# How do you simplify (14g^3h^2)/(42gh^3) and find the excluded values?

Jan 23, 2017

$\frac{14 {g}^{3} {h}^{2}}{42 g {h}^{3}} = {g}^{2} / \left(3 h\right) , h \ne 0$

#### Explanation:

In order to simplify, we must remove the indices from the variables where they appear in the numerator and denominator.

$\frac{14}{42} = \frac{1}{3}$

We know that ${a}^{3} / {a}^{2} = {a}^{3 - 2} = {a}^{1} = a$, so we can apply this to $g$ and $h$ in the fraction.

${g}^{3} / g = {g}^{3 - 1} = {g}^{2}$

${h}^{2} / {h}^{3} = {h}^{2 - 3} = {h}^{-} 1 = \frac{1}{h}$

We can now simplify the fraction:

$\frac{1 \cancel{14} {g}^{2 \cancel{3}} {h}^{- 1 \cancel{2}}}{3 \cancel{42} \cancel{g} \cancel{{h}^{3}}} = \frac{1}{3} {g}^{2} {h}^{-} 1 = {g}^{2} / \left(3 h\right)$

Since this is a fraction, the denominator can't equal $0$, so the only excluded value is $h = 0$