[1/(c-2)] - [3c/(c^2 + 2c - 8)] = -4/ (c+4) Thanks
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Originally Posted by gnarlycarly227 [1/(c-2)] - [3c/(c^2 + 2c - 8)] = -4/ (c+4) Thanks You basically want to factorise c^2 + 2c - 8 to (c+4)(c-2) then cancel out.
1/(c-2) = (c+4)/((c-2)(c+4)) -4/(c+4) = -4(c-2)/((c-2)(c+4)) 3c/(c^2 + 2c - 8) = 3c/((c-2)(c+4)) then you have.... (c+4)-3c+4(c-2)/((c-2)(c+4)) = 0 2c-4/((c-2)(c+4)) = 0 2(c-2)/((c-2)(c+4)) = 0 2/(c+4) = 0
Wait, did you want to solve or to simplify... oops. OK, take the line (c+4)-3c+4(c-2)/((c-2)(c+4)) = 0 (c+4)-3c+4(c-2) = 0 2c-4 = 0 2c = 4 c = 2
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