a) \( 2|3-x|+3 x>3 \) b) \( \frac{1}{x+2} \geq \frac{1}{3-x} \) c) \( 2 x^{2} \geq 3 x+9 \)

Let’s dive into these inequalities and unravel their mysteries! First up, for the first inequality \( 2|3-x| + 3x > 3 \), you can break it into two cases based on the absolute value. If \( x < 3 \), you'll have \( |3-x| = 3-x \), and if \( x \geq 3 \), it becomes \( |3-x| = x-3 \). Solve each case separately by isolating x; it’s like solving a little puzzle! Next, the second inequality \( \frac{1}{x+2} \geq \frac{1}{3-x} \) can be approached by finding a common denominator and cross-multiplying to eliminate the fractions. Be wary of the values where the denominators equal zero, though, because they could lead to undefined results; think of them as the “no-go zones” of your solution space! Lastly, in \( 2x^2 \geq 3x + 9 \), you can rearrange that to form a standard quadratic inequality: \( 2x^2 - 3x - 9 \geq 0 \). Using techniques like factoring or the quadratic formula will reveal the roots, which can then be analyzed to find where the expression is greater than or equal to zero. It’s like being a detective looking for the safe zones where the inequality holds true!