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Solve the inequality [tex]-3 \ \textless \ 2x + 1 \leq 11 - 3x[/tex].

A. [tex]-1 \ \textless \ x \leq 2[/tex]
B. [tex]-2 \leq x \leq 1[/tex]
C. [tex]2 \ \textless \ x \leq 2[/tex]
D. [tex]-2 \ \textless \ x \leq 2[/tex]
E. [tex]-2 \ \textless \ x \leq 3[/tex]


Sagot :

To solve the inequality [tex]\(-3 < 2x + 1 \leq 11 - 3x\)[/tex], we will break it down into two separate parts and solve each part step by step.

### Step 1: Solving [tex]\(-3 < 2x + 1\)[/tex]

1. Subtract 1 from both sides:
[tex]\[ -3 - 1 < 2x \][/tex]
Simplifying the left-hand side, we get:
[tex]\[ -4 < 2x \][/tex]

2. Divide both sides by 2:
[tex]\[ \frac{-4}{2} < x \][/tex]
Simplifying, we obtain:
[tex]\[ -2 < x \][/tex]

### Step 2: Solving [tex]\(2x + 1 \leq 11 - 3x\)[/tex]

1. Subtract [tex]\(1\)[/tex] from both sides:
[tex]\[ 2x + 1 - 1 \leq 11 - 3x - 1 \][/tex]
Simplifying, we get:
[tex]\[ 2x \leq 10 - 3x \][/tex]

2. Add [tex]\(3x\)[/tex] to both sides to gather all [tex]\(x\)[/tex] terms on one side:
[tex]\[ 2x + 3x \leq 10 \][/tex]
Simplifying, we have:
[tex]\[ 5x \leq 10 \][/tex]

3. Divide both sides by 5:
[tex]\[ x \leq \frac{10}{5} \][/tex]
Simplifying, we get:
[tex]\[ x \leq 2 \][/tex]

### Combining the Results

Now, we combine the results from both inequalities:
- From the first inequality, we have [tex]\( -2 < x \)[/tex].
- From the second inequality, we have [tex]\( x \leq 2 \)[/tex].

Combining these gives us the final solution:
[tex]\[ -2 < x \leq 2 \][/tex]

### Conclusion

The solution to the inequality [tex]\( -3 < 2x + 1 \leq 11 - 3x \)[/tex] is:
[tex]\[ D: -2 < x \leq 2 \][/tex]
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