if $f,g$ are continuous and increasing, $\exists! (u,v) \in \mathbb{R}$ so that $u+f(u-v)=v+g(v-u)=0$

I'm going in circles to show the unicity

$h: u \mapsto u + f(u - v)$ is strictly increasing with v fixed and by IVT $\exists! u \in \mathbb{R}$ such that $h(u)=0$ (easy to show)

Same is true with $d: u \mapsto v + g(v - u)$ with $u \in \mathbb{R}$ fixed

To my mind, the unicity comes from monotony and injectiv functions, but I struggle to showit.

If I choose $u$, then $v$ has to be chosen, but $u$ too and here's my circle...

I tried to suppose that there exist 2 couples $(u,v)$ that are solutions and find a contradiction, but it didn't work.

Do you have any advice?