X+5 - x-5

x-5 X+5

X+5 - x-5

x-5 X+5

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Hallandale, FL

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)

------------------------ =

x

x

--------------------------------------- =

x

x

------------------------------------ =

x

after combine like terms in numerator:

20x

-------- =

x

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

Middletown, CT

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}]

(x^{2}+5x+5x+25)-(x^{2}-5x-5x+25)

x^{2}-x^{2} 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)]

[(x+5)

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)=x^{2}

OUTER...(x)(5)=5x

INNER...(5)(x)=5x

LAST...(5)(5)=25

x^{2}+5x+5x+25

x^{2}+10x+25

The second bracketed equation...

(x-5)(x-5)

FOIL...

FIRST...(x)(x)=x^{2}

OUTER...(x)(-5)=-5x

INNER...(-5)(x)=-5x

LAST...(-5)(-5)=25

x^{2}-5x-5x+25

x^{2}-10x+25

(x^{2}+10x+25)-(x^{2}-10x+25)

x^{2}-x^{2} 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.

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

Gilant P.

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