## Calculus with Applications (10th Edition)

$(-\infty$, $\frac{50}{9})$
$\frac{8}{3}(z-4)$$\leq$$\frac{2}{9}(3z+2)$ 1. Multiply through $\frac{8}{3}z$-$\frac{32}{3}$$\leq$$\frac{2}{3}z$+$\frac{4}{9}$ 2. Subtract/add from each side to get any polynomials with a variable on one side and any without a variable on the other $\frac{6}{3}z$$\leq$$\frac{100}{9}$ 3. Simplify any possible fractions 2z$\leq$$\frac{100}{9} 4. Multiply by \frac{1}{2} on both sides to get the variable by itself z\leq$$\frac{50}{9}$ 5. Plug $\frac{50}{9}$ back into the original problem for z to get $\frac{112}{27}$ 6. Test any number between $\frac{112}{27}$ and z$\leq$$\frac{50}{9}$ into the original equation 7. Since any number you test will be negative, your interval starts at $-\infty$ and ends with the interval found earlier: $\frac{50}{9}$