+0

# is it right?it it is so how do i need to continue from here?

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340
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sabi92  Jun 20, 2015

#2
+91773
+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
Sort:

#1
+91773
+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
+91773
+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
#3
+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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