Notice that
[tex]x^2 - 4x - 21 = (x - 7) (x + 3)[/tex]
so on the left side, a factor of x - 7 in the numerator and denominator can cancel
[tex]\dfrac{x-7}{x^2-4x-21} = \dfrac{x-7}{(x-7)(x+3)} = \dfrac1{x+3}[/tex]
But we can only cancel them out as long as x ≠ 7 (because otherwise we'd have the indeterminate form 0/0 on the left side).
So, the equation is true for all x except x = 7. In set notation,
[tex]\{x \in \mathbb R \mid x \neq 7\}[/tex]
In interval notation,
[tex](-\infty,7)\cup(7,\infty)[/tex]
