1. **State the problem:** We need to find the range of values of $x$ that satisfy the compound inequality $$5x - 7 \leq 9x - 1 < 8x + 1.$$\n\n2. **Break the compound inequality into two parts:**\n\nPart 1: $$5x - 7 \leq 9x - 1$$\nPart 2: $$9x - 1 < 8x + 1$$\n\n3. **Solve Part 1:**\n\n$$5x - 7 \leq 9x - 1$$\nSubtract $5x$ from both sides:\n$$\cancel{5x} - 7 \leq \cancel{5x} + 4x - 1$$\nSimplifies to:\n$$-7 \leq 4x - 1$$\nAdd 1 to both sides:\n$$-7 + 1 \leq 4x$$\n$$-6 \leq 4x$$\nDivide both sides by 4 (positive, so inequality direction stays):\n$$\frac{-6}{4} \leq x$$\nSimplify fraction:\n$$-\frac{3}{2} \leq x$$\n\n4. **Solve Part 2:**\n\n$$9x - 1 < 8x + 1$$\nSubtract $8x$ from both sides:\n$$9x - 8x - 1 < \cancel{8x} + 1 - \cancel{8x}$$\nSimplifies to:\n$$x - 1 < 1$$\nAdd 1 to both sides:\n$$x < 2$$\n\n5. **Combine the two results:**\n\n$$-\frac{3}{2} \leq x < 2$$\n\n**Final answer:** The values of $x$ that satisfy the inequality are $$x \in \left[-\frac{3}{2}, 2\right).$$