## Calculus with Applications (10th Edition)

$k<1\\$
Applying the properties of inequality, we can P1. add any number to both sides, P2. multiply (or divide) both sides with a positive $\quad$ number to arrive at a valid inequality. $\quad$ If we P3. multiply multiply (or divide) both sides with a negative number, we must change the direction of the inequality sign, to arrive at a valid inequality.. Our goal is to, step by step, isolate the unknown on one side and interpret the result (which, if any, will be an interval) ----------------------------- $6k-4<3k-1\qquad$P1: ...$/$+4 $6k<3k+3\qquad$P1: ...$/-3k$ $3k<3\qquad$P2: ...$/\div 3$ $k<1$ In interval notation: $(-\infty,1)$.