Can I produce $1 - y$ with $y x_0 + x_1$ I want to know whether it is possible to calculate $1 - y$ using only a multiplication then an addition of two numbers which are not a function of $y$. I suspect it isn't possible but I would like to know how to prove this. 
What I've tried so far:
I tried manipulating the equation $1 - y = y  x_0 + x_1$, attempting to solve for $x_0$ and $x_1$. Through this I found two equations:
$y = \frac{1 - x_1}{x_0 + 1}$
$x_1 = y (-x_0) + y + 1$
By plugging the first equation into the second and I get the following
$x_1 = (\frac{1 - x_1}{x_0 + 1})(-x_0) + \frac{1 - x_1}{x_0 + 1} + 1$
With some manipulation I end up with $(x_0 + 1)(x_1 - 1) = x_0x_1-x_0 + x_1 - 1$ which if you do the multiplication on the left side causes all the terms to cancel leaving me with $0 = 0$ and no information about $x_0 $ or $x_1$.
I tried solving a similar equation $ y + 1 = y * x_0 + x_1$ that I know has the solution $x_0 = 1, x_ 1 = 1$ and I also ended up with all the terms cancelling so I know that the fact that happened doesn't prove that there is no solution.
How can I prove one way or the other whether there exist values of $x_0$ and $x_1$ that solve my problem?

Rewrite as
$$1 - y = 1 + (-1)\cdot y = 1 + y \cdot (-1) = y\cdot(-1) + 1$$
Now equate the coefficients to get
$$ x_0 = -1, x_1 = 1$$
