\( \begin{array}{ll}\text { (a) } \frac{x}{3}-\frac{x}{2}>1 & \text { (b) } \frac{3 x}{4}-x \leq 1 \frac{1}{2} \\ \text { (c) } \frac{y+5}{3}+y \leq 1 & \text { (d) } \frac{3 y+2}{4}-\frac{y-6}{3}>0 \\ \text { (c) }(x+3)(x-4) \leq(x-3)(x+4) & \end{array} \)

To solve these inequalities, let's break them down step by step. For (a), \(\frac{x}{3}-\frac{x}{2}>1\): Start by finding a common denominator (which is 6). Rewrite it as \(\frac{2x - 3x}{6} > 1\) leading to \(-\frac{x}{6} > 1\). Multiply through by -6 (and flip the inequality) to get \(x < -6\). For (b), \(\frac{3x}{4}-x \leq 1 \frac{1}{2}\): Rewrite \(1 \frac{1}{2}\) as \(\frac{3}{2}\). This becomes \(\frac{3x - 4x}{4} \leq \frac{3}{2}\) which simplifies to \(-\frac{x}{4} \leq \frac{3}{2}\). Multiplying by -4 (flipping the inequality) gives \(x \geq -6\). Now for (c), \(\frac{y+5}{3}+y \leq 1\): Multiply everything by 3 to eliminate the fraction: \(y + 5 + 3y \leq 3\) simplifies to \(4y + 5 \leq 3\). From here, subtract 5 to result in \(4y \leq -2\), leading to \(y \leq -\frac{1}{2}\). Finally for (d), \(\frac{3y+2}{4}-\frac{y-6}{3}>0\): Clear the denominators by multiplying through by 12, which gives us \(9y + 6 - 4y + 8 > 0\). Combine terms to have \(5y + 14 > 0\), which means \(y > -\frac{14}{5}\). And as a bonus point, for (e), \((x+3)(x-4) \leq (x-3)(x+4)\): Expand both sides to \(x^2 - x - 12 \leq x^2 + x - 12\). Simplifying gives \(-x \leq x\) or \(0 \leq 2x\), leading to \(x \geq 0\). In summary: - For \(x\), it needs to satisfy \(x < -6\) or \(x \geq 0\). - For \(y\), it ranges from \(y \leq -\frac{1}{2}\) and \(y > -\frac{14}{5}\). Inequalities can be puzzling, but remember: always flip that sign when multiplying or dividing by a negative!