Question

1. solve the compound inequality: $3x + 1 \geq -14$ and $5x + 3 \leq 18$

a. $x \geq 2$
b. $-5 \leq x \leq 3$
c. $x > -4$
d. $x \leq 10$

Explanation:

Step1: Solve \( 3x + 1 \geq -14 \)

Subtract 1 from both sides: \( 3x + 1 - 1 \geq -14 - 1 \)
Simplify: \( 3x \geq -15 \)
Divide both sides by 3: \( \frac{3x}{3} \geq \frac{-15}{3} \)
Result: \( x \geq -5 \)

Step2: Solve \( 5x + 3 \leq 18 \)

Subtract 3 from both sides: \( 5x + 3 - 3 \leq 18 - 3 \)
Simplify: \( 5x \leq 15 \)
Divide both sides by 5: \( \frac{5x}{5} \leq \frac{15}{5} \)
Result: \( x \leq 3 \)

Step3: Find the intersection

The solution to the compound inequality is the overlap of \( x \geq -5 \) and \( x \leq 3 \), which is \( -5 \leq x \leq 3 \).

Answer:

B. \( -5 \leq x \leq 3 \)