Angel H.

asked • 11/17/13

How do I solve x+5/x-5-x-5/x+5

X+5 - x-5
x-5    X+5
 

Vivian L.

Hi again Angel;
Please see below for the expansion of the numerator.  It results in 20x, as per the correct answer your provided.  I apologize, but I did not know you needed that too.  Do you also need an explanation of how FOIL is executed?
Please let me know.  I feel terribly about this.
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11/17/13

Angel H.

yes if you can set up the foil step by step for me that would also help
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11/17/13

Vivian L.

Hi again Angel;
The FOIL is set-up below.  I updated my answer.  Thank you for letting me know.  I love this stuff!
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11/17/13

Nataliya D.

Angel, please, post the your questions separately, because there is limit of space, or I have to erase everything, what was posted earlier.
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11/17/13

2 Answers By Expert Tutors

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Nataliya D. answered • 11/17/13

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(a ± b)2 = a2 ± 2ab + b2
a2 - b2 = (a - b)(a + b)
~~~~~~~~~~~~~~~  

 x + 5         x - 5
--------  −  --------  = 
 x - 5          x + 5

LCD of (x - 5) and (x + 5) is (x - 5)(x + 5)

(x + 5)(x + 5)        (x - 5)(x - 5) 
-----------------  −  ----------------  = 
(x - 5)(x + 5)         (x - 5)(x + 5)  

 (x + 5)2  −  (x - 5)2 
------------------------  =
          x2 - 52  

x2 + 10x + 25 − (x2 - 10x + 25)
--------------------------------------- =
                x2 - 25  

x2 + 10x + 25 - x2 + 10x - 25
------------------------------------  =
                 x2 - 25

after combine like terms in numerator:

   20x
-------- =
x2 - 25

       20x
---------------- .
(x + 5)(x - 5)    

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Angel H.

the step by steps are very helpful thanks so much i like being able to understand what I'm doing
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11/17/13

Vivian L. answered • 11/17/13

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Hi Angel;
 
Let's first take...
(x+5)/(x-5)
Let's multiply this by (x+5)/(x+5)
[(x+5)/(x-5)][(x+5)/(x+5)]
This is now...
   (x+5)2   
(x-5)(x+5)
 
Let's next take...
(x-5)/(x+5)
and multiply this by (x-5)/(x-5)
[(x-5)/(x+5)][(x-5)/(x-5)]
This is now...
  (x-5)2    
(x+5)(x-5)
 
The result is...
[(x+5)2]-[(x-5)2]
     (x+5)(x-5)
 
EXPANDED, THE NUMERATOR IS...
[(x+5)2]-[(x-5)2]
(x2+5x+5x+25)-(x2-5x-5x+25)
x2-x2 cancel.
(10x+25)-(-10x+25)
10x+25+10x-25
20x
 
I WOULD BE DELIGHTED TO SET-UP THE FOIL...
[(x+5)2]-[(x-5)2]
As you already know, this is the numerator only.
[(x+5)(x+5)]-[(x-5)(x-5)]
Let's begin with the first bracketed equation...
(x+5)(x+5)
FOIL...
FIRST...(x)(x)=x2
OUTER...(x)(5)=5x
INNER...(5)(x)=5x
LAST...(5)(5)=25
x2+5x+5x+25
x2+10x+25
 
The second bracketed equation...
(x-5)(x-5)
FOIL...
FIRST...(x)(x)=x2
OUTER...(x)(-5)=-5x
INNER...(-5)(x)=-5x
LAST...(-5)(-5)=25
x2-5x-5x+25
x2-10x+25
 
(x2+10x+25)-(x2-10x+25)
x2-x2 cancels
25-25 cancels
10x-(-10x)
As you probably already know, subtracting a negative number is the same thing as adding a positive number...
10x+10x
20x
 
This is the result for the numerator you were looking for.
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Angel H.

I'm sorry but that is not the right answer it is
      20x     
(x+5)(x-5)
i just need the step to know how they got this answer
 
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11/17/13

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