To determine whether a solution to the equation \(\frac{3}{a+2} + \frac{2}{a} = \frac{4a - 4}{a^2 - 4}\) is extraneous, we first need to find the solutions for \(a\) and then test them.
Given the equation: \[ \frac{3}{a+2} + \frac{2}{a} = \frac{4a - 4}{a^2 - 4} \]
Notice that \(a^2 - 4\) can be factored: \[ a^2 - 4 = (a - 2)(a + 2) \]
Now, rewrite the right-hand side: \[ \frac{4a - 4}{(a-2)(a+2)} = \frac{4(a - 1)}{(a-2)(a+2)} \]
This tells us that the denominator is zero when \(a = 2\) or \(a = -2\). Hence, \(a = -2\) is a value to check, as it makes the left-side undefined.
Next, let's find the common denominator for the left side: The common denominator for \(\frac{3}{a+2}\) and \(\frac{2}{a}\) is \(a(a + 2)\).
Rewriting the left side: \[ \frac{3a}{a(a + 2)} + \frac{2(a + 2)}{a(a + 2)} = \frac{3a + 2a + 4}{a(a + 2)} = \frac{5a + 4}{a(a + 2)} \]
Now, we equate both sides: \[ \frac{5a + 4}{a(a + 2)} = \frac{4(a - 1)}{(a - 2)(a + 2)} \]
Cross-multiplying gives: \[ (5a + 4)(a - 2) = 4(a - 1)a \]
Expanding both sides: \[ 5a^2 - 10a + 4a - 8 = 4a^2 - 4a \] \[ 5a^2 - 6a - 8 = 4a^2 - 4a \] \[ 5a^2 - 4a^2 - 6a + 4a - 8 = 0 \] \[ a^2 - 2a - 8 = 0 \]
Factoring: \[ (a - 4)(a + 2) = 0 \]
So, we have potential solutions: \[ a = 4 \quad \text{or} \quad a = -2 \]
Now let's check if either solution makes the original equation invalid due to division by zero:
- For \(a = -2\), the denominators in the original equation lead to undefined expressions (specifically \(\frac{3}{0}\)). Therefore, \(a = -2\) is an extraneous solution.
- For \(a = 4\), it does not make any denominator zero, so it is a valid solution.
Thus the answer to the question of which solution is extraneous is: \[ \text{a = -2} \]