Notice that
![x^2 - 4x - 21 = (x - 7) (x + 3)](https://img.qammunity.org/2022/formulas/mathematics/high-school/kowrualay2q1iddxa9okxplhw23ro38nsj.png)
so on the left side, a factor of x - 7 in the numerator and denominator can cancel
![(x-7)/(x^2-4x-21) = (x-7)/((x-7)(x+3)) = \frac1{x+3}](https://img.qammunity.org/2022/formulas/mathematics/high-school/977us6akl1n6lcz85dtys3iwoazunykiwt.png)
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,
![\{x \in \mathbb R \mid x \\eq 7\}](https://img.qammunity.org/2022/formulas/mathematics/high-school/ssyyw40glia8ekcw8vh875w9cv4cxga8on.png)
In interval notation,
![(-\infty,7)\cup(7,\infty)](https://img.qammunity.org/2022/formulas/mathematics/high-school/4mwzo3c6culikgypazu37sfbt2fezx5wzw.png)