How many integers satisfy $-4 \le 2x+1 \le 6$?
Subtracting $1$ from each expression, we have $-5\le 2x\le 5$. Dividing by $2$, we get $-\frac52\le x\le \frac52$. The integers satisfying this chain of inequalities are $-2,-1,0,1,2$. There are $\boxed{5}$ numbers in this list.
5
Level 4
Prealgebra
Problem: How many integers satisfy $-4 \le 2x+1 \le 6$?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve the inequality -4 ≤ 2x + 1 ≤ 6, we need to break it down into two separate inequalities: 

1. -4 ≤ 2x + 1
2. 2x + 1 ≤ 6

Starting with the first inequality:

-4 ≤ 2x + 1
Subtract 1 from both sides:
-4 - 1 ≤ 2x
-5 ≤ 2x
Now, divide both sides by 2:
-5/2 ≤ x
This simplifies to:
x ≥ -2.5

Next, we solve the second inequality:

2x + 1 ≤ 6
Subtract 1 from both sides:
2x ≤ 6 - 1
2x ≤ 5
Now, divide both sides by 2:
x ≤ 5/2
This simplifies to:
x ≤ 2.5

Now we combine the results from both inequalities:
-2.5 ≤ x ≤ 2.5

Next, we determine the integers that satisfy this combined inequality. The integers within this range are:
-2, -1, 0, 1, 2

Counting these integers gives us a total of 5 integers. Thus, the final answer is 5.<|END_OF_LATENT|> The final answer is $5$.