## Calculus: Early Transcendentals 8th Edition

$x\geq c(\frac{a+b}{ab})$
Given: $a(bx-c)\geq bc$ The solution set for $x$ can be calculated as follows: $bx\geq \frac{bc}{a}+c$ $x\geq c(\frac{1}{a}+\frac{1}{b})$ $x\geq c(\frac{a+b}{ab})$ Hence, the solution set is $x\geq c(\frac{a+b}{ab})$.