(2) Solve \( \sqrt{x+7}-\sqrt{3 x-2}=1 \)

To solve the equation \( \sqrt{x+7} - \sqrt{3x-2} = 1 \), we can start by isolating one of the square roots. Rearranging gives: \[ \sqrt{x+7} = \sqrt{3x-2} + 1 \] Next, we square both sides to eliminate the square roots: \[ (\sqrt{x+7})^2 = (\sqrt{3x-2} + 1)^2 \] This expands to: \[ x + 7 = (3x - 2) + 2\sqrt{3x-2} + 1 \] Simplifying, we get: \[ x + 7 = 3x - 1 + 2\sqrt{3x-2} \] Combining like terms results in: \[ x + 8 = 3x + 2\sqrt{3x-2} \] Rearranging yields: \[ 8 = 2x + 2\sqrt{3x-2} \] Next, divide everything by 2: \[ 4 = x + \sqrt{3x-2} \] Now isolate the square root again: \[ \sqrt{3x-2} = 4 - x \] We square both sides again: \[ (\sqrt{3x-2})^2 = (4 - x)^2 \] This produces: \[ 3x - 2 = 16 - 8x + x^2 \] Rearranging leads us to a quadratic equation: \[ x^2 - 11x + 18 = 0 \] We can factor this equation: \[ (x - 2)(x - 9) = 0 \] Hence, the solutions are \( x = 2 \) and \( x = 9 \). It's crucial to check these solutions in the original equation to ensure they are valid. For \( x = 2 \): \[ \sqrt{2+7} - \sqrt{3(2)-2} = \sqrt{9} - \sqrt{4} = 3 - 2 = 1 \quad \text{(valid)} \] For \( x = 9 \): \[ \sqrt{9+7} - \sqrt{3(9)-2} = \sqrt{16} - \sqrt{25} = 4 - 5 = -1 \quad \text{(not valid)} \] Thus, the only solution is \( \boxed{2} \).