my lesson doesnt do a good job at explaining this part of the chapter, can someone show me how to complete this?

- anonymous

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- anonymous

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- anonymous

There's a lot going on here, but we can do it!

- anonymous

i would need like denominators right?

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- anonymous

That is a great way to start! :)

- anonymous

What would the LCD be for this rational equation?

- anonymous

2

- anonymous

No... not 2.

- anonymous

1 ? 4 ?

- anonymous

\[\frac{2}{x}- \frac{2}{x-1} - \frac{2}{x-2} \]
To find the LCD first we would look to factor the denominators. However, these don't factor.
What is the denominator of the first fraction?

- anonymous

x

- anonymous

That's right - so our LCD must have an "x" in it. Now, what is the denominator of the second fraction?

- anonymous

x-1

- anonymous

So our LCD will need the "x-1" as well.
What is the third denominator?

- anonymous

x-2

- anonymous

Yes, so our LCD will need to have an "x-2" in it also.

- anonymous

So, what we have right now is:
\[\frac{2}{x}- \frac{2}{x-1} - \frac{2}{x-2} \]
LCD: x(x-1)(x-2)

- anonymous

Do you have any questions on anything we've done so far?

- anonymous

for the LCD would i distribute the x into the (x-1)(x-2) or would i just leave it ?

- anonymous

Excellent question - you would leave it. We're hoping to cancel things later (if possible) and it's easier to see how to cancel when the denominator is already factored.

- anonymous

\[\frac{2}{x}- \frac{2}{x-1} - \frac{2}{x-2} \]
LCD: x(x-1)(x-2)
Now, looking at the LCD and our 3 fractions, which pieces of the LCD does the first fraction need?

- anonymous

it needs the other half

- anonymous

Which pieces specifically?

- anonymous

the -1 and -2

- anonymous

Well, we can't just give it a -1 and a -2. It would need the full terms "x-1" and "x-2"

- anonymous

yeah thats what i ment lol

- anonymous

Ok, good. :)

- anonymous

Let's give it those pieces right now, then we'll decide what the other two fractions need, and we'll give them the pieces they need as well.

- anonymous

We decided this piece \[\frac{2}{x} \]Needs an (x-1) and (x-2)
So we will give it those needed parts by multiplying by a special form of 1.
\[\frac{2}{x}\frac{(x-1)(x-2)}{(x-1)(x-2)} \]

- anonymous

Can you see how that big fraction I just made is a special form of 1 because it has the same numerator and same denominator?

- anonymous

yeah so then would i distribute the 2?

- anonymous

Distribute and FOIL along the top - leave the bottom the way it is. BUT, before you do that - let's do the same for the other pieces.

- anonymous

Here is the second fraction\[\frac{2}{x-1}\]What does it need from our LCD?
LCD: x(x-1)(x-2)

- anonymous

it needs a (x-2)

- anonymous

What else? It needs one more thing.

- anonymous

the x?

- anonymous

i foiled the top and i got 4x-6.

- anonymous

Yes, the x. You don't want to forget that. :)
Imaging you're trying to add:
\[\frac{1}{3} and \frac{2}{37}\]
You wouldn't say that they have a 3 in common - even though you see a "3" in both denominators.

- anonymous

yeah thats true

- anonymous

Well, don't foil anything just yet... We'll do the FOILing in just a sec. First we have to set up what we want to FOIL.

- anonymous

Now, how about that third fraction? What does it need from the LCD?

- anonymous

it need the x and (x-1)

- anonymous

OUTSTANDING!! XD

- anonymous

So right now - here's what you want to have written down somewhere:
\[\frac{2}{x} \frac{(x-1)(x-2)}{(x-1)(x-2)}- \frac{2}{x-1} \frac{x(x-2)}{x(x-2)} + \frac{2}{x-2} \frac{x(x-1)}{x(x-1)}\]

- anonymous

so would we foil the numerators now?

- anonymous

Yes, now we get to do that. ALSO - notice that your denominators are all the same. Once you reach this point, when they're all the same, they cancel and you only have the numerators left.

- anonymous

Wait - forget that!

- anonymous

No canceling! thought I saw an equal sign...

- anonymous

Just foil the tops and leave the bottoms the same.

- anonymous

for the first one i got 4x-6

- anonymous

4x^2-6 i mean

- anonymous

There shouldn't be a 6 there.

- anonymous

\[\frac{2}{x} \frac{(x-1)(x-2)}{(x-1)(x-2)}- \frac{2}{x-1} \frac{x(x-2)}{x(x-2)} - \frac{2}{x-2} \frac{x(x-1)}{x(x-1)}\]
If you like, instead of writing the denominator over and over again, you can just write "LCD"
\[\frac{2(x-1)(x-2)}{LCD} - \frac{2x(x-2)}{LCD} +\frac{2x(x-1)}{LCD} \]

- anonymous

i dont know what i did wrong then.

