If x is real show that value of the expression $\frac{{\mathrm{x}}^2+34\mathrm{x}-71}{{\mathrm{x}}^2+2\mathrm{x}-7}$ does not lie between 5 and 9

Let $\mathrm{y}=\frac{{\mathrm{x}}^2+34\mathrm{x}-71}{{\mathrm{x}}^2+2\mathrm{x}-7}$
$\begin{array}{l}\mathrm{y}\left({\mathrm{x}}^2+2\mathrm{x}-7\right)={\mathrm{x}}^2+34\mathrm{x}-71\\ {\mathrm{yx}}^2+2\mathrm{xy}-7\mathrm{y}={\mathrm{x}}^2+34\mathrm{x}-71\\ \Rightarrow {\mathrm{yx}}^2+2\mathrm{xy}-7\mathrm{y}-{\mathrm{x}}^2-34\mathrm{x}+71=0\\ (\mathrm{y}-1){\mathrm{x}}^2+2(\mathrm{y}-17)\mathrm{x}+(-7\mathrm{y}+71)=0\end{array}$
Since $\mathrm{x}\in \mathrm{R}$ , discriminent $\ge$ 0
{Compare with ax2+bx+c}
$\begin{array}{l}\Rightarrow {\mathrm{b}}^2-4\mathrm{ac}\ge 0.\\ \Rightarrow \left[2\right[\mathrm{y}-17]{]}^2-4(\mathrm{y}-1\left)\right(-7\mathrm{y}+71)\ge 0\\ \left[2\right(\mathrm{y}-17){]}^2-4(\mathrm{y}-1\left)\right(-7\mathrm{y}+71)\ge 0\\ 4\left[{\mathrm{y}}^2-34\mathrm{y}+289\right]-4\left[-7{\mathrm{y}}^2+78\mathrm{y}-71\right]\ge 0\\ 4\left[{\mathrm{y}}^2-34\mathrm{y}+289+7{\mathrm{y}}^2-78\mathrm{y}+71\right]\ge 0\\ \Rightarrow 8{\mathrm{y}}^2-112\mathrm{y}+360\ge 0;8\left[{\mathrm{y}}^2-14\mathrm{y}+45\right]\ge 0\\ {\mathrm{y}}^2-14\mathrm{y}+45\ge 0;(\mathrm{y}-5)(\mathrm{y}-9)\ge 0\\ \mathrm{y}\le 5\text{ or }\mathrm{y}\ge 9\end{array}$
[If $\mathrm{\alpha }<\mathrm{\beta }\text{ then }(\mathrm{x}-\mathrm{\alpha })(\mathrm{x}-\mathrm{\beta })\ge 0\Rightarrow \mathrm{x}\le \mathrm{\alpha }\text{ or }\mathrm{x}\ge \mathrm{\beta }]$ y does not lie between 5 and 9