Question

Solve $\frac{6}{7x-21}-\frac{1}{x^2-6x+9}+\frac{1}{x^2-9}=0$

Answer 1

Given equation appears to be a non-quadratic equation. But when we simplify the equation, it will reduce to a quadratic equation.
Now, $\frac{6}{7(x-3)}-\frac{1}{(x-3{)}^2}+\frac{1}{(x+3)(x-3)}$
$\to \frac{6(x^2-9)-7(x+3+7(x-3\left)\right)}{7(x-3{)}^2(x+3)}=0$
$\to 6x^2-54-42=0\to x^2-16=0$
The equation $x^2=16$ is quadratic and hence we have two values x = 4 and x= -4.
$\therefore$ Solution set is $\{-4,4\}$