Question

solve each inequality. then graph the solution.

1. $4a + 3 > 15$

$4a + 3 - 3 > 15 - 3$
$4a > 12$
$\frac{4a}{4} > \frac{12}{4}$
$a > 3$
(number line graph with open circle at 3, arrow to the right)

1. $1 \geq \frac{m}{4} - 3$

(handwritten: $4 \geq m - 12$)
(hand-drawn number line with numbers 1,2,3,4 and an arrow)

1. $5 \geq -\frac{s}{12} - 3$
2. $3x - 2 > -10 + 29$
3. $\frac{n}{9} + 3 + 1 \leq 10 + 2$
4. $8p + 13 + 2p < 103$

Explanation:

Problem 2: \( 1 \geq \frac{m}{4} - 3 \)

Step 1: Add 3 to both sides

To isolate the term with \( m \), we add 3 to both sides of the inequality.
\( 1 + 3 \geq \frac{m}{4} - 3 + 3 \)
\( 4 \geq \frac{m}{4} \)

Step 2: Multiply both sides by 4

To solve for \( m \), we multiply both sides of the inequality by 4.
\( 4 \times 4 \geq \frac{m}{4} \times 4 \)
\( 16 \geq m \) or \( m \leq 16 \)

Step 1: Add 3 to both sides

Add 3 to both sides to isolate the term with \( s \).
\( 5 + 3 \geq -\frac{s}{12} - 3 + 3 \)
\( 8 \geq -\frac{s}{12} \)

Step 2: Multiply both sides by -12 (and reverse inequality)

Multiply both sides by -12. Since we multiply by a negative number, the inequality sign reverses.
\( 8 \times (-12) \leq -\frac{s}{12} \times (-12) \)
\( -96 \leq s \) or \( s \geq -96 \)

Step 1: Simplify the right side

Simplify \( -10 + 29 \) to get 19.
\( 3x - 2 > 19 \)

Step 2: Add 2 to both sides

Add 2 to both sides to isolate the term with \( x \).
\( 3x - 2 + 2 > 19 + 2 \)
\( 3x > 21 \)

Step 3: Divide both sides by 3

Divide both sides by 3 to solve for \( x \).
\( \frac{3x}{3} > \frac{21}{3} \)
\( x > 7 \)

Answer:

\( m \leq 16 \) (Graph: A closed circle at 16 and an arrow pointing to the left on the number line)

Problem 3: \( 5 \geq -\frac{s}{12} - 3 \)