solve simultaneously

Solve simultaneously:

\(x+y=2\)

and

\(y+1={-3 \over x-1}\)

 

I did this:

\(x+y=2\)

\(x=2-y.....(1)\)

\(y+1={-3 \over x-1}.....(2)\)

 

(1) into (2)

\(y+1={-3 \over (2-y)-1}\)

\(y+1={-3 \over -2+y}\)

\((y+1)(y-2)=-3\)

\(y^2-y=-1\)

\(y(y-1)=-1\)

\(y=-1\)  or  \(y-1=-1\)

                     \(y=0\)

Then I substituted the y values into (1), getting the x values to be 3 and 2 respectively.

 

Then another question follows:

Hence, or otherwise, determine the value of:

\(({1 \over x}+{1 \over y})\)

 

The teacher says the answer should be \(-{1 \over4}\),  I do not get that.

 

where did I go wrong?...please help..Thank you all very much indeed.