Question

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

Answer 1

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

Answer 2

i would need like denominators right?

Answer 3

That is a great way to start! :)

Answer 4

What would the LCD be for this rational equation?

Answer 5

2

Answer 6

No... not 2.

Answer 7

1 ? 4 ?

Answer 8

\[\frac{2}{x}- \frac{2}{x-1} - \frac{2}{x-2} \]
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?

Answer 9

x

Answer 10

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

Answer 11

x-1

Answer 12

So our LCD will need the "x-1" as well.

What is the third denominator?

Answer 13

x-2

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

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.

Answer 19

\[\frac{2}{x}- \frac{2}{x-1} - \frac{2}{x-2} \]
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?

Answer 20

it needs the other half

Answer 21

Which pieces specifically?

Answer 22

the -1 and -2

Answer 23

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

Answer 24

yeah thats what i ment lol

Answer 25

Ok, good. :)

Answer 26

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.

Answer 27

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.

\[\frac{2}{x}\frac{(x-1)(x-2)}{(x-1)(x-2)} \]

Answer 28

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?

Answer 29

yeah so then would i distribute the 2?

Answer 30

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.

Answer 31

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

Answer 32

it needs a (x-2)

Answer 33

What else? It needs one more thing.

Answer 34

the x?

Answer 35

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

Answer 36

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.

Answer 37

yeah thats true

Answer 38

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.

Answer 39

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

Answer 40

it need the x and (x-1)

Answer 41

OUTSTANDING!! XD

Answer 42

So right now - here's what you want to have written down somewhere:

\[\frac{2}{x} \frac{(x-1)(x-2)}{(x-1)(x-2)}- \frac{2}{x-1} \frac{x(x-2)}{x(x-2)} + \frac{2}{x-2} \frac{x(x-1)}{x(x-1)}\]

Answer 43

so would we foil the numerators now?

Answer 44

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.

Answer 45

Wait - forget that!

Answer 46

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

Answer 47

Just foil the tops and leave the bottoms the same.

Answer 48

for the first one i got 4x-6

Answer 49

4x^2-6 i mean

Answer 50

There shouldn't be a 6 there.

Answer 51

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

Answer 52

i dont know what i did wrong then.

Answer 53

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?

Answer 54

2x-2

Answer 55

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

Answer 56

What do you get when you do that?

Answer 57

2x^3 -4x-4

Answer 58

It should be this: