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I got this question and I'm stuck on one of the steps.

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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?

Community
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Grimestock
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  • The right hand side needs a common denominator. Multiplying that last fraction by $\frac {2(a-2)}{2(a-2)}$ will take care of that. Then you can add those two fractions together. – Doug M Jun 13 '17 at 22:22
  • your idea is correct and note that denominator should be $a\neq \pm 2$ – haqnatural Jun 13 '17 at 22:23

2 Answers2

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$$\frac{2}{(a-2)(a+2)} = \frac{1}{2(a-2)(a+2)} - \frac{1}{a+2}$$

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.

Zubin Mukerjee
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Yes, and that would give you $$2(2)=1-2(a-2)$$ $$4=5-2a$$ $$a=\frac12$$

Akababa
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