Question

Lowkey need help with this.

Answer 1

9514 1404 393

Answer:

c = 14

no extraneous solutions

Explanation:

You can subtract the right-side expression, combine fractions, and set the numerator to zero.

`(c-4)/(c-2)-\left((c-2)/(c+2)-(1)/(2-c)\right)=0\\\\(c-4)/(c-2)-(1)/(c-2)-(c-2)/(c+2)=0\\\\((c-5)(c+2)-(c-2)^2)/((c-2)(c+2))=0\\\\((c^2-3c-10)-(c^2-4c +4))/((c-2)(c+2))=0\\\\(c-14)/((c-2)(c+2))=0\\\\\boxed{c=14}`

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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.