anonymous
  • anonymous
my lesson doesnt do a good job at explaining this part of the chapter, can someone show me how to complete this?
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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anonymous
  • anonymous
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anonymous
  • anonymous
There's a lot going on here, but we can do it!
anonymous
  • anonymous
i would need like denominators right?

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anonymous
  • anonymous
That is a great way to start! :)
anonymous
  • anonymous
What would the LCD be for this rational equation?
anonymous
  • anonymous
2
anonymous
  • anonymous
No... not 2.
anonymous
  • anonymous
1 ? 4 ?
anonymous
  • 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
  • anonymous
x
anonymous
  • anonymous
That's right - so our LCD must have an "x" in it. Now, what is the denominator of the second fraction?
anonymous
  • anonymous
x-1
anonymous
  • anonymous
So our LCD will need the "x-1" as well. What is the third denominator?
anonymous
  • anonymous
x-2
anonymous
  • anonymous
Yes, so our LCD will need to have an "x-2" in it also.
anonymous
  • 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
  • anonymous
Do you have any questions on anything we've done so far?
anonymous
  • anonymous
for the LCD would i distribute the x into the (x-1)(x-2) or would i just leave it ?
anonymous
  • 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
  • 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
  • anonymous
it needs the other half
anonymous
  • anonymous
Which pieces specifically?
anonymous
  • anonymous
the -1 and -2
anonymous
  • 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
  • anonymous
yeah thats what i ment lol
anonymous
  • anonymous
Ok, good. :)
anonymous
  • 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
  • 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
  • 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
  • anonymous
yeah so then would i distribute the 2?
anonymous
  • 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
  • anonymous
Here is the second fraction\[\frac{2}{x-1}\]What does it need from our LCD? LCD: x(x-1)(x-2)
anonymous
  • anonymous
it needs a (x-2)
anonymous
  • anonymous
What else? It needs one more thing.
anonymous
  • anonymous
the x?
anonymous
  • anonymous
i foiled the top and i got 4x-6.
anonymous
  • 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
  • anonymous
yeah thats true
anonymous
  • 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
  • anonymous
Now, how about that third fraction? What does it need from the LCD?
anonymous
  • anonymous
it need the x and (x-1)
anonymous
  • anonymous
OUTSTANDING!! XD
anonymous
  • 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
  • anonymous
so would we foil the numerators now?
anonymous
  • 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
  • anonymous
Wait - forget that!
anonymous
  • anonymous
No canceling! thought I saw an equal sign...
anonymous
  • anonymous
Just foil the tops and leave the bottoms the same.
anonymous
  • anonymous
for the first one i got 4x-6
anonymous
  • anonymous
4x^2-6 i mean
anonymous
  • anonymous
There shouldn't be a 6 there.
anonymous
  • 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
  • anonymous
i dont know what i did wrong then.
anonymous
  • 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
  • anonymous
2x-2
anonymous
  • anonymous
Good! Now, you're going to foil that with the second factor: (2x-2)(x-2)
anonymous
  • anonymous
What do you get when you do that?
anonymous
  • anonymous
2x^3 -4x-4
anonymous
  • anonymous
It should be this: |dw:1332653149065:dw|
anonymous
  • 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
  • anonymous
ok so i would do the same for the 2nd right?
anonymous
  • anonymous
Yes, and the third. :)
anonymous
  • anonymous
the 2nd would be 2x^2-4x and the thrid one would be 2x^2-2x
anonymous
  • 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
  • anonymous
Now, there's a little trick here... notice that your second term is being subtracted...
anonymous
  • anonymous
We can combine these all into one fraction as long as we remember to use parenthesis... like this...
anonymous
  • anonymous
\[\frac{(2x^2-6x+4)-(2x^2-4x) +(2x^2-2x)}{LCD} \]
anonymous
  • anonymous
Because - we need to remember to distribute that negative sign.
anonymous
  • anonymous
so the second numerator would be -2x^2+4x
anonymous
  • anonymous
Yes, that's right!
anonymous
  • anonymous
and once you distribute that negative sign, you can combine like terms
anonymous
  • anonymous
would i still keep the subtraction sign in between the first and second equations?
anonymous
  • anonymous
Only if you're using it to indicate that the 2x^2 is negative. It should look like this right now...
anonymous
  • anonymous
\[ \frac{2x^2-6x+4-2x+4x+2x^2-2x}{LCD} \]
anonymous
  • anonymous
wait.. I forgot the squared on the -2x
anonymous
  • anonymous
\[ \frac{2x^2-6x+4-2x^2+4x+2x^2-2x}{LCD} \]
anonymous
  • anonymous
So from here, we combine like terms on the top - and it looks like some things will cancel each other out too.
anonymous
  • anonymous
what is the LCD again?
anonymous
  • anonymous
x(x-1)(x-2) right?
anonymous
  • anonymous
LCD: x(x-1)(x-2) <--- although you know to check the original question to find that, right? ;)
anonymous
  • anonymous
Yes, good! :)
anonymous
  • 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
  • anonymous
they both have the 2 in common
anonymous
  • 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
  • anonymous
i only see 2 pieces not three
anonymous
  • anonymous
There should be 3 pieces: \[2x^2 -4x+4\]
anonymous
  • anonymous
oh i thought you were talking about the numerator and the denominator
anonymous
  • anonymous
Oh - no. Just the numerator. :)
anonymous
  • 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
  • anonymous
they all have a 2 in common. i dont see anyother number
anonymous
  • anonymous
Right! So factor that 2 out. What are we left with on top?
anonymous
  • anonymous
we are left with x^2-2x+2
anonymous
  • 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
  • anonymous
Now, we want to factor what we have in parenthesis.
anonymous
  • anonymous
Well - on the top I mean. :)
anonymous
  • anonymous
So, how would you factor \[x^2-2x+2\]
anonymous
  • anonymous
Oh - nevermind - it doesn't factor. We're done!
anonymous
  • 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
  • anonymous
your've honestly done a better job at teaching me than my teacher has. thank you so much!
anonymous
  • anonymous
You're welcome! Better than your teacher, huh? :) That means a lot to me!
anonymous
  • anonymous
That means I'm getting a metal for this one, right? ;)
anonymous
  • anonymous
lol yeah. and a fan :)
anonymous
  • anonymous
Right on! Thanks! :)
anonymous
  • anonymous
Well, I'm going to head to bed! Good luck with your math!! :)
anonymous
  • anonymous
thanks !

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