Solving a system of two linear PDE: $u_x+v_x +u_y=0$ and $v_x+u_y-{1\over 2} v_y=0$

Question

trying to solve the following cauchy problem: $$u_x+v_x +u_y=0\\v_x+u_y-{1\over 2} v_y=0\\u(x,0)=1-x,v(x,0)=x$$ my solution is: 1. multiply each equation by $t_1,t_2$ and sum the two equations like that: $$t_1u_x +v_x(t_1+t_2)+u_y(t_1+t_2)-0.5t_2v_y\\{t_1 \over t_1+t_2}={t_1+t_2\over-0.5t_2}=\lambda$$ then found eigen values and eigen vectors for the new system: $$\lambda_1=-1,(1,-2)^T\\\lambda_2=2,(-2,1)^T$$ rewriting the system as: $$-1*(-u+v)_x+(-u+v)_y=0\\2*(u+0.5v)_x+(u+0.5v)_y=0$$ then, define Q and R: $$R=-u+v,Q=u+0.5v$$ then I got: $$-R_x+R_y=0 =>y+x=c_1\\2Q_x +Q_y=0 => y-0.5x=c_2$$ and now I'm stuck. don't know how to find Q and R.