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

Nov 14, 2015

I found no real value of $c$.

#### Explanation:

I would choose $c \left(c - 2\right)$ as common denominator and write:
$\frac{c + c - 2}{c \left(c - 2\right)} = \frac{2}{c \left(c - 2\right)}$
where on the left I arranged the numerators to the new common denominator.

So you can now simplify the two denominators:

$\frac{c + c - 2}{\cancel{\left(c \left(c - 2\right)\right)}} = \frac{2}{\cancel{\left(c \left(c - 2\right)\right)}}$
$2 c - 2 = 2$
$2 c = 4$
$c = \frac{4}{2} = 2$
BUT
if you use $c = 2$ into your original equation you get a division by zero that cannot be evaluated.