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

X+5 - x-5
x-5    X+5

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?
yes if you can set up the foil step by step for me that would also help
Hi again Angel;
The FOIL is set-up below.  I updated my answer.  Thank you for letting me know.  I love this stuff!
Angel, please, post the your questions separately, because there is limit of space, or I have to erase everything, what was posted earlier.

### 2 Answers by Expert Tutors

Nataliya D. | Patient and effective tutor for your most difficult subject.Patient and effective tutor for your mos...
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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)

the step by steps are very helpful thanks so much i like being able to understand what I'm doing
Vivian L. | Microsoft Word/Excel/Outlook, essay composition, math; I LOVE TO TEACHMicrosoft Word/Excel/Outlook, essay comp...
3.0 3.0 (1 lesson ratings) (1)
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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.