The equation provided in the image is:

$|x - 1| = 7x - 13$

Step-by-step solution:

1. Split into two cases:

The absolute value equation can be solved by considering two cases:

• Case 1: $x - 1 \geq 0$ (so $|x - 1| = x - 1$)
• Case 2: $x - 1 < 0$ (so $|x - 1| = -(x - 1)$)

Case 1: $x - 1 \geq 0$

Here, $|x - 1| = x - 1$. Substituting into the equation:

$x - 1 = 7x - 13$

Solve for $x$:

$x - 7x = -13 + 1$

$-6x = -12$

$x = 2$

Check the condition: For this case, $x - 1 \geq 0$, so $x \geq 1$. Since $x = 2$ satisfies $x \geq 1$, this solution is valid.

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Case 2: $x - 1 < 0$

Here, $|x - 1| = -(x - 1)$. Substituting into the equation:

$-(x - 1) = 7x - 13$

Simplify:

$-x + 1 = 7x - 13$

Solve for $x$:

$-x - 7x = -13 - 1$

$-8x = -14$

$x = \frac{7}{4}$

Check the condition: For this case, $x - 1 < 0$, so $x < 1$. Since $x = \frac{7}{4} \approx 1.75$ does NOT satisfy $x < 1$, this solution is not valid.

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Final Solution:

The only valid solution is:

$\boxed{x = 2}$

Would you like a detailed explanation of the checking process or further clarification?

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Related Questions:

1. How do you graph absolute value equations such as $|x - 1| = 7x - 13$?
2. How can you check for extraneous solutions in absolute value equations?
3. What are the key differences between linear and absolute value equations?
4. Can you have no solutions to an absolute value equation? If so, when?
5. How do you solve more complex absolute value equations with multiple terms?

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Tip:

Always verify solutions of absolute value equations by substituting them back into the original equation to avoid extraneous solutions!