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$$\frac{7a}{6a^2-15a} + \frac{12a}{4a^2-25}$$

I determined the LCD of the denominators: $(3a)(2a-5)(2a+5)$. I then multiplied all nominators by the LCM, combined the terms and got:

$$\frac{a(86a+95)}{(3a)(2a-5)(2a+5)}$$

Where did I go wrong?

jvdhooft
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Grimestock
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    What happened to the $a$ in the denominator? Note: you can reduce the first fraction by cancelling a factor of $a$ . – lulu Jun 12 '17 at 19:53
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    You went wrong somewhere in the part that you haven't shown: "Multiply everything by LCM". That should rather be:

    $$\require{cancel} \frac{7 \cancel{a}}{3\cancel{a}(2a-5)} + \frac{12a}{(2a-5)(2a+5)}=\frac{7 \cdot (2a+5)+12a\cdot 3}{3(2a-5)(2a+5)}$$

     – dxiv Jun 12 '17 at 19:56
  • somewhere between the first line and the last line. But you don't show us no "between" lines, so how can we point at the error? – miracle173 Jun 12 '17 at 20:01

1 Answers1

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$$\frac{7a}{6a^2-15a} + \frac{12a}{4a^2-25} = \frac{7}{3(2a-5)} + \frac{12a}{(2a-5)(2a+5)} = \frac{7(2a+5) + 12a3}{3(2a-5)(2a+5)} = \frac{50a+35}{3(2a-5)(2a+5)} = \frac{5(10a+7)}{3(2a-5)(2a+5)}$$

jvdhooft
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