Question

x-6/1998+x-4/2000=x-2000/4+x-1998/6

Answer 1

          \(\dfrac{x-6}{1998}\) + \(\dfrac{x-4}{2000}\) = \(\dfrac{x-2000}{4}\) + \(\dfrac{x-1998}{6}\)

\(\dfrac{x-6}{1998}\) - 1 + \(\dfrac{x-4}{2000}\) - 1 = \(\dfrac{x-2000}{4}\) - 1 + \(\dfrac{x-1998}{6}\) - 1

\(\dfrac{x-6-1998}{1998}\) + \(\dfrac{x-4-2000}{2000}\)  = \(\dfrac{x-2000-4}{4}\) + \(\dfrac{x-1998-6}{6}\)

\(\dfrac{x-2004}{1998}\)  + \(\dfrac{x-2004}{2000}\) = \(\dfrac{x-2004}{4}\) + \(\dfrac{x-2004}{6}\)

(\(x-2004\)).[\(\dfrac{1}{1998}\) + \(\dfrac{1}{2000}\) - \(\dfrac{1}{4}\) - \(\dfrac{1}{6}\)] = 0

\(x\) - 2004 = 0

\(x\)             = 2004