- anonymous

2(x-1)(x-2) To multiply this, first let's multiply the 2 times the first factor.
2(x-1) <--- what do you get when you distribute here?

- anonymous

2x-2

- anonymous

Good! Now, you're going to foil that with the second factor:
(2x-2)(x-2)

- anonymous

What do you get when you do that?

- anonymous

2x^3 -4x-4

- anonymous

It should be this:
|dw:1332653149065:dw|

- anonymous

Remember: You multiply the first terms, the outside terms, the inside terms, and the last terms.
Then, combine like terms (usually the inside and outside terms are like terms).

- anonymous

ok so i would do the same for the 2nd right?

- anonymous

Yes, and the third. :)

- anonymous

the 2nd would be 2x^2-4x and the thrid one would be 2x^2-2x

- anonymous

\[\frac{2(x-1)(x-2)}{LCD} - \frac{2x(x-2)}{LCD} +\frac{2x(x-1)}{LCD} \]
\[\frac{2x^2-6x+4}{LCD} - \frac{2x^2-4x}{LCD} +\frac{2x^2-2x}{LCD} \]
Very good!

- anonymous

Now, there's a little trick here... notice that your second term is being subtracted...

- anonymous

We can combine these all into one fraction as long as we remember to use parenthesis... like this...

- anonymous

\[\frac{(2x^2-6x+4)-(2x^2-4x) +(2x^2-2x)}{LCD} \]

- anonymous

Because - we need to remember to distribute that negative sign.

- anonymous

so the second numerator would be -2x^2+4x

- anonymous

Yes, that's right!

- anonymous

and once you distribute that negative sign, you can combine like terms

- anonymous

would i still keep the subtraction sign in between the first and second equations?

- anonymous

Only if you're using it to indicate that the 2x^2 is negative. It should look like this right now...

- anonymous

\[ \frac{2x^2-6x+4-2x+4x+2x^2-2x}{LCD} \]

- anonymous

wait.. I forgot the squared on the -2x

- anonymous

\[ \frac{2x^2-6x+4-2x^2+4x+2x^2-2x}{LCD} \]

- anonymous

So from here, we combine like terms on the top - and it looks like some things will cancel each other out too.

- anonymous

what is the LCD again?

- anonymous

x(x-1)(x-2) right?

- anonymous

LCD: x(x-1)(x-2) <--- although you know to check the original question to find that, right? ;)

- anonymous

Yes, good! :)

- anonymous

\[ \frac{4-4x+2x^2}{LCD} \]And, we can write the numerator in standard form.
\[ \frac{2x^2-4x+4}{LCD} \]And now, replace back the LCD.
\[ \frac{2x^2-4x+4}{x(x-1)(x-2)} \]From here, factor the numerator (can you see the number all 3 terms in the numerator have in common?)

- anonymous

they both have the 2 in common

- anonymous

Well, all 3 of the top pieces would need to have a 2 in common to factor it out. Do all 3 of them have a 2 in common?

- anonymous

i only see 2 pieces not three

- anonymous

There should be 3 pieces: \[2x^2 -4x+4\]

- anonymous

oh i thought you were talking about the numerator and the denominator

- anonymous

Oh - no. Just the numerator. :)

- anonymous

Our focus right now, is factoring the numerator. See, if we would have FOILed the denominator way back at the beginning of this question, we would have to factor it out now. We saved ourselves that step. So all we have to do now is factor the numerator and see if anything will cancel with the denominator.

- anonymous

they all have a 2 in common. i dont see anyother number

- anonymous

Right! So factor that 2 out. What are we left with on top?

- anonymous

we are left with x^2-2x+2

- anonymous

Yes, although the 2 should still be visible on the outside of a set of parenthesis, like this:
\[ \frac{2(x^2-2x+2)}{x(x-1)(x-2)} \]

- anonymous

Now, we want to factor what we have in parenthesis.

- anonymous

Well - on the top I mean. :)

- anonymous

So, how would you factor \[x^2-2x+2\]

- anonymous

Oh - nevermind - it doesn't factor. We're done!

- anonymous

This is how you would leave the answer, in factored form.\[ \frac{2(x^2-2x+2)}{x(x-1)(x-2)} \]That way, when the teacher sees the answer, he/she can tell that it doesn't simplify any further.

- anonymous

your've honestly done a better job at teaching me than my teacher has. thank you so much!

- anonymous

You're welcome! Better than your teacher, huh? :) That means a lot to me!

- anonymous

That means I'm getting a metal for this one, right? ;)

- anonymous

lol yeah. and a fan :)

- anonymous

Right on! Thanks! :)

- anonymous

Well, I'm going to head to bed! Good luck with your math!! :)

- anonymous

thanks !

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