# Can you tell me if i solved this right

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• Jul 8th 2013, 10:54 AM
Urpes
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'.
• Jul 8th 2013, 11:09 AM
ebaines
Re: Can you tell me if i solved this right
You got this far:

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

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

$x^2 -3x -2 = 2$

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.
• Jul 8th 2013, 12:43 PM
Urpes
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) ?
• Jul 8th 2013, 01:02 PM
ebaines
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 $\frac {xx}{yy}$ and $\sqrt {abc}$. Here's a good introductory explanation: mimetextutorial.html
• Jul 8th 2013, 01:35 PM
Urpes
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.' ?
• Jul 8th 2013, 01:48 PM
ebaines
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?
• Jul 8th 2013, 01:57 PM
Urpes
Re: Can you tell me if i solved this right
Quote:

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?
• Jul 8th 2013, 02:03 PM
ebaines
Re: Can you tell me if i solved this right
The original equation is what they gave you to work with:

$\frac x {x-2} + \frac 1 {x-4} = \frac 2 {x^2-6x+8}$
• Jul 9th 2013, 08:13 AM
Urpes
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..?
• Jul 9th 2013, 08:28 AM
ebaines
Re: Can you tell me if i solved this right
Quote:

Originally Posted by Urpes
and I can't divide by zero right..?

Right. So x=4 is NOT a solution to the original equation.
• Jul 9th 2013, 08:42 AM
Urpes
Re: Can you tell me if i solved this right
Quote:

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?
• Jul 9th 2013, 08:54 AM
ebaines
Re: Can you tell me if i solved this right
Quote:

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?
• Jul 9th 2013, 09:10 AM
Urpes
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
• Jul 9th 2013, 09:16 AM
ebaines
Re: Can you tell me if i solved this right
Substituting -1 for x, for the left hand side you have:

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

And for the right hand side:

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

Looks good to me!
• Jul 9th 2013, 09:27 AM
Urpes
Re: Can you tell me if i solved this right
Quote:

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

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

And for the right hand side:

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

Looks good to me!

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

Where did those two come from?
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