## College Algebra (10th Edition)

$-8\leq x\lt -3$ Or, in interval notation: $[-8,\ -3)$
We solve: $14\lt 5-3x\leq 29$ We subtract 5: $14-5\lt -3x\leq 29-5$ $9\lt -3x\leq 24$ And divide by $-3$: $-3\gt x\geq-8$ $-8\leq x\lt -3$ Or, in interval notation: $[-8,\ -3)$