# Find the least common denominator of algebraic fractions

Having trouble with one of my math homework problems. I need to find the least common denominator (LCD) to solve the problem. I'm not sure how to figure this out one. Thanks in advance.

$$\frac{3}{j^2+6j} + \frac{2j}{j+6} - \frac{2}{3j} .$$

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The first term you listed. Thank you. I actually forgot the J after the 6. For the denominator of the first term, it should be j^2+6j, sorry. –  Mark Sep 14 '11 at 0:38
And the second should also be $\frac{2j}{j+6}$? And the third $\frac{2}{3j}$? –  TMM Sep 14 '11 at 0:40
Thanks, that makes it much easier to read. It is now displayed exactly as the problem is. Thanks. –  Mark Sep 14 '11 at 0:42
Mark: instead of just "not sure how to figure this one out", how about sharing what you've tried so far. We can help steer you in the right direction and/or point out missteps. You'd learn more than just reading someone else's solution too. :) –  Fixee Sep 14 '11 at 0:44
Here goes: note that if you take the LCD of the last two fractions, you get $3j^2+18j$. So... –  Guess who it is. Sep 14 '11 at 0:44

Hint: First try finding the LCD of some integers. For example, evaluate $$\frac{3}{4} + \frac{7}{10} + \frac{13}{25}$$ and be very conscious of how you're doing it when you get a common denominator. Next try adding these: $$\frac{1}{x^2} + \frac{1}{x}$$ and $$\frac{1}{x+3} + \frac{1}{x^2-9}$$ It's the same thing each time, but you need to use a bit of algebra. Now try your homework problem. Good luck!
Hint: The denominator $j^2 + 6j$ factors as $j(j + 6)$. Now, look at all the factors that appear in the various denominators: $$j, j + 6, 3.$$ So, a good common denominator might be $3j(j+6)$. How can you make all the fractions have this denominator?