Solve the equation, check for extraneous solutions. |4x - 1| = 2x + 13

To solve the equation \(|4x - 1| = 2x + 13\), we need to consider both cases for the absolute value:

### Case 1: \(4x - 1 = 2x + 13\)

Simplify the equation:

\[
4x - 2x = 13 + 1
\]

\[
2x = 14
\]

\[
x = 7
\]

### Case 2: \(4x - 1 = -(2x + 13)\)

Simplify the equation:

\[
4x - 1 = -2x - 13
\]

\[
4x + 2x = -13 + 1
\]

\[
6x = -12
\]

\[
x = -2
\]

### Checking for Extraneous Solutions:

Substitute the solutions back into the original equation to check if they are valid.

1. For \(x = 7\):

\[
|4(7) - 1| = 2(7) + 13
\]

\[
|28 - 1| = 14 + 13
\]

\[
|27| = 27 \quad \text{(True)}
\]

2. For \(x = -2\):

\[
|4(-2) - 1| = 2(-2) + 13
\]

\[
|-8 - 1| = -4 + 13
\]

\[
|-9| = 9 \quad \text{(True)}
\]

Both \(x = 7\) and \(x = -2\) satisfy the original equation, so they are not extraneous solutions.

### Solutions:
\[
x = 7 \quad \text{and} \quad x = -2
\]

Would you like more details or have any questions?

Here are some related questions you might be interested in:

1. How do you solve absolute value equations with more complex expressions?
2. What are the properties of absolute value functions?
3. How can you graph the solution to absolute value equations?
4. What are extraneous solutions, and why do they occur?
5. How do you solve inequalities involving absolute values?

**Tip:** Always check potential solutions in the original equation, especially when dealing with absolute values, to avoid extraneous solutions!

Solve the absolute value equation |4x - 1| = 2x + 13 step by step, considering both cases of the absolute value and checking for extraneous solutions. Learn key algebra concepts and formulas, suitable for students in Grades 8-10.

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