Question

The following rational equation has denominators that contain variables. For this​ equation, a. Write the value or values of the variable that make a denominator zero. These are the restrictions on the variable. b. Keeping the restrictions in​ mind, solve the equation.
StartFraction 2 Over x plus 2 EndFraction plus StartFraction 4 Over x minus 2 EndFraction equals StartFraction 16 Over left parenthesis x plus 2 right parenthesis left parenthesis x minus 2 right parenthesis EndFraction
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Part 1
a. What​ is/are the value or values of the variable that​ make(s) the denominators​ zero?
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Answer 1

To determine the values of the variable that make the denominators zero, we need to identify the denominators in the given rational equation:

\[
\frac{2}{x + 2} + \frac{4}{x - 2} = \frac{16}{(x + 2)(x - 2)}
\]

In this equation, the denominators are \( x + 2 \), \( x - 2 \), and \( (x + 2)(x - 2) \).

### Part 1: Finding restrictions

1. **Set each denominator to zero**:
- For \( x + 2 = 0 \):
\[
x = -2
\]
- For \( x - 2 = 0 \):
\[
x = 2
\]

### Restrictions:
The values that make the denominators zero are:
- \( x = -2 \)
- \( x = 2 \)

Thus, the restrictions on the variable \( x \) are:
\[
x = -2 \quad \text{and} \quad x = 2
\]

### Part 2: Solving the equation (assuming restrictions are kept in mind)

Now, we can proceed to solve the equation while keeping these restrictions in mind. To solve the equation, we can eliminate the fractions by multiplying both sides by \( (x + 2)(x - 2) \):

\[
(x + 2)(x - 2) \left( \frac{2}{x + 2} + \frac{4}{x - 2} \right) = (x + 2)(x - 2) \left( \frac{16}{(x + 2)(x - 2)} \right)
\]

Simplifying this, we get:

\[
2(x - 2) + 4(x + 2) = 16
\]

Expanding both sides:

\[
2x - 4 + 4x + 8 = 16
\]

Combine like terms:

\[
6x + 4 = 16
\]

Now, isolate \( x \):

\[
6x = 16 - 4
\]
\[
6x = 12
\]
\[
x = 2
\]

However, \( x = 2 \) is one of our restrictions (it makes the denominator zero), meaning this solution is not valid.

### Conclusion:
Since the only potential solution \( x = 2 \) is restricted, there are no valid solutions for the equation within the allowed domain.

Thus, the answer for Part 1 regarding restrictions is:
\[
x = -2, \quad x = 2
\]

And for Part 2, **there are no valid solutions** due to the restrictions.