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

Answer 1

Nathan Axel

#-1/(3c-5)#

The first step in handling algebraic fractions is to factorize the numerator and/or denominator, if at all possible.

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

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

Align the common factors on the numerator and denominator now.

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

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

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Answer 2

Eva Abramson

#1/(5-3c)#

First, determine whether algebraic fractions 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

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Answer 3

Eli Assouline

To simplify the expression (-7c)/(21c^2-35c), you can factor out -7c from the numerator and the denominator. This gives us (-7c)/(7c(3c-5)). Next, we can cancel out the common factor of 7c, resulting in the simplified expression -1/(3c-5).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.