Solve the simultaneous equations 2^x−y = 8 and 23x−y = 128.

The first equation is $2^{x-y} = 8$.
Since $8 = 2^3$, we can rewrite the equation as:

$$2^{x-y} = 2^3$$

Equating the exponents gives our first linear equation:

$$x - y = 3 \quad (1)$$

The second equation is $2^{3x-y} = 128$.
Since $128 = 2^7$, we can rewrite the equation as:

$$2^{3x-y} = 2^7$$

Equating the exponents gives our second linear equation:

$$3x - y = 7 \quad (2)$$

We now have a system of two linear equations:

1. $x - y = 3$
2. $3x - y = 7$

Subtract equation (1) from equation (2) to eliminate $y$:

$$(3x - y) - (x - y) = 7 - 3$$ $$3x - y - x + y = 4$$ $$2x = 4$$ $$x = \frac{4}{2}$$ $$x = 2$$

Using equation (1):

$$x - y = 3$$ $$2 - y = 3$$ $$-y = 3 - 2$$ $$-y = 1$$ $$y = -1$$

The solution to the system of equations is $x=2$ and $y=-1$.

The values are:
$x = \boxed{2}$
$y = \boxed{-1}$