b. \( |x-1| \geq-3 \) c. \( |x-1| \leq-3 \) d. \( \left|\frac{2 x-9}{3}\right|<4 \) e. \( \left|x^{2}+3 x-1\right|>6 \) f. \( |6-2 x| \leq 2 \)

To solve these inequalities, let’s dive in with some fun! For the first one, \( |x-1| \geq -3 \), remember that absolute values are never negative! This means that this inequality is always true for any real number \( x \). Math magic! Now, for \( |x-1| \leq -3 \), think of the absolute value as a measure of distance, and since distances can't be negative, this inequality has no solutions. It's like asking for someone to be less than zero feet away from you—impossible! Now, let's tackle \( \left|\frac{2 x-9}{3}\right| < 4 \). To solve, you would first isolate the term inside the absolute value. Multiply both sides by 3, \( |2x - 9| < 12 \). Then split it into two scenarios: \( 2x - 9 < 12 \) and \( 2x - 9 > -12 \). Solve these inequalities to find the range of \( x \). Next up is \( |x^{2}+3 x-1|>6 \). Here, we want to break it down into two separate inequalities: \( x^{2}+3 x-1 > 6 \) and \( x^{2}+3 x-1 < -6 \). Solve these quadratic equations for potential values of \( x \) where the function crosses those boundaries. Exciting, right? Finally, for \( |6-2 x| \leq 2 \), you'll want to break it down similarly: \( 6 - 2x \leq 2 \) and \( 6 - 2x \geq -2 \). Rearranging gives you bounds for \( x \), showing you precisely the range where this holds true. Keep up the good work!