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$$ \frac{3a+2}{36-a^2} - \frac{a-4}{a^2-8a+12} $$

Got stuck on this equation. $$ \frac{3a+2}{(6-a)(6+a)} - \frac{a-4}{(a-6)(a-2)} $$

For this part, what do I do about the $(6-a)(6+a)$? Do I take away a $-1$ or something else?

Edit: I think I figured out the answer, thanks to all of your help. enter image description here

Grimestock
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  • You will want to combine both fractions under a common denominator. What do you think this denominator would be? Hint: you need three factors multiplied together. – fractal1729 Jun 12 '17 at 20:45
  • Between this and your prior question it looks like you are just posting your homework here for us to do for you. What are your thoughts on this problem? – lulu Jun 12 '17 at 20:46
  • Sorry. I finished the majority of my homework and just had three questionsI was stuck on. I think I might have to flip the second fraction just like I would if I had to divide? I also think I might have to take a -1 out of the first fraction in order to make the denominator easier to manage? But I'm not 100% sure. – Grimestock Jun 12 '17 at 21:03

2 Answers2

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Hint: $$ -\frac{a-4}{(a-6)(a-2)}=+\frac{a-4}{(6-a)(a-2)} $$

Emilio Novati
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  • Thanks. I think I figured out the answer. [IMG]http://i66.tinypic.com/rws7b5.png[/IMG]

    Edit: Sorry, I don't know how to post photos in comments.

     – Grimestock Jun 12 '17 at 21:13
  • OK ! The result is correct. To learn how to write math formulas, study this: https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference.. And don't forget to accept a question when it is good for you. – Emilio Novati Jun 12 '17 at 21:23
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HINT Note that $-(a-6) = 6-a$, so $$ \frac{3a+2}{(6-a)(6+a)} - \frac{a-4}{(a-6)(a-2)} = \frac{1}{6-a} \left[ \frac{3a+2}{6+a} + \frac{a-4}{a-2}\right] $$ and now just bring the fractions inside the bracket to a common denominator.

gt6989b
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