Find the distinct real number X and Y such that X+Y=XY=Xsquared - Ysquared, where X=0 and Y=0?

Question

Real Numbers. I am in calculus but I dont understand Please Help Me!!!

I entered it wrong it should say where X and Y are not equal to 0. Sorry!



My teacher will not help. or explain. Thats why i am asking all of you for help...

Answer 1

I'll give you a start:

x^2 - y^2 can be factored (x + y)(x - y)
so one equation you have is
(x + y)(x - y) = (x + y)
so
(x - y) = 1 or x = y + 1
(you know you can divide by (x + y) since it cannot be
zero. If it was xy would also be zero, but x would also
= -y contradicting that they are not both zero).
Now plug x = y + 1 into your other equation:
x + y = xy
so
1 + y + y = (1 + y)y
then
y^2 - y - 1 = 0
solve that to get y, then x = 1 + y gives you x.

Answer 2

I'll assume you mean't X!=0 and Y!=0.

Since X+Y = XY, we know that X+Y is not zero.

Since X+Y = X^2 - Y^2 = (X-Y)(X+Y)

and X+Y is not zero, we know that:

1 = X-Y

or:

X = Y+1.

Putting this into the equation:

XY = X+Y

and solve the resulting quadratic equation for Y. You should get two values for Y.

Answer 3

u said X=0 & Y=0
0 is a real number u know
otherwise u have 2 equations in 2 unknowns:
X+Y = XY
X^2 - Y^2 = XY (OR X+Y of course)

Answer 4

Is this a trick question? You just said that X=0 and Y=0

Answer 5

Isn't the answer 0?

Answer 6

You're in calculus but you don't understand? Am I missing something? Ask your teacher, that's their job!