We have the following system of inequalities:

1. $ 2 - 3x \leq 14 $
2. $ 3x - 1 \leq 14 $

### Solving the first inequality: 
\[
2 - 3x \leq 14
\]
- Subtract 2 from both sides:
\[
-3x \leq 12
\]
- Divide by $-3$, remembering to flip the inequality sign:
\[
x \geq -4
\]

### Solving the second inequality:
\[
3x - 1 \leq 14
\]
- Add 1 to both sides:
\[
3x \leq 15
\]
- Divide by 3:
\[
x \leq 5
\]

### Final solution:
From both inequalities, we get:
\[
-4 \leq x \leq 5
\]

So, the solution is $ x \in [-4, 5] $.

Would you like more details, or do you have any questions?

Here are 5 related questions you could explore:

1. How would the solution change if the inequalities were strict (i.e., $ < $ instead of $ \leq $)?
2. What happens if one of the inequalities had a negative coefficient in front of $x$?
3. Can you solve a system of inequalities involving quadratic terms?
4. How would you graph the solution to this system of inequalities on a number line?
5. What happens if you add an equality constraint to the system (e.g., $ x = 2 $)?

**Tip:** When dividing by a negative number in inequalities, always remember to reverse the inequality sign.

2-3x $$ \leq 14 \text{ and } 3x-1 \leq 14 $$

Learn how to solve the system of linear inequalities 2 - 3x ≤ 14 and 3x - 1 ≤ 14 with a step-by-step guide. This problem covers key algebraic concepts, such as manipulating inequalities and finding the solution range for x. Suitable for students in Grades 7-9.

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