How do you simplify #8/(c-1)+c/(1-c)#?

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Answer 1 · 2017-03-30T04:35:52

#(c-8)/(1-c)# #[c!=1]#

Expression #=8/(c-1)+c/(1-c)#

#= (8(1-c)+c(c-1))/((c-1)(1-c))#

#=(8-8c+c^2-c)/((c-1)(1-c))#

#=(8-9c+c^2)/((c-1)(1-c))#

#= ((c-8)(c-1))/((c-1)(1-c))#

If #c!=1#

Expression #=((c-8)cancel((c-1)))/(cancel((c-1))(1-c))#

#=(c-8)/(1-c)#

Alternatively,
Note that #8/(c-1)+c/(1-c) = (-8)/-(c-1)+c/(1-c)#
#= (-8)/(1-c)+c/(1-c)#

#=(c-8)/(1-c)#