hard fraction

**Ackbeet** · November 12th 2010, 06:45 AM

Not quite. What did you do with the (2) and the (3) to get its LCM? That's what you have to do with the (2) and the (x+2) in this situation.

**andyboy179** · November 12th 2010, 06:52 AM

okay, would it be 2x+4?!?!?!?

**Ackbeet** · November 12th 2010, 07:05 AM

There you go. I'd probably write it as 2(x+2) to emphasize how you got it. Now that you know what the common denominator is, can you add the fractions together?

**andyboy179** · November 12th 2010, 07:16 AM

YES!!!! so would the overall answer =x+9?

**Ackbeet** · November 12th 2010, 07:20 AM

No, I don't think so. Do this problem step-by-step. After you've found the common denominator, you add the fractions. What do you get when you do that?

**andyboy179** · November 12th 2010, 07:25 AM

when i add them i get x-1-4 over 2x+4
than i get X-5over 2x+4
then, x-5=2x+4
then, -5=x+4
then, =x+9

**Ackbeet** · November 12th 2010, 07:41 AM

Aha. You're missing a very important step. Let's say you're adding the fractions

$\dfrac{1}{2}+\dfrac{1}{3}.$

We've already determined that the LCM of the denominators, which is the same thing as the common denominator, is 6. What would your next step be in this addition problem?

**andyboy179** · November 12th 2010, 08:21 AM

1x1=1 and 2x3=6
=1/6

**Ackbeet** · November 12th 2010, 08:29 AM

No, no, no. You've found the common denominator. Now you have to get the common denominator. You do that by multiplying fractions by fancy ways of writing 1 in order to obtain the common denominator in each denominator. Let me illustrate:

$\displaystyle \frac{1}{2}+\frac{1}{3}=\frac{1}{2}\left(\frac{3}{ 3}\right)+\frac{1}{3}\left(\frac{2}{2}\right).$

So far, you would agree I haven't changed anything, right? I've multiplied each fraction by something that's equal to 1 (the fractions that are in parentheses). But I've chosen those numbers carefully so that when I multiply out the denominators, I'll get the common denominator that I need. Thus, the next step is to multiply out the fractions, which you can do the way you want to do it:

$\displaystyle\frac{1}{2}\left(\frac{3}{3}\right)+\ frac{1}{3}\left(\frac{2}{2}\right)=\frac{3}{6}+\fr ac{2}{6}.$

Do you see how this is done? And do you know what the next steps are?

**andyboy179** · November 12th 2010, 08:32 AM

yes i understand! after that would i add then and get 6/6=6?

**Ackbeet** · November 12th 2010, 08:34 AM

I'm not so sure you understand as well as you think you do. Yes, you would add, but the steps would look more like this:

$\dfrac{3}{6}+\dfrac{2}{6}=\dfrac{3+2}{6}=\dfrac{5} {6}.$

Make sense? I think you multiplied instead of adding.

**andyboy179** · November 12th 2010, 08:35 AM

oh i get it now

**Ackbeet** · November 12th 2010, 08:37 AM

Ok. So, picking back up from where we left off at post # 21, you're trying to add the two fractions

$\dfrac{x-1}{2}-\dfrac{4}{x+2}.$

You've determined that the least common denominator is $2(x+2).$ Now you have to get the common denominator. What do you get?

**andyboy179** · November 12th 2010, 08:38 AM

now for my question, i have worked it out on a piece of paper and i get the answer as 2x^2 - 8x +12, would i then do quadratic equation to get the answer?

**Ackbeet** · November 12th 2010, 08:41 AM

I think you might be skipping steps there. That's not the quadratic that I get. Post your work, please.

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