Systems of inequations with one variable

We will solve both inequations independently and next we will intersect the regions of solutions: $$ x(x-2) < 0 \Rightarrow \left\{ \begin{array}{l} x < 0 \\ x-2 > 0 \end{array}\right. \ \text{ o bé } \ \left\{ \begin{array}{l} x > 0 \\ x-2 < 0 \end{array}\right.$$ we must take the second option since the first one has no solution. Then : $0 < x < 2$.

On the other hand: $$ x^2+x > -(1+x) \Rightarrow x^2+2x+1 > 0$$

We solve the quadratic equation $$ x^2+ 2x +1=0 \Rightarrow x=\dfrac{-2\pm\sqrt{4-4}}{2}=-1$$

Consequently we have: $$ x^2+ 2x +1=(x+1)^2=(x+1)(x+1) > 0 \Rightarrow x+1 > 0 \ \ \text{ o bé } \ \ x+1 > 0$$

we deduce that the solutions will be: $x+1 > 0$ and $x+1 > 0$

Now we intersect the regions of both solutions and obtain the region: $$ 0 < x < 2$$