Solve for x in the following two equations. Please show your work.

a not equal to 0  and

1)      2^x+y=0

         y= -2^x

2)    2^x-y=0

        y= 2^x

Am I not understanding the problem ?

There are no simultaneous solutions.

Solve for x:
(2^x+y)/(2^x-y) = 0

Simplify and substitute z = 1/(2^x-y):
(2^x+y)/(2^x-y)  =  1+(2 y)/(2^x-y)  =  1+2 y z  =  0:
1+2 y z = 0

Subtract 1 from both sides:
2 y z = -1

Divide both sides by 2 y:
z = -1/(2 y)

Substitute back for z = 1/(2^x-y):
1/(2^x-y) = -1/(2 y)

Substitute u = 2^x-y:
1/u = -1/(2 y)

Take the reciprocal of both sides:
u = -2 y

Substitute back for u = 2^x-y:
2^x-y = -2 y

Add y to both sides:
2^x = -y

Take the logarithm base 2 of both sides:
Answer: | x = (log(-y))/(log(2))+((2 i) pi n)/(log(2))  for  n element Z