Question

Solve $\frac{x^2+6x-7}{x^2+1}\le 2$

Answer 1

As $x^2+1$ is always greater than $0$, we can cross-multiply and say that

$\Rightarrow$ $x^2+6x-7\le 2x^2+2$
$\Rightarrow$ $x^2-6x+9\ge 0$
$\Rightarrow$ ${(x-3)}^2\ge 0$
$\Rightarrow$ $xϵR$