- Simplify the equation \(-2|8-x|-6=-12\) to isolate the absolute value, obtaining \(|8-x| = 3\).
- Split the absolute value equation into two cases: \(8-x = 3\) and \(8-x = -3\).
- Calculate the solutions: \(x = 5\) and \(x = 11\).
- Conclude that Johan is incorrect due to a sign error.
- Final Answer: \(\boxed{\text{He is not correct because he made a sign error.}}\)
Explanation
- Analyzing and Solving the Equation
The given equation is \(-2|8-x|-6=-12\). We aim to find the set of values for \(x\) that satisfy this equation and verify if Johan's claimed solutions of \(x=5\) and \(x=-11\) are correct. First, we isolate the absolute value term by adding 6 to both sides of the equation:
\[-2|8-x| = -12 + 6\]
\[-2|8-x| = -6\]
Next, divide both sides by \(-2\):
\[|8-x| = \frac{-6}{-2}\]
\[|8-x| = 3\]
Now, we have an absolute value equation of the form \(|A| = c\), where \(c \ge 0\). This leads to two separate linear equations:
- \(8 - x = 3\)
- \(8 - x = -3\)
Solving the first equation:
\(8 - 3 = x\)
\(x = 5\)
Solving the second equation:
\(8 + 3 = x\)
\(x = 11\)
The correct solutions are \(x = 5\) and \(x = 11\). Comparing these to Johan's stated solutions of \(x=5\) and \(x=-11\), we see that Johan correctly identified \(x=5\) but made a sign error for the second solution, yielding \(-11\) instead of \(11\). Therefore, Johan is not correct because he made a sign error.
Examples
Understanding how to solve absolute value equations is crucial in many real-world applications where only the magnitude of a deviation matters, regardless of whether it is positive or negative. For example, if a quality control process in a factory requires a part's weight to be within \(3\) grams of the target weight of \(8\) grams, we can model the acceptable weight range as \(|8 - x| = 3\). Solving this informs us that the weight \(x\) must be between \(5\) grams and \(11\) grams, helping the manufacturer maintain consistency and reliability in their products.