## Precalculus (6th Edition)

$\color{blue}{[-1 , +\infty)}$
Add $2$ and subtract $x$ to both sides, then combine like terms to obtain: $-2x-2+2 -x\le 1+x+2-x \\-3x \le 3$ Divide $-3$ to both sides. Since a negative number was divided to both sides of the inequality, the inequality symbol will be reversed. $\dfrac{-3x}{-3} \ge \dfrac{3}{-3} \\x \ge -1$ In interval notation, the solution set is: $\color{blue}{[-1 , +\infty)}$