High School

Solve [tex]|2x + 1| \leq 15[/tex].

A. [tex]-7 \leq x \leq 7[/tex]

B. [tex]x \leq 7[/tex]

C. [tex]-8 \geq x \geq 7[/tex]

D. [tex]-8 \leq x \leq 7[/tex]

Answer :

To solve the inequality [tex]\(|2x + 1| \leq 15\)[/tex], we can break it down into two separate inequalities, because the absolute value inequality [tex]\(|a| \leq b\)[/tex] can be rewritten as:

[tex]\[
-b \leq a \leq b
\][/tex]

So for our expression, we have:

[tex]\[
-15 \leq 2x + 1 \leq 15
\][/tex]

We'll solve these two inequalities one at a time.

### Solving the First Inequality:

[tex]\[
2x + 1 \geq -15
\][/tex]

1. Subtract 1 from both sides to isolate the term with [tex]\(x\)[/tex]:
[tex]\[
2x \geq -15 - 1
\][/tex]

[tex]\[
2x \geq -16
\][/tex]

2. Divide both sides by 2 to solve for [tex]\(x\)[/tex]:
[tex]\[
x \geq -8
\][/tex]

### Solving the Second Inequality:

[tex]\[
2x + 1 \leq 15
\][/tex]

1. Subtract 1 from both sides:
[tex]\[
2x \leq 15 - 1
\][/tex]

[tex]\[
2x \leq 14
\][/tex]

2. Divide both sides by 2:
[tex]\[
x \leq 7
\][/tex]

### Combining the Solutions:

Now we combine the two solutions:

- From the first inequality, [tex]\(x \geq -8\)[/tex]
- From the second inequality, [tex]\(x \leq 7\)[/tex]

Putting these together, we get:

[tex]\[
-8 \leq x \leq 7
\][/tex]

So the solution to the inequality [tex]\(|2x + 1| \leq 15\)[/tex] is:

[tex]\[
-8 \leq x \leq 7
\][/tex]

This means that [tex]\(x\)[/tex] can take any value between [tex]\(-8\)[/tex] and [tex]\(7\)[/tex], inclusive.

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