which solution to the equation $\frac{3}{a + 2}+\frac{2}{a}=\frac{4a - 4}{a^{2}-4}$ is extraneous?\n$a=-2$\n$…

Explanation:

Step1: Factor the denominator

Factor $a^{2}-4=(a + 2)(a - 2)$. The given equation $\frac{3}{a + 2}+\frac{2}{a}=\frac{4a-4}{a^{2}-4}$ has a common - denominator of $a(a + 2)(a - 2)$. Multiply each term by $a(a + 2)(a - 2)$: $3a(a - 2)+2(a + 2)(a - 2)=(4a - 4)a$.

Step2: Expand the expressions

Expand the left - hand side: $3a^{2}-6a+2(a^{2}-4)=4a^{2}-4a$. $3a^{2}-6a + 2a^{2}-8=4a^{2}-4a$. Combine like terms: $5a^{2}-6a-8=4a^{2}-4a$.

Step3: Rearrange to form a quadratic equation

Move all terms to one side: $5a^{2}-4a^{2}-6a + 4a-8=0$. $a^{2}-2a-8=0$.

Step4: Solve the quadratic equation

Factor the quadratic equation: $(a - 4)(a+2)=0$. Set each factor equal to zero: $a - 4=0$ gives $a = 4$; $a + 2=0$ gives $a=-2$.

Step5: Check for extraneous solutions

The original equation has denominators $a + 2$, $a$, and $a^{2}-4=(a + 2)(a - 2)$. When $a=-2$, the denominators $a + 2$ and $a^{2}-4$ are equal to 0. When $a = 4$, the denominators $a+2$, $a$, and $a^{2}-4$ are non - zero. So $a=-2$ is an extraneous solution.

Answer:

A. $a=-2$