67th Czech and Slovak Mathematical Olympiad

We rewrite the system as $$F(x)>0\wedge G(x)<0,$$ where $F(x)=x^2-(a-1)x-b$ and $G(x)=x^2-ax-b+1$. Observe that $F(x)-G(x)=x-1$. The condition $a+b>2$ implies that $$F(1)=G(1)=2-a-b<0,$$ hence $x=1$ is not a solution. However, $G(1)<0$ implies that the quadratic equation $G(x)=0$ has a root $x_0>1$. Then $$F(x_0)=F(x_0)-0=F(x_0)-G(x_0)=x_0-1>0.$$ From $F(1)<0$ and $F(x_0)>0$ we deduce that there exists a root $x_1$ of $F(x)=0$ that belongs to the open interval $(1,x_0)$. Since $$F(1)<0\wedge F(x_1)=0,\text{and}G(1)<0\wedge G(x_0)=0,$$ any $x\in (x_1,x_0)$ is a solution to the original system.