Appendix A - Review - A.6 Rational Expressions - A.6 Assess Your Understanding - Page A53: 43

$\dfrac{4-x}{x-2}, \quad x\neq 2$

The denominators are additive inverses of each other so we factor out factoring $-1$ out of the denominator of the second rational as follows: \begin{align*} \frac{4}{x-2}+\frac{x}{2-x}&=\frac{4}{x-2}+\frac{x}{-(x-2)}\\ \\&=\frac{4}{x-2}+\frac{-x}{x-2}\\ \\&=\frac{4}{x-2}-\frac{x}{x-2}\\ \\&=\frac{4-x}{x-2}, \quad x\ne2 \end{align*}