# Thread: Can you tell me if i solved this right

1. ## Can you tell me if i solved this right

Got one more question; is this correct? can it be simplified further

I just uploaded a picture quick because im not so good at writing equations outside of word

http://img580.imageshack.us/img580/9407/b3ku.png

The first line is the question. It justs says ' Solve'.

2. ## Re: Can you tell me if i solved this right

You got this far:

$\displaystyle \frac {x^2-3x-2}{(x-2)(x-4)} = \frac {2}{(x-2)(x-4)}$

Since the denminators are the same, you can simply multiply both sides by (x-2)(x-4) to get

$\displaystyle x^2 -3x -2 = 2$

Now rearrange and factor to solve for x. One thing to be aware of however: note that because you had factors (x-2) and (x-4) in the denominators to start with, x cannot be equal to either 2 or 4.

3. ## Re: Can you tell me if i solved this right

Thank you for bearing with me and my terrible looking equations. If I may ask, what are you using to make your equations so clean?

x2 -3x -2 -2 = 2 -2

x2- 3x - 4 = 0

(x-4) (x+1) ?

4. ## Re: Can you tell me if i solved this right

OK, so far so good. So what values for x do you get?

As for equation formatting we use a tool called "LaTeX." It uses commands between start and end delimiters of [ tex] and [/tex]. Inside those delimiters the symbol "^" specifies "raised to the value of" and functions with a leading backslash and arguments within pairs of {} squiggly brackets such as \frac {xx}{yy} and \sqrt{abc} yield $\displaystyle \frac {xx}{yy}$ and $\displaystyle \sqrt {abc}$. Here's a good introductory explanation: mimetextutorial.html

5. ## Re: Can you tell me if i solved this right

Ooh that will come in handy thank you.

I'm not sure how to go about finding the values for x is there some kind of equation I should be following to get the answer?

Is it: 'The values you get for x are: x= 4. -1.' ?

6. ## Re: Can you tell me if i solved this right

Don't forget what I said about x = 4 back in post #2! As a matter of course you should always double-check your math, in this case by verifying that if x = -1 and x = 4 the original equation is satisfied. So - what happens if you substitute x = 4 into the original equation?

7. ## Re: Can you tell me if i solved this right

Originally Posted by ebaines
Don't forget what I said about x = 4 back in post #2! As a matter of course you should always double-check your math, in this case by verifying that if x = -1 and x = 4 the original equation is satisfied. So - what happens if you substitute x = 4 into the original equation?
Yes yes thats right my mistake. By original equation do you mean the very first one or x2-3x-4?

8. ## Re: Can you tell me if i solved this right

The original equation is what they gave you to work with:

$\displaystyle \frac x {x-2} + \frac 1 {x-4} = \frac 2 {x^2-6x+8}$

9. ## Re: Can you tell me if i solved this right

Im pretty sure I made a mistake

but

if I sub the x=4 in the original equation I'm going to get

1
__
(4) -4

and I can't divide by zero right..?

10. ## Re: Can you tell me if i solved this right

Originally Posted by Urpes
and I can't divide by zero right..?
Right. So x=4 is NOT a solution to the original equation.

11. ## Re: Can you tell me if i solved this right

Originally Posted by ebaines
Right. So x=4 is NOT a solution to the original equation.
Now I'm completely lost so I can't use 2, or 4. What do I do to solve it then? Sub in -1?

12. ## Re: Can you tell me if i solved this right

Originally Posted by Urpes
Now I'm completely lost so I can't use 2, or 4. What do I do to solve it then? Sub in -1?
Back in post#5 you came up with two possible answers: 4 and -1. You have now ruled out 4 as a possibility. That leaves -1 as a possible answer. So - does it work in the original equation?

13. ## Re: Can you tell me if i solved this right

Not when I put it in it doesn't.

(-1) 1 2
____ + ____ = ______
(-1)-2 (-1)-4 (-1)2-6x+8

-1 1 2
___ + ___ = __
-3 -5 15

14. ## Re: Can you tell me if i solved this right

Substituting -1 for x, for the left hand side you have:

$\displaystyle \frac {-1}{-3} + \frac 1 {-5} = \frac 5 {15} + \frac {-3}{15} = \frac 2 {15}$

And for the right hand side:

$\displaystyle \frac 2 {(-1)^2 - 6(-1) + 8} = \frac 2 {1 + 6 + 8} = \frac 2 {15}$

Looks good to me!

15. ## Re: Can you tell me if i solved this right

Originally Posted by ebaines
Substituting -1 for x, for the left hand side you have:

$\displaystyle \frac {-1}{-3} + \frac 1 {-5} = \frac 5 {15} + \frac {-3}{15} = \frac 2 {15}$

And for the right hand side:

$\displaystyle \frac 2 {(-1)^2 - 6(-1) + 8} = \frac 2 {1 + 6 + 8} = \frac 2 {15}$

Looks good to me!
All of that makes sense to me except for the $\displaystyle \frac 5 {15} + \frac {-3}{15}$ part

Where did those two come from?

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