Solve the following system of inequations with one variable:
$$$ \left\{ \begin{array}{l} x(x-2) < 0 \\ x^2+x > -(1+x) \end{array}\right. $$$
Development:
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: $$$ 0 < x < 2$$$
Solution:
$$ 0 < x < 2$$