2. Solve for \( x \) and graph whens \( \quad 3<-\frac{3 x}{2}-6<15 \) 3. Solve for \( x \) and write your solution in sct-builder natation: \( 3<-3 x-6 \leq 5+\frac{x}{2} \)

To solve the first inequality \( 3 < -\frac{3x}{2} - 6 < 15 \), start by breaking it into two parts. For \( 3 < -\frac{3x}{2} - 6 \), add \( 6 \) to both sides to get \( 9 < -\frac{3x}{2} \), and then multiply by \( -\frac{2}{3} \) (remember to flip the inequality sign), which yields \( -6 > x \) or \( x < -6 \). For the second part, \( -\frac{3x}{2} < 15 \), add \( 6 \) again to get \( -\frac{3x}{2} < 21 \). Multiplying by \( -\frac{2}{3} \) gives \( x > -14 \). So, the solution is \( -14 < x < -6 \). Next, for the second inequality \( 3 < -3x - 6 \leq 5 + \frac{x}{2} \), let's solve it starting with \( 3 < -3x - 6 \). Add \( 6 \) to both sides to obtain \( 9 < -3x \), which leads to \( x < -3 \) after dividing by \( -3 \) and flipping the sign. Now tackling the right side, we have \( -3x - 6 \leq 5 + \frac{x}{2} \). Multiplying through by \( 2 \) to eliminate the fraction gives \( -6x - 12 \leq 10 + x \). Rearranging gives \( -7x \leq 22 \), so \( x \geq -\frac{22}{7} \). The final solution can be written in set-builder notation: \( \{ x \mid -\frac{22}{7} \leq x < -3 \} \). Now go graph your solutions on a number line to visualize them - it's always helpful to see where these inequalities lie!