Question

graph the solution set.

1. $n + 2 \leq -5$ and $n + 6 \geq -6$
2. $t + 14 \geq 15$ or $t - 9 < -10$
3. $-3 \leq 7c + 4 < 18$

Explanation:

Problem 2: \( n + 2 \leq -5 \) and \( n + 6 \geq -6 \)

Step1: Solve \( n + 2 \leq -5 \)

Subtract 2 from both sides: \( n \leq -5 - 2 \) → \( n \leq -7 \)

Step2: Solve \( n + 6 \geq -6 \)

Subtract 6 from both sides: \( n \geq -6 - 6 \) → \( n \geq -12 \)

Step3: Find the intersection

The solution is \( -12 \leq n \leq -7 \) (since it's "and", we take the overlap of \( n \leq -7 \) and \( n \geq -12 \))

Step1: Solve \( t + 14 \geq 15 \)

Subtract 14 from both sides: \( t \geq 15 - 14 \) → \( t \geq 1 \)

Step2: Solve \( t - 9 < -10 \)

Add 9 to both sides: \( t < -10 + 9 \) → \( t < -1 \)

Step3: Find the union

Since it's "or", the solution is all \( t \) such that \( t < -1 \) or \( t \geq 1 \) (combine both solution sets)

Step1: Subtract 4 from all parts

\( -3 - 4 \leq 7c + 4 - 4 < 18 - 4 \) → \( -7 \leq 7c < 14 \)

Step2: Divide all parts by 7

\( \frac{-7}{7} \leq \frac{7c}{7} < \frac{14}{7} \) → \( -1 \leq c < 2 \)

Answer:

\( -12 \leq n \leq -7 \)

Problem 4: \( t + 14 \geq 15 \) or \( t - 9 < -10 \)