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

Multiply throughout by `x(x+2)` to remove the denominators (bottoms) of the fractions:

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

Cancelling gives:

`2x(x+2)-(x+2)=3x`

Expanding the brackets:

`2x^2+4x-x-2=3x`

`2x^2-2=0`

`x^2-1=0`

Factoring gives:

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

So `x = -1` or `x = 1`.

CHECK: Substituting `x = -1` into both the left hand side and right hand side of the question gives:

`"LHS"=2-1/x=2-1/-1=3`

`"RHS"=3/(x+2)=3/(-1+2)=3="LHS" `

Likewise, for `x = +1`,

LHS `= 2 - 1 = 1`

RHS `= 3/3 = 1 =` LHS