# How do you solve 8-z/3>=11?

How do you solve 8-z/3>=11?

ReportAnswer 1

$z \le - 9$

#### Explanation:

We are given:

$8 - \frac{z}{3} \ge 11$

Now, we can solve it with the following steps.

$- \frac{z}{3} \ge 11 - 8$

$- \frac{z}{3} \ge 3$

Multiplying by $3$ on both sides, we get

$- z \ge 9$

We now need $z$ instead of $- z$, so we divide by $- 1$.

When dividing by a negative number, we must inverse the inequality sign.

$z \le - 9$

See a solution process below:

#### Explanation:

First, subtract $\textcolor{red}{8}$ from each side of the inequality to isolate the $z$ term while keeping the inequality balanced:

$8 - \textcolor{red}{8} - \frac{z}{3} \ge 11 - \textcolor{red}{8}$

$0 - \frac{z}{3} \ge 3$

$- \frac{z}{3} \ge 3$

Now, multiply each side of the inequality by $\textcolor{b l u e}{- 3}$ to solve for $z$ while keeping the inequality balanced. However, because we are multiplying or dividing an inequality by a negative number we must reverse the inequality operator:

$\textcolor{b l u e}{- 3} \times - \frac{z}{3} \textcolor{red}{\le} \textcolor{b l u e}{- 3} \times 3$

$\textcolor{b l u e}{- 3} \times \frac{z}{-} 3 \textcolor{red}{\le} - 9$

$\cancel{\textcolor{b l u e}{- 3}} \times \frac{z}{\textcolor{b l u e}{\cancel{\textcolor{b l a c k}{- 3}}}} \textcolor{red}{\le} - 9$

$z \le - 9$

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