How do you solve #1/(c-2) + 1/c = 2/(c(c-2))#?

1 Answer
Nov 14, 2015

Answer:

I found no real value of #c#.

Explanation:

I would choose #c(c-2)# as common denominator and write:
#(c+c-2)/(c(c-2))=2/(c(c-2))#
where on the left I arranged the numerators to the new common denominator.

So you can now simplify the two denominators:

#(c+c-2)/cancel((c(c-2)))=2/cancel((c(c-2)))#
#2c-2=2#
#2c=4#
#c=4/2=2#
BUT
if you use #c=2# into your original equation you get a division by zero that cannot be evaluated.