please help: getting lost somewhere in complex fractions

layer3d

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Apr 17, 2008
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I am working on a stand-alone algebra book here and I can't seem to find the solution on the final answer provided in the back of the book. Onto the problem,

simplify the following equation:
(x-1)/[(x-5)(x-1)]

my solution:
(x-1)/[x-5)(x-1)]

= (x-1)
-------------
x^2-x-5x+5
= (x-1)
-----------
x^2-4x+5

this is the part where I get lost after that unfinished sequence, and if I invert them into multiplication, it would be something like this: (x-1) * (x^2-4x+5), which I believe wouldn't be the correct next step since the answer to that is nothing like what the book provided. Can anyone please help me determine the right solution? I would very much appreciate...
 
(x-1)/[(x-5)(x-1)] means...

\(\displaystyle \frac{x-1}{(x-5)(x-1)}\)

That is not an equation. It is an expression. You can't solve an expression. You might turn it into an equation if you set it equal to something, possibly zero. At any rate, you should be able to simplify the fraction by dividing both numerator and denominator by x-1, with the understanding that x cannot equal zero. If you do that, you would end up with the fraction (expression) 1/(x-5).

(x-1)
-----------
x^2-4x+5

You mention that if you invert it becomes whatever. You must realize that what you have here is really
\(\displaystyle \frac{\frac{x-1}{1}}{\frac{x^2-4x+5}{1}}\)

which when converted to multiplication by "inverting" becomes..

\(\displaystyle \frac{x-1}{1}\cdot \frac{1}{x^2-4x+5}\)

which puts you right back where you started.
 
layer3d said:
simplify the following equation:
(x-1)/[(x-5)(x-1)]

my solution:
(x-1)/[x-5)(x-1)]

= (x-1)
-------------
x^2-x-5x+5
= (x-1)
-----------
x^2-4x+5

this is the part where I get lost....

How would you simplify:

\(\displaystyle \frac{2}{3\cdot 2}\)

you would eliminate the common factor (2) - and get:

\(\displaystyle =\, \frac{1}{3}\)

Correct???

Follow the exact same method. To help you visualize, put a parenthesis around the numerator.
 
Thanks so much for the reply guys, I really appreciate it. This book I am working on explains things in a story-style explanation instead of a direct-approach manner. I'm thinking of looking for another Algebra book that explains things in a clearer method.

Loren, your solution is the one I'm looking for to get the right answer. In that case, could you please explain to me why, x-1, should be divided on both the numerator and denominator? Is it because it's the LCM or something? If pinpointing out the LCM then I should be fine from here on and work on the other expressions in this same manner.

edit: ok, I've written the missing step here.

= (x-1)/(x-1)
------------------
(x^2-4x+5/x-1)

= 1
------
x-4-5

Although that is the wrong answer, I'm wondering what happened to the -4x/x part? Wouldn't that give you a -4 and should be included? Please explain.. :(
 
Subhotosh Khan said:
How would you simplify:

\(\displaystyle \frac{2}{3\cdot 2}\)

you would eliminate the common factor (2) - and get:

\(\displaystyle =\, \frac{1}{3}\)

Correct???

Follow the exact same method. To help you visualize, put a parenthesis around the numerator.

Thanks for the reply there but if I work on it that way it will give me a different answer from the one the book says heheh.
 
layer3d said:
(x-1)/[(x-5)(x-1)]

my solution:
(x-1)/[x-5)(x-1)]

common factor in the numerator and the denominator is (x-1)

so if you "cancel" those out (like I showed you with numerical example) - you get

\(\displaystyle \frac{1}{x-5}\)

If your book gives you different answer - then you would need to talk to your teacher.
 
simplify the following equation:
(x-1)/[(x-5)(x-1)]

\(\displaystyle \frac{x-1}{(x-5)(x-1)}=\frac{1\cdot (x-1)}{(x-5)(x-1)}=\frac{1}{x-5}\cdot \frac{x-1}{x-1}= \frac{1}{x-5}\cdot 1=\frac{1}{x-5}\)
 
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