Question: I can solve easy problems such as (x/2)+3=2+(3x/4). That is easy because the Lowest Common Denominator is 4. But what really gets me stuck is a problem like this one.
(6/x-2) = ( 21/(x-2)(x+2) )+ 1.

Please help me understand this!!!!!

Hi Zach,

I want to change the problem to

 ⁶/_(x-2)= ²¹/_(x-2)(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)/_(x-2)(x+2)= ²¹/_(x-2)(x+2) + ^(x-2)(x+2)/_(x-2)(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

x² - 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/_(x-2)(x+2) + 1

Then the least common denominator is x(x - 2)(x + 2) and the equation becomes

^6(x-2)(x+2)/_x(x-2)(x+2) - ^2x(x-2)(x+2)/_x(x-2)(x+2) = ^21x/_x(x-2)(x+2) + ^x(x-2)(x+2)/_x(x-2)(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/_(x-2)= ²¹/_(x-2)(x+2) + 1

Penny