Resposta :
Olá, Adilson.
Seja x e y os valores aplicados procurados.
[tex]x(1+\underbrace{0,03}_{3\%})=y(1+\underbrace{0,05}_{5\%}) \Rightarrow 1,03x=1,05y \ \ (\times 100)\Rightarrow\\\\ 103x=105y \Rightarrow x=\frac{105}{103}y\\\\ x+y=2000 \Rightarrow \frac{105}{103}y+y=2000 \Rightarrow 105y+103y=103\cdot2000 \Rightarrow\\\\ y=\frac{206000}{208}\Rightarrow \boxed{y\approx R\$\ 990,38} \Rightarrow \boxed{x\approx R\$\ 1.009,62}[/tex]