jmc

We can write $9x=10\lfloor x\rfloor .$ Since $x=\lfloor x\rfloor +\{x\},$ $$9\lfloor x\rfloor +9\{x\}=10\lfloor x\rfloor .$$Then $9\{x\}=\lfloor x\rfloor .$ Since $\{x\}<1,$ $\lfloor x\rfloor =9\{x\}<9.$ Thus, $\lfloor x\rfloor \le 8.$

If $\lfloor x\rfloor =8,$ then $\{x\}=\frac{8}{9},$ so the largest possible value of $x$ is $8+\frac{8}{9}=\boxed{\frac{80}{9}}.$