Hi Zach,
I want to change the problem to
^{6}/_{(x2)}= ^{21}/_{(x2)(x+2)} + 1
The denominators here are (x  2) and (x  2)(x + 2) so the least common denominator is (x  2)(x + 2). Thus the problem becomes
^{6(x+2)}/_{(x2)(x+2)}= ^{21}/_{(x2)(x+2)} + ^{(x2)(x+2)}/_{(x2)(x+2)}
Thus, as long as (x  2)(x + 2) is not zero,
6(x + 2) = 21 + (x  2)(x + 2)
Expanding and simplifying I got
x^{2}  6x + 5 = 0
This factors as
(x  1)(x  5) = 0
so x = 1 or x = 5.
If the problem is as you sent,
^{6}/_{x}  2 = ^{21}/_{(x2)(x+2)} + 1
Then the least common denominator is x(x  2)(x + 2) and the equation becomes
^{6(x2)(x+2)}/_{x(x2)(x+2)}  ^{2x(x2)(x+2)}/_{x(x2)(x+2)} = ^{21x}/_{x(x2)(x+2)} + ^{x(x2)(x+2)}/_{x(x2)(x+2)}
If x(x  2)(x + 2) is not zero then
6(x  2)(x + 2)  2x(x  2)(x + 2) = 21 x + x(x  2)(x + 2)
Simplifying this expression gives a cubic that doesn't seem to factor so I think that the intended problem was
^{6}/_{(x2)}= ^{21}/_{(x2)(x+2)} + 1
Penny
