (x+4)(x+3)$

Now, set each part equal to zero.

$\displaystyle x+4=0$

$\displaystyle x+3=0$

Now, solve for $\displaystyle x$.

$\displaystyle x=-4$

$\displaystyle x=-3$

To find the excluded values, find the $\displaystyle x$-values that make the denominator equal zero. 

To do this, set the denominator equal to zero:

$\displaystyle x^{2}-16=0$

Next, solve for x.

$\displaystyle x^{2}=16$

$\displaystyle x=4, -4$