Solve for \( d \) \( \begin{array}{l}3+d<3-d \\ d<6 \\ d<-6 \\ d<0 \\ d<-3\end{array} \)

Let's tackle the inequalities step by step! First inequality: \[ 3 + d < 3 - d \] To solve it, we'll rearrange it: \[ d + d < 3 - 3 \] This simplifies to: \[ 2d < 0 \] Dividing both sides by 2 gives: \[ d < 0 \] The second inequality is: \[ d < 6 \] The third inequality: \[ d < -6 \] The fourth inequality states: \[ d < 0 \] The fifth inequality is: \[ d < -3 \] Now let's combine these inequalities. The strictest condition here is \( d < -6 \) because it is more limiting than \( d < 0 \) and \( d < -3 \). The main solution that satisfies all of these inequalities is: \[ d < -6 \] To sum it up, the solution for \( d \) is \( d < -6 \).