Last one (I hope)

krisolaw

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Jun 5, 2005
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45
[(2)+ 3/(x + 1)]/(1/x) + 1

Perhaps I should have written it this way:
(2) + 3/(x + 1)/(all over) (1/x) + 1 or
numerator: 2 + 3/(x+1) denominator: (1/x) + 1

Does that clarify it?

I get the LCD of x(x+1) but I am not sure that is right. Could someone please show me, I am lost on this one. Thanks again!
 
G

Guest

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krisolaw said:
[(2 + 3/(x + 1)]/(1/x) + 1
There's something wrong with your brackets: you have four left brackets and only three right brackets.

What are you trying to do? "Simplify"?
 

krisolaw

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Jun 5, 2005
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Yes, we are suppose to simplify each complex fraction. I have been working on my mid-term study guide all day and this is the last problem that I can't seem to solve correctly. I have the LCD as x(x+1) and when I plug it in I can get the numerator to come out correctly according to the book, but no matter how I plug it in the denominator, I can't get the same answer. I have walked away from this problem several times now, but seeing as how I have completed the other 70 or so problems on my study guide, this is the last one I am having difficulties with. Please explain this one to me. Thanks in advance.

The book says the answer is (2x^2 + 5x)/(x^3 + x^2 + x + 1)
 

Denis

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Feb 17, 2004
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krisolaw said:
Yes, we are suppose to simplify each complex fraction. I have been working on my mid-term study guide all day and this is the last problem that I can't seem to solve correctly. I have the LCD as x(x+1) and when I plug it in I can get the numerator to come out correctly according to the book, but no matter how I plug it in the denominator, I can't get the same answer. I have walked away from this problem several times now, but seeing as how I have completed the other 70 or so problems on my study guide, this is the last one I am having difficulties with. Please explain this one to me. Thanks in advance.

The book says the answer is (2x^2 + 5x)/(x^3 + x^2 + x + 1)
Agree with numerator; denominator definitely wrong: bad book!

2 + 3/(x+1) = (2x+5) / (x+1)

1/x + 1 = (x+1) / x

[(2x+5) / (x+1)] / [(x+1) / x]
= x(2x+5) / [(x+1)(x+1)]
= (2x^2 + 5x) / (x + 1)^2
 

Gene

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Oct 8, 2003
Messages
1,904
The edited description helps. You could have typed it as
[2 + 3/(x + 1)]/[1/x + 1]
or hit the C(ode) button twice and typed between the boxes as
Code:
   2 + 3/(x + 1)
   --------------
    1/x + 1
anyhow start with

2+3/(x+1) = (2(x+1)+3)/(x+1) and
1/x+1 = (x+1)/x
The (x+1)/x inverts to the numerator as a multiplier to get
x(2x+2+3)/(x+1)^2 =
(2x^2+5x)/(x+1)^2

I don't see any more to do with it and I don't agree with the books answer. Something is wrong. If you try x=2 the original gives 2 but the book answer gives 1.2 but whatever you plug in for x you should get the same thing. That's a way of checking the validity of your answers. Try a few different numbers though. The book agrees if x=1.

BTW if you look under
math lessons
at Order of Operations
you will see why your going astray in youe typing.

PS. I had this typed before I saw Denis' post.
 

tkhunny

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Apr 12, 2005
Messages
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The "Code" button?
Code:
 Numerator
-----------
Denominator
I guess I should have read that sooner. Thanks for the upgrade.
 

soroban

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Jan 28, 2005
Messages
5,588
Hello, krisolaw!

Denis and Gene are correct . . . the book is wrong.

. . . . . .3
. 2 + ------
. . . . x + 1
--------------
. . .1
. . --- + 1
. . .x
When I get a "complex fraction" (one with more than two 'levels'),
. . I multiply top and bottom by the LCD.
. . This immediately eliminates the 'complex' part.


. . . . . . . . . . . . . 3
. x(x + 1) . 2 + ------
. . . . . . . . . . . . x + 1 . . . . . . . . 2x(x + 1) + 3x
. . . . . . . . -------------- . . = . . ---------------------
. . . . . . . . . . 1 . . . . . . . . . . . . x + 1 + x(x + 1)
. x(x + 1) . . --- + 1
. . . . . . . . . . x


. . . . . . . 2x<sup>2</sup> + 5x . . . . . x(2x + 5)
. . . = . --------------- . = . ------------
. . . . . . x<sup>2</sup> + 2x + 1 . . . . .(x + 1)<sup>2</sup>


I've always learned/invented the Easy Way for everything.
. . . . . . My philosophy: no pain . . . no pain.
 
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