Answer
$y = \frac{1}{x-4}$, where $x\geq 5$
Work Step by Step
$x = t^2+5$
$t = \sqrt{x-5}$
We can replace this value in the equation with $y$:
$y = \frac{1}{t^2+1}$
$y = \frac{1}{(\sqrt{x-5})^2+1}$
$y = \frac{1}{(x-5)+1}$
$y = \frac{1}{x-4}$
Since $x = t^2+5$, then $x\geq 5$
$y = \frac{1}{x-4}$, where $x\geq 5$
