\( \frac{4}{3} x+11 \geq 7 \) or \( -4 x-3 \leq-23 \) Solution Interval Notation \( \square \)

To solve the compound inequality: \[ \frac{4}{3}x + 11 \geq 7 \quad \text{or} \quad -4x - 3 \leq -23 \] we'll solve each inequality separately and then combine the solutions using interval notation. ### 1. Solving \(\frac{4}{3}x + 11 \geq 7\): \[ \frac{4}{3}x + 11 \geq 7 \] Subtract 11 from both sides: \[ \frac{4}{3}x \geq 7 - 11 \] \[ \frac{4}{3}x \geq -4 \] Multiply both sides by \(\frac{3}{4}\) to solve for \(x\): \[ x \geq -4 \times \frac{3}{4} \] \[ x \geq -3 \] ### 2. Solving \(-4x - 3 \leq -23\): \[ -4x - 3 \leq -23 \] Add 3 to both sides: \[ -4x \leq -23 + 3 \] \[ -4x \leq -20 \] Divide both sides by \(-4\) (remember to reverse the inequality sign when dividing by a negative number): \[ x \geq 5 \] ### Combining the Solutions: The solutions to the individual inequalities are: - \(x \geq -3\) - \(x \geq 5\) Since the original compound inequality uses "or," we take the union of these solutions. The interval \(x \geq -3\) already includes \(x \geq 5\), so the combined solution is: \[ x \geq -3 \] ### Interval Notation: \[ [-3, \infty) \] **Answer:** \([-3, \infty)\)