Question

Solve, graph and give interval notation for the inequality 9-2|4x+1|>3

Answer 1

To solve the inequality \( 9 - 2|4x + 1| > 3 \), we start by isolating the absolute value.

1. Subtract 9 from both sides:

\[
-2|4x + 1| > 3 - 9
\]
\[
-2|4x + 1| > -6
\]

2. Divide both sides by -2. Remember to reverse the inequality sign when dividing by a negative number:

\[
|4x + 1| < 3
\]

3. Now, interpret this absolute value inequality. The inequality \( |A| < B \) can be rewritten as:

\[
-B < A < B
\]

For our case, \( A = 4x + 1 \) and \( B = 3 \), thus we have:

\[
-3 < 4x + 1 < 3
\]

4. We now split this into two inequalities:

\[
-3 < 4x + 1
\]
and
\[
4x + 1 < 3
\]

5. Solve the first inequality:

\[
-3 < 4x + 1
\]
Subtract 1 from both sides:

\[
-4 < 4x
\]
Divide by 4:

\[
-1 < x \quad \text{or} \quad x > -1
\]

6. Now solve the second inequality:

\[
4x + 1 < 3
\]
Subtract 1 from both sides:

\[
4x < 2
\]
Divide by 4:

\[
x < \frac{1}{2}
\]

7. Combine the two results:

\[
-1 < x < \frac{1}{2}
\]

In interval notation, the solution is:

\[
(-1, \frac{1}{2})
\]

8. To graph the solution on a number line, draw an open interval (not including the endpoints) between \(-1\) and \(\frac{1}{2}\).

So the final solution is:

**Interval notation:** \((-1, \frac{1}{2})\)