Verify whether the given value of $x$ is a solution of the quadratic equation or not: $\frac{x+1}{x-1} - \frac{x-1}{x+1} = \frac{5}{6}, x = 5$

(YES) To verify if $x = 5$ is a solution,we substitute $x = 5$ into the left-hand side $(LHS)$ of the equation.
$LHS$ = $\frac{x+1}{x-1} - \frac{x-1}{x+1}$
Substituting $x = 5$:
$LHS$ = $\frac{5+1}{5-1} - \frac{5-1}{5+1}$
$LHS$ = $\frac{6}{4} - \frac{4}{6}$
Simplifying the fractions:
$LHS$ = $\frac{3}{2} - \frac{2}{3}$
Finding a common denominator (which is $6$):
$LHS$ = $\frac{3 \times 3}{6} - \frac{2 \times 2}{6} = \frac{9}{6} - \frac{4}{6} = \frac{5}{6}$
Since $LHS$ = $RHS$ (which is $\frac{5}{6}$),the given value $x = 5$ is a solution of the quadratic equation.