$$ \frac{\left\vert y-10\right\vert}{-2} $$=-6 solve for both equations check for extraneous solutions

We are given the equation:

\[
\frac{\left| y - 10 \right|}{-2} = -6
\]

### Step 1: Eliminate the denominator by multiplying both sides by -2.

\[
\left| y - 10 \right| = (-6)(-2)
\]

\[
\left| y - 10 \right| = 12
\]

### Step 2: Set up two equations based on the absolute value definition.

The absolute value equation $ \left| y - 10 \right| = 12 $ gives us two possibilities:

1. $ y - 10 = 12 $
2. $ y - 10 = -12 $

### Step 3: Solve both equations.

For the first equation:
\[
y - 10 = 12
\]
\[
y = 12 + 10
\]
\[
y = 22
\]

For the second equation:
\[
y - 10 = -12
\]
\[
y = -12 + 10
\]
\[
y = -2
\]

### Step 4: Check for extraneous solutions.

We must substitute both solutions into the original equation to verify.

#### Checking $ y = 22 $:

\[
\frac{\left| 22 - 10 \right|}{-2} = \frac{\left| 12 \right|}{-2} = \frac{12}{-2} = -6
\]

This is correct.

#### Checking $ y = -2 $:

\[
\frac{\left| -2 - 10 \right|}{-2} = \frac{\left| -12 \right|}{-2} = \frac{12}{-2} = -6
\]

This is also correct.

### Conclusion:
Both $ y = 22 $ and $ y = -2 $ are valid solutions, and there are no extraneous solutions.

Would you like further details on this? Here are 5 related questions:

1. How does the absolute value affect solving equations?
2. What is an extraneous solution, and why do we check for it?
3. Can an absolute value equation ever have no solution?
4. How would this process change if the equation was a fraction with a positive denominator?
5. Can absolute value equations have more than two solutions?

**Tip:** Always isolate the absolute value expression first before proceeding with solving the equation.

Learn how to solve the absolute value equation |y - 10|/-2 = -6. This solution explains the steps to eliminate the denominator, solve for two equations, and check for extraneous solutions. Suitable for students in Grades 9-11.

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