9514 1404 393
Answer:
c = 14
no extraneous solutions
Step-by-step explanation:
You can subtract the right-side expression, combine fractions, and set the numerator to zero.
[tex]\dfrac{c-4}{c-2}-\left(\dfrac{c-2}{c+2}-\dfrac{1}{2-c}\right)=0\\\\\dfrac{c-4}{c-2}-\dfrac{1}{c-2}-\dfrac{c-2}{c+2}=0\\\\\dfrac{(c-5)(c+2)-(c-2)^2}{(c-2)(c+2)}=0\\\\\dfrac{(c^2-3c-10)-(c^2-4c +4)}{(c-2)(c+2)}=0\\\\\dfrac{c-14}{(c-2)(c+2)}=0\\\\\boxed{c=14}[/tex]
__
Check
(14 -4)/(14 -2) = (14 -2)/(14 +2) -1/(2 -14) . . . . substitute for c
10/12 = 12/16 -1/-12
5/6 = 3/4 +1/12 . . . . true
There is one solution (c=14) and it is a solution to the original equation. There are no extraneous solutions.
