Posted by **kenneth** on Wednesday, August 25, 2010 at 5:47pm.

(2/x^2-4)+(3/x^2)+(x-6+1/x+3)

someone please show me how to work this problem with the steps, so I can see if I can get it. This is due tomorrow in class. Thank you.

- high school algerbra class -
**MathMate**, Wednesday, August 25, 2010 at 7:20pm
See comments at:

http://www.jiskha.com/display.cgi?id=1282707251

The parentheses that have been added do not help, because they are implicit with the rules of priorities of multiplication/division over addition/subtraction.

If the original problem is a fraction with a denominator of 2 and x²-4 as the denominator, you *need to insert parentheses around all numerators and denominators that contain more than one term*.

2/x²-4 = (2/x²) - 4

while

2/(x²-4) = 2/((x+2)(x-2))

Your expression as is translates to:

(2/x^2-4)+(3/x^2)+(x-6+1/x+3)

=(2/x²) - 3 + (3/x²) + x - 6 + (1/x) + 3

which I doubt is the same that was printed in the book.

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