Question

Solve: (x)/(x-2) + (1)/(x-4) = (2)/(x^2 - 6x + 8)

Answer 1

multiply with LCM

this time it's (x-2)(x-4) that is equal to x^2 -6x +8

x(x-4) + x-2 = 2
x^2 - 4x + x -2 -2 =0
x^2 - 3x -4 =0
(x-4)(x+1) = 0
x = 4
x=-1

Answer 2

What is the LCD? Let's start there.

Answer 3

Least Common Denominator?

Answer 4

That's right - first we want to identify the Least Common Denominator.

Answer 5

Can you tell what it is?

Answer 6

Well I don't really know how to tell D:

Answer 7

Alright - I'll explain how to find it. :)

Answer 8

When you are asked to solve rational equations like this, the first thing you want to do is factor everything.

Answer 9

Factor them by what? LCD?

Answer 10

Factor each polynomial, all the numerators and denominators. I'll show you:

Answer 11

This is what you were given:
\[\frac{x}{x-2} + \frac{1}{x-4} = \frac{2}{x^2 - 6x + 8} \]Now, first identify 3 fractions. Each fraction has a numerator and denominator, but only one of these pieces can be factored. The denominator on the right-most fraction.

Answer 12

To factor it, we're looking specifically at this:
\[x^2-6x+8\]Here, you want to ask yourself, what two numbers multiply to +8 and add to -6.

Answer 13

is it because it is a trinomial?

Answer 14

That's right! :)
Thanks for all the medals, by the way.

Answer 15

Can you see how it would factor?

Answer 16

It would factor like this:
We're looking for numbers that multiply to 8 and add to -6. So the numbers we are looking for are -4 and -2. So this trinomial would factor like this:
(x-4)(x-2)

And you can always FOIL those back out to check to see that you get what you started with.

Answer 17

I was eating late dinner sorry lol and your welcome. thanks for the tutor! (:

Answer 18

Happy to help! You ready for the next step?

Answer 19

Yes!

Answer 20

Alright. :)

Answer 21

you make it easier than my online class really is :P

Answer 22

So we can rewrite the original question in factored form, like this:\[\frac{x}{x-2} + \frac{1}{x-4} = \frac{2}{(x-4)(x-2)} \]

Answer 23

The reason we do this is because we are eventually going to multiply both sides of this by the LCD in order to completely clear the fractions. But, we have to know what factors all the pieces are made from in order to do that first.

Answer 24

So at this point we can see that the LCD is: (x-4)(x-2)

Answer 25

alright so what do i do with the LCD

Answer 26

From here, you're going to multiply each fraction by the LCD over 1 - like this:

Answer 27

\[\frac{(x-4)(x-2)}{1}\frac{x}{x-2} + \frac{(x-4)(x-2)}{1}\frac{1}{x-4} = \frac{(x-4)(x-2)}{1}\frac{2}{(x-4)(x-2)} \]

Answer 28

It's a long process, but now comes the good part - you get to cancel.

Answer 29

So, looking at the first piece specifically, what is going to cancel?
\[\frac{(x-4)(x-2)}{1}\frac{x}{x-2}\]

Answer 30

I cancelled in my head xD thats my favorite part of this once i ACTUALLY get to this part without making a mistake :P

Answer 31

Hey - even better! :)

Answer 32

so you would cance the (x-2)

Answer 33

Exactly!

Answer 34

then your left with (x-4)x/1 right?

Answer 35

Yes, although because the denominator is 1, we don't need to write it anymore.

Answer 36

So then because you were able to cancel in your head, you see how we're left with:
\[x(x-4)+1(x-2)=2\]

Answer 37

The canceling is the payoff for picking the right LCD. If you've done your job right, at this point, there should be no fractions left.

Answer 38

oh yah forgot that second piece :P

Answer 39

Oh, be sure to remember all of them. We have 3 pieces to this equation. :)

Answer 40

So from here, what do we do?

Answer 41

Bring the equation together?

Answer 42

That's right - distribute, combine like terms... What will that leave you with?

Answer 43

(x-4)((x-2)(x-2)

Answer 44

I'm not sure what you did there.

Answer 45

Here's where you want to go with that:

Answer 46

First, use the distributive properto to get rid of those parenthesis:\[x(x-4)+1(x-2)=2\]
Then combine the -4x with the +x\[x^2-4x+x-2=2\]
Now, recognize that this thing is going to end up quadratic, so we want to set it equal to zero.
\[x^2-3x-2=2\]You can subtract 2 from both sides to do that.
\[x^2-3x-4=0\]

Answer 47

WOW!

Answer 48

Do you need me to explain any of those steps a little better?

Answer 49

I think that was the best you could explain mate xD

Answer 50

Excellent!

Answer 51

Alright, so we're left looking at this:
\[x^2-3x-4=0\]
Now you want to factor that. What does it leave you with?

Answer 52

(x-4)(x+3)
x=4
x= -3
?

Answer 53

Well, it can't be (x-4)(x-3) because if you tried to FOIL that out, you'd get:
\[x^2-7x+12\]

Answer 54

We're looking for numbers that multiply to -4 and add to -3

Answer 55

Oops I meant (x-4)(x+1)

Answer 56

That's right! That's perfect! :)

Answer 57

So we're left with:\[(x-4)(x+1)=0\]

Answer 58

Now we can use the Zero-Product Property:
\[x-4=0\]&\[x+1=0\]

Answer 59

This is GREAT xD

Answer 60

So, what do you get for the answers?

Answer 61

x = 4 and x = -1 ?

Answer 62

There's a little trick to this one. Because these are rational equations, we have to check to see if those are good answers. According to the equation, those are our answers, but they may turn out to be bad x's. We have to plug them in to the original equation and make sure they don't cause division by zero.

Answer 63

Only one of those answers is correct. The other is a bad x.

Answer 64

So, my question for you, which is the right one, and which is the bad one?

Answer 65

how would i know D: lol

Answer 66

Well, only one of them will cause division by zero with our original equation.

Answer 67

am i suppsed to plug it in and find out?

Answer 68

You could plug them in and find out that way, or you can just eyeball it. Have a look:

When we factored the original equation we were left with this:\[\frac{x}{x-2}+\frac{1}{x-4}=\frac{2}{(x-4)(x-2)}\]

Answer 69

Our two solutions are:
x=4 and x=-1

Answer 70

Now, the -1 won't cause any problems. But that 4 will because look at the second fraction. If we plug a 4 in for x, what does the bottom become?

Answer 71

0

Answer 72

Exactly - and we can't ever divide by 0. So that makes x=4 a "bad x". Because x=4 causes division by 0 in our original equation, we have to throw it out. So this question only has 1 answer.

x=-1 :)

Answer 73

So you always want to check your answers for these to make sure they're not bad x's.

Answer 74

Sometimes there are no bad x's, sometimes (like this time) there is 1 bad x, and other times, both solutions are bad x's.

Answer 75

And that's how they're done! :) I hope that helped!!

Answer 76

THANKS!! xD

Answer 77

You're welcome!!