I got this question and I'm stuck on one of the steps.
And this is where I'm stuck.I'm not sure what to do next. Do I multiply everything by 2(a-2)(a+2) over 1 or something entirely different?
I got this question and I'm stuck on one of the steps.
And this is where I'm stuck.I'm not sure what to do next. Do I multiply everything by 2(a-2)(a+2) over 1 or something entirely different?
$$\frac{2}{(a-2)(a+2)} = \frac{1}{2(a-2)(a+2)} - \frac{1}{a+2}$$
You can arrange the equation so that all terms have a common denominator. In this case, the simplest such denominator will be $2(a-2)(a+2)$.
$$\frac{4}{2(a-2)(a+2)} = \frac{1}{2(a-2)(a+2)} - \frac{2(a-2)}{2(a-2)(a+2)}$$
Now, if $a \neq \pm 2$, we can multiply both sides by the denominator to get
$$4 = 1 - 2(a-2)$$
$$4 = 1 - 2a + 4$$ $$2a = 1$$
$$a = \boxed{\frac{1}{2}\,}$$
Since the $a = \pm 2$ possibilities would not work with your original equation, $1/2$ is the only solution.