I have $v_x + v_y = 1$.
It's obvious that $v(x,y) = \frac{1}{2}(x+y)$ works as a solution, but I need to find a general solution to this PDE. What can I do?
I have tried doing a substitution, to no avail. Perhaps I am over thinking this problem and there is some really simple solution to a non-homogenous PDE like this one that my book doesn't discuss. Any help is greatly appreciated.
That is an inhomogeneous transport equation. You need to use the method of characteristics to solve it. Basically, this method seeks curves along which the PDE becomes an ordinary differential equation.
Assume your solution is of the form $v(x,y)=v(x,Y(x))=V(x)$, where $Y$ is a characteristic curve of the solution. Then, $$\frac{d}{dx}V(x)=v_x(x,Y(x))+v_y(x,Y(x))\ Y'(x).$$ So, if you pick $Y$ such that $Y'(x)=1$, you get that $$\frac{d}{dx}V(x)=v_x(x,Y(x))+v_y(x,Y(x)) = 1.$$ And that is a much easier ODE you can solve.
Therefore, you just need to follow three steps:
Find the characteristic by solving $Y'(X)=1$ with the final condition $Y(x)=y$.
Determine the solution along a characteristic by solving $V'(X)=1$ with the initial condition $V(0)=V(0,Y(0))=0$ (or any other initial condition you might have).
Find the solution at the endpoint of the characteristic, $v(x,y)=V(x)$, which is the solution of the PDE at $(x,y)$.
You can find plenty of notes on this topic by googling. For instance, check here.
Note that Lagrange's form $P(x,y,v)p+Q(x,y,v)q=R(x,y,v)$ for which the subsidiary equations are
$$\frac{dx}{P}=\frac{dy}{Q}=\frac{dv}{R}$$
In your case, $P=Q=R=1$, so it is easy to obtain $v-x=c_1$ and $y-x=c_2$ where $c_1$ and $c_2$ are constants of integration. Combining these two, you get $v-x=f(y-x)$ or $v=f(y-x)+x$ as the general solution for an arbitrary function $f$.