Let \( A = \sqrt{x+y} \) and \( B = \sqrt{y+z} \).\nFrom the first given equation, we have:\n\[ 2A - 3B = 2 \]\n\nNow, manipulate the second equation to express it in terms of \( A \) and \( B \):\n\[ 4x - 5y - 9z = 4(x+y) - 9(y+z) \]\nThus, \( 4A^2 - 9B^2 = 8 \).\n\nFactor the difference of squares:\n\[ (2A - 3B)(2A + 3B) = 8 \]\n\nSince \( 2A - 3B = 2 \), substituting this gives:\n\[ 2(2A + 3B) = 8 \implies 2A + 3B = 4 \]\n\nSolving the linear system for \( A \) and \( B \):\n\[ (2A - 3B) + (2A + 3B) = 2 + 4 \implies 4A = 6 \implies A = \frac{3}{2} \]\n\[ 2A + 3B = 4 \implies 3 + 3B = 4 \implies 3B = 1 \implies B = \frac{1}{3} \]\n\nSo, \( x+y = A^2 = \frac{9}{4} \) and \( y+z = B^2 = \frac{1}{9} \).\n\nNow evaluate the expression inside the square root:\n\[ \frac{20x+38y+18z+1}{9y+9z+2} \]\n\nRewrite the numerator in terms of \( (x+y) \) and \( (y+z) \):\n\[ 20x+38y+18z+1 = 20(x+y) + 18(y+z) + 1 \]\n\[ = 20\left(\frac{9}{4}\right) + 18\left(\frac{1}{9}\right) + 1 = 45 + 2 + 1 = 48 \]\n\nRewrite the denominator:\n\[ 9(y+z) + 2 = 9\left(\frac{1}{9}\right) + 2 = 1 + 2 = 3 \]\n\nCalculate the full fraction and its square root:\n\[ \sqrt{\frac{48}{3}} = \sqrt{16} = 4 \]