Solve the equation. \( \begin{aligned} \sqrt{3 x-4} & =2 \sqrt{x-5} \\ x & =[?]\end{aligned} \)

To solve the equation: \[ \sqrt{3x - 4} = 2\sqrt{x - 5} \] **Step 1: Determine the Domain** Both square roots must have non-negative arguments: 1. \( 3x - 4 \geq 0 \) \( 3x \geq 4 \) \( x \geq \frac{4}{3} \) 2. \( x - 5 \geq 0 \) \( x \geq 5 \) The more restrictive condition is \( x \geq 5 \). So, the domain is \( x \geq 5 \). **Step 2: Solve the Equation** Starting with the original equation: \[ \sqrt{3x - 4} = 2\sqrt{x - 5} \] Square both sides to eliminate the square roots: \[ (\sqrt{3x - 4})^2 = (2\sqrt{x - 5})^2 \] \[ 3x - 4 = 4(x - 5) \] Expand the right side: \[ 3x - 4 = 4x - 20 \] Bring like terms together: \[ 3x - 4x = -20 + 4 \] \[ -x = -16 \] Multiply both sides by -1: \[ x = 16 \] **Step 3: Verify the Solution** Check \( x = 16 \) in the original equation: \[ \sqrt{3(16) - 4} = \sqrt{48 - 4} = \sqrt{44} = 2\sqrt{11} \] \[ 2\sqrt{16 - 5} = 2\sqrt{11} \] Both sides are equal, so \( x = 16 \) is a valid solution. **Final Answer:** \( x = 16 \)