$\dfrac{x-5}{1990}-1+\dfrac{x-15}{1980}-1 = \dfrac{x-1980}{15}-1+\dfrac{x-1990}{5}-1$
<=> $\dfrac{x-1995}{1990}+\dfrac{x-1995}{1980} = \dfrac{x-1995}{15}+\dfrac{x-1995}{5}$
<=> $(x-1995)(\dfrac{1}{1990}+\dfrac{1}{1980}-\dfrac{1}{15}-\dfrac{1}{5}) = 0$
<=> $x = 1995$