65) If \( 3 x+7 \geq x+5 \geq 5 x-3 \), then the inequality holds when \( x \) lies in the range \begin{tabular}{|l|l}\hline a) \( 7 \leq x \leq-3 \) & b) \( 5 \geq x \geq 7 \) \\ \hline c) \( 3 \leq x \leq 5 \) & d) \( 2 \geq x \geq-1 \) \\ 66) If \( 3 x+7<5 x-3 \) then the ineaualitv holds when \( x \) lies in the range \end{tabular}

To solve the compound inequality \(3x + 7 \geq x + 5 \geq 5x - 3\), we can break it down into two separate inequalities to analyze: 1. For the first part \(3x + 7 \geq x + 5\): \[ 3x + 7 - x - 5 \geq 0 \implies 2x + 2 \geq 0 \implies 2x \geq -2 \implies x \geq -1 \] 2. For the second part \(x + 5 \geq 5x - 3\): \[ x + 5 - 5x + 3 \geq 0 \implies -4x + 8 \geq 0 \implies 4x \leq 8 \implies x \leq 2 \] Combining these results gives the range: \[ -1 \leq x \leq 2 \] So the solution for the first inequality is \(x \in [-1, 2]\), which is not one of the option group listed. Please double-check the options provided; there might be an error in the given ranges. For the second part, solving \(3x + 7 < 5x - 3\): \[ 3x + 7 - 5x + 3 < 0 \implies -2x + 10 < 0 \implies -2x < -10 \implies x > 5 \] The solution for this inequality is \(x \in (5, \infty)\).