`1/(x-2)+2=(3x)/(x+2)` Solve the equation by using the LCD. Check for extraneous solutions.

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`

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