Question

solve each inequality and graph its solution.

1. $2x + 4 \geq 24$
2. $\frac{m}{3} - 3 \leq -6$
3. $-3(p + 1) \leq -18$
4. $-4(-4 + x) > 56$
5. $-b - 2 > 8$
6. $-4(3 + n) > -32$
7. $4 + \frac{n}{3} < 6$
8. $-3(r - 4) \geq 0$
9. $-7x + 7 \leq -56$
10. $-3(p - 7) \geq 21$

Explanation:

Problem 1: \( 2x + 4 \geq 24 \)

Step 1: Subtract 4 from both sides

To isolate the term with \( x \), we subtract 4 from both sides of the inequality.
\( 2x + 4 - 4 \geq 24 - 4 \)
\( 2x \geq 20 \)

Step 2: Divide both sides by 2

To solve for \( x \), we divide both sides of the inequality by 2.
\( \frac{2x}{2} \geq \frac{20}{2} \)
\( x \geq 10 \)

Step 1: Add 3 to both sides

To isolate the term with \( m \), we add 3 to both sides of the inequality.
\( \frac{m}{3} - 3 + 3 \leq -6 + 3 \)
\( \frac{m}{3} \leq -3 \)

Step 2: Multiply both sides by 3

To solve for \( m \), we multiply both sides of the inequality by 3.
\( 3 \times \frac{m}{3} \leq -3 \times 3 \)
\( m \leq -9 \)

Step 1: Divide both sides by -3 (and reverse the inequality)

When dividing by a negative number, we reverse the inequality sign.
\( \frac{-3(p + 1)}{-3} \geq \frac{-18}{-3} \)
\( p + 1 \geq 6 \)

Step 2: Subtract 1 from both sides

To solve for \( p \), we subtract 1 from both sides of the inequality.
\( p + 1 - 1 \geq 6 - 1 \)
\( p \geq 5 \)

Answer:

\( x \geq 10 \) (Graph: A closed circle at 10 and an arrow pointing to the right on the number line.)

Problem 2: \( \frac{m}{3} - 3 \leq -6 \)