# How do you simplify (-7c)/(21c^2-35c)?

Sep 25, 2016

$- \frac{1}{3 c - 5}$

#### Explanation:

When dealing with algebraic fractions the first step is to factorise the numerator/denominator, if possible.

Here, the denominator has a $\textcolor{b l u e}{\text{common factor}}$ of 7c which can be taken out.

$\Rightarrow \frac{- 7 c}{7 c \left(3 c - 5\right)}$

Now cancel, common factors on numerator/denominator.

$\Rightarrow \frac{- {\cancel{7 c}}^{1}}{{\cancel{7 c}}^{1} \left(3 c - 5\right)}$

$= - \frac{1}{3 c - 5}$

Sep 25, 2016

$\frac{1}{5 - 3 c}$

#### Explanation:

With algebraic fractions, check first if they can be factored,

$\frac{- 7 c}{21 {c}^{2} - 35 c} = \text{ " (-7c)/(7c(3c-5))" } \leftarrow$ with factors - cancel!

$= \frac{- \cancel{7 c}}{\cancel{7 c} \left(3 c - 5\right)} \text{ } \leftarrow$ the minus sign can be anywhere ...

= (color(red)(-)1)/(3c-5) = color(red)(-)1/(3c-5) = 1/(color(red)(-)(3c-5)

$= \frac{1}{5 - 3 c} \text{ } \leftarrow$ note the switch-round