**LCD** is an acronym for **least common denominator.** It is the product of distinct factors on the denominator side. Basically, find LCD is the same as finding the LCM (least common multiple) of the denominators.

For the given equation `1/(x-2)+2=(3x)/(x+2)` , the denominators are `(x-2)` and `(x+2)` . Both are distinct factors.

Thus,

`LCD = (x-2)(x+2) or (x^2-4)`

To simplify the equation,we multiply each term by the LCD.

`1/(x-2)*(x-2)(x+2)+2*(x-2)(x+2)=(3x)/(x+2)*(x-2)(x+2)`

Cancel out common factors to get rid of the factor form.

`1*(x+2)+2*(x^2-4)=(3x) *(x-2)`

Apply distribution property.

`x+2+2x^2-8= 3x^2-6x`

Combine like terms.

`2x^2+x-8+2= 3x^2-6x`

`2x^2+x-6= 3x^2-6x`

Subtract `2x^2` from both sides of the equation.

`2x^2+x-6-2x^2= 3x^2-6x-2x^2`

`x-6= x^2-6x`

Subtract `x` from both sides of the equation.

`x-6-x= x^2-6x-x`

`-6= x^2-7x`

Add `6` on both sides of the equation.

`-6+6= x^2-7x+6`

`0 = x^2-7x+6`

Apply factoring on the trinomial: `x^2-7x+6` .

`0 = (x-1)(x-6)`

Apply zero-factor property to solve for `x` by equating each factor to 0.

`x-1=0`

`x-1+1=0+1`

`x=1`

and

`x-6=0`

`x-6+6=0+6`

`x=6`

To check for extraneous solution, plug-in each `x ` on `1/(x-2)+2=(3x)/(x+2)` .

Let `x=1` on `1/(x-2)+2=(3x)/(x+2).`

`1/(1-2)+2=?(3*1)/(1+2)`

`1/(-1)+2=?3/3`

`-1+2 =?1`

`1=1` **TRUE**

Let `x=6` on `1/(x-2)+2=(3x)/(x+2)` .

`1/(6-2)+2=?(3*6)/(6+2)`

`1/4+2=?18/8`

`1/4+8/4=?9/4`

`9/4=9/4` **TRUE**

Therefore, there are *no extraneous solutions*.

**Real exact solutions** of the given equation `1/(x-2)+2=(3x)/(x+2)` :

`x=1` and `x=6`