Check this out!

Let $3^x = m$ and $2^{y-1} = n$

Then the equations are

$3m + 4n = 7$

$m+n = 2$

Whose solutios is $m = 1,n=1$

$\Rightarrow 3^x = 1 \Rightarrow x = 0$

and likewise $\Rightarrow 2^{y-1} = 1 \Rightarrow y-1 = 0 \Rightarrow y = 1$

Therefore $x - y = 0 - 1 = \boxed{-1}$

multiply 2nd equation by 3 and do elimination

$3^{x+1} = 3 \times 3^{x}$

$2^{y+1} = 2 \times 2^{y}$

$2^{y-1} = \frac {2^{y}} {2}$

So we have:

[1.0] $3 \times 3^{x} + 2 \times 2^{y} = 7$

[2.0] $3^{x} + \frac {2^{y}} {2} = 2$ that we can simplify multiplying it by 2:

[2.1] $2 \times 3^{x} + 2^{y} = 4$

substracting [1.0] and [2.1] we obtain:

[3.0] $(3 \times 3^{x} - 2 \times 3^{x}) + ( 2 \times 2^{y} - 2^{y}) = 7 - 4 \Rightarrow 3^{x} + 2^{y} = 3$

substracting [2.1] and [3.0] we obtain:

$(2 \times 3^{x} - 3^{x}) + (2^{y} - 2^{y}) = 4 - 3 \Rightarrow 3^{x} = 1 \Rightarrow x = 0$

substituting x = 0 in [2.1] we finally obtain y:

$2 \times 3^{0} + 2^{y} = 4 \Rightarrow 2 + 2^{y} = 4 \Rightarrow 2^{y} = 2 \Rightarrow y = 1$

In fact:

$3^{0+1} + 2^{1+1}= 3 + 4 = 7$

$3^{0} + 2^{1-1}= 1 + 1 = 2$