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

- anonymous

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

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- katieb

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- lgbasallote

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

- anonymous

What is the LCD? Let's start there.

- anonymous

Least Common Denominator?

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## More answers

- anonymous

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

- anonymous

Can you tell what it is?

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

Factor them by what? LCD?

- anonymous

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

- anonymous

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.

- anonymous

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.

- anonymous

is it because it is a trinomial?

- anonymous

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

- anonymous

Can you see how it would factor?

- anonymous

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.

- anonymous

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

- anonymous

Happy to help! You ready for the next step?

- anonymous

Yes!

- anonymous

Alright. :)

- anonymous

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

- anonymous

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

- anonymous

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.

- anonymous

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

- anonymous

alright so what do i do with the LCD

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

Hey - even better! :)

- anonymous

so you would cance the (x-2)

- anonymous

Exactly!

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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.

- anonymous

oh yah forgot that second piece :P

- anonymous

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

- anonymous

So from here, what do we do?

- anonymous

Bring the equation together?

- anonymous

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

- anonymous

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

- anonymous

I'm not sure what you did there.

- anonymous

Here's where you want to go with that:

- anonymous

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.
\[x^2-3x-4=0\]

- anonymous

WOW!

- anonymous

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

- anonymous

I think that was the best you could explain mate xD

- anonymous

Excellent!

- anonymous

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?

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

This is GREAT xD

- anonymous

So, what do you get for the answers?

- anonymous

x = 4 and x = -1 ?

- anonymous

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.

- anonymous

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

- anonymous

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

- anonymous

how would i know D: lol

- anonymous

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

- anonymous

am i suppsed to plug it in and find out?

- anonymous

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

- anonymous

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

- anonymous

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?

- anonymous

0

- anonymous

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 :)

- anonymous

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

- anonymous

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.

- anonymous

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

- anonymous

THANKS!! xD

- anonymous

You're welcome!!

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