Question

(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.

Answer 1

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 <i>need to insert parentheses around all numerators and denominators that contain more than one term</i>.

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.