Question

when is the equation $\frac{x-7}{x^2-4x-21} =\frac{1}{x+3}$ true?

Answer 1

Notice that

$x^2 - 4x - 21 = (x - 7) (x + 3)$

so on the left side, a factor of x - 7 in the numerator and denominator can cancel

$\dfrac{x-7}{x^2-4x-21} = \dfrac{x-7}{(x-7)(x+3)} = \dfrac1{x+3}$

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,

$\{x \in \mathbb R \mid x \neq 7\}$

In interval notation,

$(-\infty,7)\cup(7,\infty)$