Solve for z.

Question

z+3≥18–2z or z+8–12z≥19
Write your answer as a compound inequality with integers.

Answer 1

To solve the compound inequality \( z + 3 \geq 18 - 2z \) or \( z + 8 - 12z \geq 19 \), we will solve each inequality separately.

Inequality 1: \( z + 3 \geq 18 - 2z \)

1. Add \( 2z \) to both sides: \[ z + 3 + 2z \geq 18 \] This simplifies to: \[ 3z + 3 \geq 18 \]

2. Subtract 3 from both sides: \[ 3z \geq 15 \]

3. Divide by 3: \[ z \geq 5 \]

Inequality 2: \( z + 8 - 12z \geq 19 \)

1. Combine like terms: \[ -11z + 8 \geq 19 \]

2. Subtract 8 from both sides: \[ -11z \geq 11 \]

3. Divide by -11 (remember to flip the inequality sign): \[ z \leq -1 \]

Combine the Results

The solution from the first inequality is \( z \geq 5 \), and from the second inequality, it is \( z \leq -1 \). The overall solution, combining both inequalities, is:

\[ z \leq -1 \quad \text{or} \quad z \geq 5 \]

In interval notation, this can be expressed as:

\[ (-\infty, -1] \cup [5, \infty) \]

Thus, the final answer is:

\[ \text{The solution is } (-\infty, -1] \cup [5, \infty). \]