we know the line passes through those points, so let's find the EQUATion of it firstly.

so, that's its EQUATion, now, we know the (0,0) is a solution of this inequality, namely the (0,0) point lies in the "true region" or will be the "shaded region" of the inequality, let's plug those values,

well, clearly 0 is not equals to -14.... we know 0 is greater, recall for the negative values, the farther from zero the smaller.
we also know that the line is dashed, meaning the borderline values are not included, so is either a > or <, thus since 0 > -14, then
