We start with the following three equations:

1. y+1 =λ (6x-2y-2)

2. x = λ (-2x+6y+6)

3. (x+y+1)^2+2(y-x+1)^2 = 4

We notice that if we add the first two equations we get the following:

x+y+1 = λ (6x-2y-2) + λ (-2x+6y+6) = λ(6x-2y-2-2x+6y+6) = λ(4x+4y+4) = 4λ(x+y+1)

x+y+1 = 4λ(x+y+1)  Therefore, λ = .25

Now you have two equations and two unknowns.

· y+1 =.25(6x-2y-2)

· x = .25 (-2x+6y+6)

We can simplify this to two equations that are basically the same (so we really have one equation and two unknowns):

· 0=  x-y-1

· 0= -x+y+1

We can use the above to simplify equation #3, which reduces to (x+y+1)^2 = 4, or x+y+1 = 2, or x+y-1=0.

So now we have two equations and two unknowns:

· x-y-1=0

· x+y-1=0

If we add these two equations we get 2x – 2 = 0

Therefore, x = 1.

If x = 1, then y = 0.