I think I got one. (0,1)

*edit*

got another one (1, 2)

*edit*

um is the other one (2, 3) ?

I really don't know the entire "proper" method.

but I just used simple substitution.

(x^(x+1)) + 1 = (x+1)^x

this worked out to get me to this equation

(x + 1)^x - ((x)^x)x = 1

I originally tried to solve setting equal to zero, but I noticed that if x is 0 then anything in the left side is going to come out to 1 - 0 which satisfies the equation.

Then I was like, what the hell lets see if one works for x as well. Sure enough, it came out to 2 - 1 = 1.

then I was like, ok maybe the next one is two for x.......

and sure enough, 9 - 8 = 1

obviously I just plugged in the x solutions to get the Y values.

Guessing and checking is not how I was taught to solve equations. I just haven't done math in 4 years.................

I used to be an ace at a calculus.... Got an A in the course with out hardly any work.... DX

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