Question

solve the compound inequality $3z - 2 > 7$ or $z + \frac{1}{2} \leq \frac{3}{2}$. (fill each blank with an integer or inequality symbol.)
$z \leq \square$ or $z \square \square$

Explanation:

Step1: Solve $3x - 2 > 7$

$3x - 2 > 7 \\
3x > 7 + 2 \\
3x > 9 \\
x > \frac{9}{3} \\
x > 3$

Step2: Solve $x + \frac{1}{2} \leq 1$

$x + \frac{1}{2} \leq 1 \\
x \leq 1 - \frac{1}{2} \\
x \leq \frac{1}{2}$

Answer:

$x \leq \frac{1}{2}$ or $x > 3$