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)   (Show Source): You can put this solution on YOUR website!


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 and the second fraction by .

It should look like this:
-

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.




This can be factored, but it does NOT reduce the fraction, so factoring is optional:


R^2 at SCC


Answer by AnlytcPhil(1806)   (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
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