Find all solutions to equation in $\mathbb{R}$ Find all solutions to equation in $\mathbb{R}$
$$\sqrt{\frac{x-7}{1989}} + \sqrt{\frac{x-6}{1990}} + \sqrt{\frac{x-5}{1991}} = \sqrt{\frac{x-1989}{7}} + \sqrt{\frac{x-1990}{6}} + \sqrt{\frac{x-1991}{5}}$$
Solution: For $1991\leq x<1996$,
$\frac{x-7}{1989}>\frac{x-1989}{7}, \frac{x-6}{1990}>\frac{x-1990}{6}$ and $\frac{x-5}{1991}>\frac{x-1991}{5}$
$\sqrt{\frac{x-7}{1989}}>\sqrt{\frac{x-1989}{7}}, \sqrt{\frac{x-6}{1990}}>\sqrt{\frac{x-1990}{6}}$ and $\sqrt{\frac{x-5}{1991}}>\sqrt{\frac{x-1991}{5}}$
L.H.S. > R.H.S.
For $x>1996$,
$\frac{x-7}{1989}<\frac{x-1989}{7}, \frac{x-6}{1990}<\frac{x-1990}{6}$ and $\frac{x-5}{1991}<\frac{x-1991}{5}$
$\sqrt{\frac{x-7}{1989}}<\sqrt{\frac{x-1989}{7}}, \sqrt{\frac{x-6}{1990}}<\sqrt{\frac{x-1990}{6}}$ and $\sqrt{\frac{x-5}{1991}}<\sqrt{\frac{x-1991}{5}}$
L.H.S. < R.H.S.
$\therefore$ only solution is $x=1996$
Is there another method of solving this? More algebraic

You have a nice solution.
Anyway, consider that you are looking for the zero of function
$$f(x)=\left(\sqrt{\frac{x-7}{1989}} + \sqrt{\frac{x-6}{1990}} + \sqrt{\frac{x-5}{1991}}\right) -\left( \sqrt{\frac{x-1989}{7}} + \sqrt{\frac{x-1990}{6}} + \sqrt{\frac{x-1991}{5}}\right)$$
Perform a Taylor expansion at $x=1991$ to second order (I shall not write the coefficients). So, you face a quadratic equation in $(x-1991)$ and the solution of it is $1994.58$.
Otherwise, use Newton method and face the following iterates
$$\left(
\begin{array}{cc}
n & x_n \\
 0 & 1991.10 \\
 1 & 1992.92 \\
 2 & 1995.40 \\
 3 & 1995.98 \\
 4 & 1996.00
\end{array}
\right)$$
