[tex]\begin{gathered} 4+\text{ 12 z }\leq\text{ 17 + 4z +11z} \\ \text{step 1 : add like terms } \\ 4\text{ + 12z }\leq17\text{ + 15z } \\ 12\text{ z -15z }\leq\text{ 17 -4 } \\ \text{step 2 : solve for z } \\ -3z\leq13\ldots\ldots\ldots(\text{ then divide both sides by -3)} \\ \frac{-3z}{-3}\ge\frac{13}{-3}\ldots..\text{ ( take note that the inequality sign change when we divide by a negative number )} \\ z\text{ }\ge-\frac{13}{3\text{ }} \end{gathered}[/tex]
