Given that x and y are integers, then for the equation x+y= 2xy how many solutions are there?
2 x-1!=0, y = x/(2 x-1)
!= means doesn't equal.
so, since x and y are integers,
clearly x=y=0 is a solution
If x>0, 2x-1 > x unless x=1, so x=y=1 is a solution
For x>1, x/(2x-1) is a fraction.
For x<0, x/(2x-1) is always a fraction, so
(0,0) and (1,1) are the only integer solutions.
To determine the number of solutions for the equation x + y = 2xy, we need to analyze the equation further.
Starting with the equation x + y = 2xy, we can rearrange it as follows:
x + y - 2xy = 0
Rearranging it even further, we can factor out a common term:
x(1 - 2y) + y = 0
Now, we can solve for x in terms of y:
x = -y / (1 - 2y)
To find the number of solutions, we need to consider the possible values of y that will yield integer values for x.
At this point, we can observe that y = 0 is not a valid solution since it would result in division by zero.
Now, let's consider two cases separately:
Case 1: When y ≠ 1/2:
In this case, 1 - 2y ≠ 0. Therefore, the value of x will be a fraction or an integer depending on the value of y, which means there are infinitely many solutions.
Case 2: When y = 1/2:
If y = 1/2, then 1 - 2y = 0, and the equation becomes 0x + y = 0. In this case, x can be any integer, resulting in infinitely many solutions.
In conclusion, there are infinitely many solutions for the equation x + y = 2xy.