Question

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

Answer 1

`-1/(3c-5)`

Explanation:

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

Here, the denominator has a `color(blue)"common factor"` of 7c which can be taken out.

`rArr(-7c)/(7c(3c-5))`

Now cancel, common factors on numerator/denominator.

`rArr(-cancel(7c)^1)/(cancel(7c)^1(3c-5))`

`=-1/(3c-5)`

Answer 2

`1/(5-3c)`

Explanation:

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

`(-7c)/(21c^2-35c) = " " (-7c)/(7c(3c-5))" "larr` with factors - cancel!

`=(-cancel(7c))/(cancel(7c)(3c-5)) " "larr` the minus sign can be anywhere ...

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

`=1/(5-3c)" "larr` note the switch-round