Solve y^2+y<0

$$y ^ { 2 } + y < 0$$

Solve for y

$$y\in \left(-1,0\right)$$

Solution Steps

Factor out y.

$$y\left(y+1\right)<0$$

For the product to be negative, y+1 and y have to be of the opposite signs. Consider the case when y+1 is positive and y is negative.

$$y+1>0 y<0$$

The solution satisfying both inequalities is y\in \left(-1,0\right).

$$y\in \left(-1,0\right)$$

Consider the case when y is positive and y+1 is negative.

$$y>0 y+1<0$$

This is false for any y.

$$y\in \emptyset$$

The final solution is the union of the obtained solutions.

$$y\in \left(-1,0\right)$$