Answer
The equation has two solutions, 4 and 1.
Work Step by Step
$\frac{2}{c-2}=2-\frac{4}{c}$
$\frac{2}{c-2}+\frac{4}{c}=2$
$\frac{2c+4(c-2)}{(c-2)c}=2$
$\frac{2c+4c-8}{c^2-2c}=2$
$6c-8=2(c^2-2c)$
$6c-8=2c^2-4c$
$2c^2-4c-6c+8=0$
$2c^2-10c+8=0$
$2c^2-2c-8c+8=0$
$2c(c-1)-8(c-1)=0$
$(2c-8)(c-1)=0$
$2c-8=0$ or $c-1=0$
$c=4$ or $c=1$
Check:
$\frac{2}{4-2}=1;2-\frac{4}{4}=1$
$\frac{2}{1-2}=-2;2-\frac{4}{1}=-2$
The equation has two solutions, 4 and 1.
