## Precalculus (6th Edition) Blitzer

$[6, \infty)$
"$y$ is at least 4" $\qquad$ is written as $\quad y\geq 4.$ Substituing the expression for y, $1-(x+3)+2x\geq 4 \qquad$ ... simplify $1-x-3+2x\geq 4$ $x-2\geq 4 \qquad$ ... add 2 $x\geq 6$ Solution set = $[6, \infty)$.