lets see if there is any genius among the lot?

Question

solve x and y
2^x+y=6^y

3^x-1=2^y+1

Answer 1

a) If you meant 2^(x+y) = 6^y
and 3^(x-1) = 2^(y+1)

The first one is 2^x * 2^y = 2^y * 3^y
So 2^x = 3^y
or x*ln2 = y*ln3 ... (1)

The second one is
3^x / 3 = 2 * 2^y
x*ln3 = ln6 + y*ln2 ... (2)

Solve to get x = 2.7095, y = 1.7095
or x = ln(3) / ln(1.5)
y = ln(2) / ln(1.5)

b) If you meant
2^x + y - 6^y = 0
3^x - 2 - 2^y = 0
Then let f(x,y) and g(x,y) be defined as the above two left hand sides.
We could use a 2 variable Newton's method to solve that.
df = fx * dx + fy* dy
dg = gx*dx + gy*dy

Where fx, fy, gx and gy are partial derivatives
You could write that in matrix form to solve for dx and dy
You'll eventually get
x = 1.135625
y = 0.567560

Answer 2

2^x + y = 6^y
x = 0, y = 0

3^x-1 = 2^y+1 (I'm assuming the ±1 are also in the exponent)
x = 1, y = -1

Answer 3

Take the log of both the sides
then (x+y)*log2 = y*log6...........Log6=log(3*2) = log3 + log2
(x-1)*log3 = (y+1)*log2

x+y = y*(log3 + log2)/log2
x-1 = (y+1) * (log2base3)

x+y = y*(log3base2 + 1)
x-1 = (y+1) * (log2base3)
Let log3base2 = t
then x+y = y*(t + 1)
x-1 = (y+1) * (1/t)
Expressing first equation in terms of y,
x = y*t + y - y = y*t..........substituting in second equ. for x,
y*t = (y+1)*(1/t)
y*t^2 = y+1
y*t^2 - y =1
y(t^2-1) = 1
y = 1/(t^2 -1)
therefore x = y*t = t/(t^2-1).........where t = log3base2