Solution
= Solution
The inverse transformation is $x=2u+v$, $y=v$, with absolute <Jacobian determinant> two. The original positivity constraints become $v>0$ and $2u+v>0$. Thus
$$
\boxed{A=\{(u,v):v>\max(0,-2u)\}},\qquad
f_{U,V}(u,v)=\tfrac12e^{-u-v}\mathbf1_A.
$$
Integrating out $v$ gives
$$
f_U(u)=\frac12e^{-u}\int_{\max(0,-2u)}^\infty e^{-v}\,dv
=\frac12e^{-u-\max(0,-2u)}
=\boxed{\frac12e^{-|u|}}.
$$
This is the centred <Laplace distribution> with scale one.