[Guest]

x^2+y=x+x+y

Find x and y.

[Guest]

x^2+y=x+x+y

Find x and y.

that just simplifies to x*x=x+x

x=2, y = and number you want it to it won't make a difference

[Guest]

yay! pretty easy right?

[Guest]

Or...

x=0, y=anything

[Guest]

Or...

x=0, y=anything

anything ^0=1 so that wouldn't quit work. Nice thinking though

[unreality]

reaymond: x is squared not raised to the zero. There are two answers to the quadratic:

x = 0

x = 2

y = anything

[Guest]

reaymond: x is squared not raised to the zero. There are two answers to the quadratic:

My fault, I never really took the puzzle in and thought it was x^x (which in the case of 2 is x^2) Sorry ross!

[Guest]

actually ... wouldn't the answer be:

x^2-2x=0;

(x+2^(1/2))*(x-2^(1/2))=0;

so x=+-2^(1/2)

[Guest]

actually ... wouldn't the answer be:

x^2-2x=0;

(x+2^(1/2))*(x-2^(1/2))=0;

so x=+-2^(1/2)

that is wrong:

(x+2^(1/2))*(x-2^(1/2))=0;

it should be:

(x+(2x)^(1/2))*(x-(2x)^(1/2))=0;

But more simply:

x^2-2x=0

x(x-2)=0

x=0 or x=2

[Guest]

that is wrong:

(x+2^(1/2))*(x-2^(1/2))=0;

it should be:

(x+(2x)^(1/2))*(x-(2x)^(1/2))=0;

But more simply:

x^2-2x=0

x(x-2)=0

x=0 or x=2

x^2+y=x+x+y

- y from both sides gives

x^2=x+x

then trial and error for values of X which leaves 2 and 0

[Guest]

i was originall thinking that

x=2

y=0

but i guess there are other possiblilities