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sabi92  Jun 20, 2015

Best Answer 

 #2
avatar+91045 
+15

Sabi has told me that the answer is different from the book answer.    

 

$$\frac{9}{5a^2-11ab+2b^2}+\frac{17}{10a^2+13ab-3b^2}\\\\
=\frac{9}{5a^2-10ab-1ab+2b^2}+\frac{17}{10a^2-2ab+15ab-3b^2}\\\\
=\frac{9}{5a(a-2b)-b(a-2b}+\frac{17}{2a(5a-b)+3b(5a-b)}\\\\
=\frac{9}{(5a-b)(a-2b)}+\frac{17}{(2a+3b)(5a-b)}\\\\
=\frac{9(2a+3b)}{(5a-b)(a-2b)(2a+3b)}+\frac{17(a-2b)}{(2a+3b)(5a-b)(a-2b)}\\\\
=\frac{18a+27b}{(5a-b)(a-2b)(2a+3b)}+\frac{17a-34b}{(2a+3b)(5a-b)(a-2b)}\\\\
=\frac{18a+27b+17a-34b}{(5a-b)(a-2b)(2a+3b)}\\\\
=\frac{35a-7b}{(5a-b)(a-2b)(2a+3b)}\\\\
=\frac{7(5a-b)}{(5a-b)(a-2b)(2a+3b)}\\\\
=\frac{7}{(a-2b)(2a+3b)}\\\\$$

 

Is it the same now Sabi ?

Melody  Jun 21, 2015
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3+0 Answers

 #1
avatar+91045 
+10

Yes, I think it looks good so far, now you just have to get them both with a common denominator.

That will be  (5a-b)(a-2b)(2a+3b)

then you can add the numerators  

Melody  Jun 20, 2015
 #2
avatar+91045 
+15
Best Answer

Sabi has told me that the answer is different from the book answer.    

 

$$\frac{9}{5a^2-11ab+2b^2}+\frac{17}{10a^2+13ab-3b^2}\\\\
=\frac{9}{5a^2-10ab-1ab+2b^2}+\frac{17}{10a^2-2ab+15ab-3b^2}\\\\
=\frac{9}{5a(a-2b)-b(a-2b}+\frac{17}{2a(5a-b)+3b(5a-b)}\\\\
=\frac{9}{(5a-b)(a-2b)}+\frac{17}{(2a+3b)(5a-b)}\\\\
=\frac{9(2a+3b)}{(5a-b)(a-2b)(2a+3b)}+\frac{17(a-2b)}{(2a+3b)(5a-b)(a-2b)}\\\\
=\frac{18a+27b}{(5a-b)(a-2b)(2a+3b)}+\frac{17a-34b}{(2a+3b)(5a-b)(a-2b)}\\\\
=\frac{18a+27b+17a-34b}{(5a-b)(a-2b)(2a+3b)}\\\\
=\frac{35a-7b}{(5a-b)(a-2b)(2a+3b)}\\\\
=\frac{7(5a-b)}{(5a-b)(a-2b)(2a+3b)}\\\\
=\frac{7}{(a-2b)(2a+3b)}\\\\$$

 

Is it the same now Sabi ?

Melody  Jun 21, 2015
 #3
avatar+262 
+5

yes thank you very much  my answer was wrong because i didnt open the brachets 

sabi92  Jun 21, 2015

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