Question

which solution to the equation \\(\frac{3}{a + 2} + \frac{2}{a} = \frac{4a - 4}{a^2 - 4}\\) is extraneous?\
\\(a = -2\\)\
\\(\text{neither } a = -2 \text{ nor } a = 4\\)\
\\(a = 4\\)\
\\(a = -2 \text{ and } a = 4\\)

Explanation:

Step1: Find the domain restrictions

The original equation is \(\frac{3}{a + 2}+\frac{2}{a}=\frac{4a - 4}{a^{2}-4}\). First, factor the denominators: \(a^{2}-4=(a + 2)(a - 2)\). The domain restrictions are values that make any denominator zero. So, \(a+2
eq0\) (i.e., \(a
eq - 2\)), \(a
eq0\), and \(a - 2
eq0\) (i.e., \(a
eq2\)).

Step2: Solve the equation (simplify)

Multiply each term by the least common denominator (LCD), which is \(a(a + 2)(a - 2)\) to eliminate the fractions:

$$
LATEXBLOCK0
$$

The solutions from the factoring are \(a = 4\) and \(a=-2\).

Step3: Identify extraneous solutions

An extraneous solution is a solution that does not satisfy the domain restrictions. From the domain restrictions, we know \(a
eq - 2\) (since \(a=-2\) makes the denominator \(a + 2\) zero). The solution \(a = 4\) is in the domain (since \(4
eq-2\), \(4
eq0\), and \(4
eq2\)). So, \(a=-2\) is extraneous.

Answer:

\(a=-2\) (the option with "a = -2")