Is there any elementary way (or using Lambert-W maybe) to solve this system of the exponential equation: $$ \begin{cases} 3^{x+y}+2^{y-1}=23, \\ 3^{2x-1}+2^{y+1}=43. \end{cases} $$
I have tried to eliminate the exponent of 2 but it gets me $$ 12 \cdot 3^{x + y} + 3^{2x} = 405 $$ which is more complicated.
I have also tried to substitute $ 3^x = u $ and $ 2^y = v $ but there is still $ 3^y $.
Any advice is welcome (it's okay to use non-elementary method). Thanks :)
On
Hint:
$$(3^x)^2+(12\cdot3^y)3^x-405=0$$
The discriminant is $$(12\cdot3^y)^2+4\cdot405=16\cdot3^{2y+2}+3^4\cdot20=4\cdot3^2(4\cdot9^y+45)$$
For rational $3^x,$ we need $$(2\cdot3^y)^2+45$$ to be perfect square $=d^2, d\ge0$(say)
$$\implies45=d^2-(2\cdot3^y)^2=(d+2\cdot3^y)(d-2\cdot3^y)\le(d+2\cdot3^y)^2$$
$$\implies d+2\cdot3^y\ge\sqrt{45}>6$$
Again, $d+2\cdot3^y$ must divide $45,$ hence can be one of $$\{9,15,45\}$$
From here we can find $3^y$ and $d$ and hence $3^x$
It turns out that my Professor is making a mistake. The correct system of equations is $$ \begin{cases} 3^{x + y} + 2^{y - 1} = 239, \\ 3^{2x - 1} + 2^{y + 1} = 43. \end{cases} $$
Answer: We know that $ 239 = 243 - 4 = 3^5 - 2^2 $ and $ 43 = 27 + 16 = 3^3 + 2^4 $, so we have new system of equations that satisfy $$ \begin{cases} x + y = 5, \\ y - 1 = 2, \\ 2x - 1 = 3, \\ y + 1 = 4. \end{cases} $$ Which only satisfied for $ (x, y) = (2, 3) $.
And also, thanks for your answer. I really appreciate it.