Question

1. $40 \leq -10x$
2. $-16 \leq -3 + x$

Explanation:

Problem 11: Solve \( 40 \leq -10x \)

Step 1: Divide both sides by -10 (and reverse inequality)

When dividing an inequality by a negative number, the inequality sign reverses. So, divide both sides of \( 40 \leq -10x \) by \(-10\):
\( \frac{40}{-10} \geq \frac{-10x}{-10} \)
Simplify: \( -4 \geq x \) or \( x \leq -4 \).

Problem 13: Solve \( -16 \leq -3 + x \)

Step 1: Add 3 to both sides

Add 3 to both sides of \( -16 \leq -3 + x \) to isolate \( x \):
\( -16 + 3 \leq -3 + x + 3 \)
Simplify: \( -13 \leq x \) or \( x \geq -13 \).

Answer:

s:

1. \( x \leq -4 \)
2. \( x \geq -13 \)