Solve the compound inequality -4 <= 3x - 1 <= 8.

Step-by-step solution

1. 1

   Recognise a single chained statement, not two problems. $-4\le3x-1\le8$ asserts $-4\le3x-1$ and $3x-1\le8$ simultaneously. Because the variable appears in only the middle part, both conditions can be handled in one pass — every operation is simply applied to all three parts at once.

2. 2

   Add 1 to all three parts. Adding the same constant everywhere preserves both inequality directions:

   $-4+1\le3x-1+1\le8+1\qquad\Longrightarrow\qquad-3\le3x\le9$

3. 3

   Divide all three parts by 3. The divisor is positive, so neither $\le$ flips. (Had it been $-3$, both signs would reverse and the chain would need rewriting from largest to smallest.)

   $\frac{-3}{3}\le\frac{3x}{3}\le\frac{9}{3}\qquad\Longrightarrow\qquad-1\le x\le3$

4. 4

   State the answer with closed endpoints. In interval notation $[-1,3]$. Both ends are included because the original relations allowed equality; the bracket shape is not cosmetic, it records that fact.

5. 5

   Verify the endpoints and one interior point. At $x=-1$: $3(-1)-1=-4$, and $-4\le-4$ ✓ (left endpoint attained). At $x=3$: $3(3)-1=8$, and $8\le8$ ✓ (right endpoint attained). At $x=0$: $-1$, comfortably inside ✓. Outside, $x=4$ gives $11>8$ ✗. Testing the raw chain against $[-1,3]$ at $110$ exact rational points agrees everywhere ✓.