Solve $\dfrac{3-x}{x-1}\geq x+4$, $(x>1)$.
$-2-\sqrt{11}\leq x \leq -2+\sqrt{11}$
$x>1$
$-2-\sqrt{11}\leq x<1$
$1<x\leq-2+\sqrt{11}$
Correct: $\dfrac{3-x}{x-1} \geq x+4$, where $x>1\Leftrightarrow \dfrac{3-x}{x-1}-4-x\geq0$.
We define $f(x)=\dfrac{3-x}{x-1}-4-x$ and solve $f(x)=0$:
$$\begin{align*}
\dfrac{3-x}{x-1}-4-x=0 &\Leftrightarrow 3-x+(x-1)(-x-4)=0\\
&\Leftrightarrow -x^2-4x+7=0\\
&\Leftrightarrow x^2+4x-7=0\\
&\Leftrightarrow x=-2-\sqrt{11} \mbox{ or } x=-2+\sqrt{11}.
\end{align*}$$
Note that $x=-2-\sqrt{11}$ is outside the domain. Via a sign chart we find $f(x)\geq 0$ if $1 < x\leq-2+\sqrt{11}$.
Go on.
Wrong: Consider the domain of the function.
Try again.
Wrong: Determine the points of intersection.
See Example 2 (film).
Wrong: Consider the domain of the function.
Try again.