SOLUTION: <pre>Hi, can you help me solve 3x-2 2 ————— - ————— x+2 x-2</pre>

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Question 23358: 
Hi, can you help me solve

3x-2      2
————— - —————
 x+2     x-2
Found 2 solutions by rapaljer, AnlytcPhil:
Answer by rapaljer(4671) About Me  (Show Source):
You can put this solution on YOUR website!
%28%283x-2%29%2F%28x%2B2%29-2%2F%28x-2%29%29

Because this is a subtraction problem, you must first find a common denominator, and in this case the LCD = (x+2)(x-2)

You'll have to multiply the first fraction by %28x-2%29%2F%28x-2%29 and the second fraction by %28x%2B2%29%2F%28x%2B2%29.

It should look like this:
%28%283x-2%29%2F%28x%2B2%29%29%2A%28%28x-2%29%2F%28x-2%29%29 -%282%2F%28x-2%29%29%2A%28%28x%2B2%29%2F%28x%2B2%29%29

Place this all over the common denominator and multiply out the binomials in the numerators. HOWEVER, be CAREFUL to keep the negative between the numerators.
%28%283x-2%29%2A%28x-2%29+-+2%2A%28x%2B2%29+%29%2F%28%28x%2B2%29%2A%28x-2%29%29
%283x%5E2+-8x+%2B4+-2x+-4+%29%2F%28%28x%2B2%29%2A%28x-2%29%29
%283x%5E2+-10x+%29%2F%28%28x%2B2%29%2A%28x-2%29%29

This can be factored, but it does NOT reduce the fraction, so factoring is optional:
%28x%2A%283x+-10%29+%29%2F%28%28x%2B2%29%2A%28x-2%29%29

R^2 at SCC


Answer by AnlytcPhil(1806) About Me  (Show Source):
You can put this solution on YOUR website!
Hi, can you help me solve

3x-2      2
————— - —————
 x+2     x-2

We want to cause both those denominators to become the LCD, which is
(x+2)(x-2).  We use the principle:

If you multiply the numerator and denominator of a fraction by the
same thing you don't change the value of the fraction. You only change
its looks. 

The first fraction has denominator x+2.  It needs to become the LCD, 
which is (x+2)(x-2), so it needs to be multiplied by (x-2). 

I can do this without changing the fraction's value if I'll also 
multiply the numerator by (x-2).

So the first fraction:

3x-2     
————— 
 x+2 

becomes 

(3x-2)(x-2)    
——————————— 
 (x+2)(x-2)

FOIL the top out but DON'T FOIL the bottom!

 3x²-8x+4
——————————
(x+2)(x-2) 

Now the second fraction has denominator x-2.  It needs to become 
the LCD, which is (x+2)(x-2), so it needs to be multiplied by (x+2). 

I can do this without changing the fraction's value if I'll also 
multiply the numerator by (x+2) also.

Now the second fraction:

  2     
————— 
 x-2 

becomes 

  2(x+2)    
——————————— 
 (x-2)(x+2)

Multiply the top out but not the bottom!

   2x+4
——————————
(x-2)(x+2)

Now back to the original problem:

3x-2      2
————— - —————
 x+2     x-2

becomes

 3x²-8x+4       2x+4
—————————— - ——————————
(x+2)(x-2)   (x-2)(x+2)
 
Now the denominators are equal.

So we subtract the numerators, placed first in parentheses,
then write that over the common denominator.


(3x²-8x+4) - (2x+4)
———————————————————
    (x+2)(x-2)

Now remove the parentheses on top and collect terms. Leave
the bottom as it is:

  3x²-8x+4-2x-4
———————————————————
    (x+2)(x-2)

     3x²-10x
———————————————————
    (x+2)(x-2)

That's good enough.  You can leave it like that.  But if you
like you can factor out x on the top. 

    x(3x-10)
—————————————————
    (x+2)(x-2) 

In some problems, doing that produces something that will cancel,
but that didn't happen this time. For instance, if that (3x-10)
had been (x+2) or (x-2) instead it would have canceled. But it
wasn't so it didn't. 

Edwin
AnlytcPhil@aol.com