Solve 2^x +3^y=17,2^x+2-3^y+1=5

Answer 1

Just by inspection, and assuming integers, 2^3 + 3^2 = 8+9 = 17
But if you want to go through the algebra,
using the fact that 2^2=4 and 3^1=3, we have
2^x + 3^y = 17
4*2^x + 3*3^y = 5
Now, if you let u=2^x and v=3^y, we have
u + v = 17
4u - 3v = 5
and again we have u=8, v=9

so, what are x and y?