Solution

Let

$\sqrt{x+y}=a$

and

$\sqrt{y+z}=b$

Given,

$2\sqrt{x+y}-3\sqrt{y+z}=2$

So,

$2a-3b=2$

Also,

$x=a^2-y$

and

$z=b^2-y$

Now use the second condition:

$4x-5y-9z=8$

$4(a^2-y)-5y-9(b^2-y)=8$

$4a^2-4y-5y-9b^2+9y=8$

$4a^2-9b^2=8$

So,

$(2a-3b)(2a+3b)=8$

Since $2a-3b=2$,

$2(2a+3b)=8$

$2a+3b=4$

Now solve:

$2a-3b=2$

$2a+3b=4$

Adding both equations,

$4a=6$

$a=\frac{3}{2}$

Now,

$2a+3b=4$

$3+3b=4$

$3b=1$

$b=\frac{1}{3}$

Now,

$20x+38y+18z+1$

$=20(a^2-y)+38y+18(b^2-y)+1$

$=20a^2+18b^2+1$

Also,

$9y+9z+2=9(y+z)+2=9b^2+2$

Now substitute $a=\frac{3}{2}$ and $b=\frac{1}{3}$.

Numerator:

$20a^2+18b^2+1=20\left(\frac{3}{2}\right)^2+18\left(\frac{1}{3}\right)^2+1$

$=20\cdot \frac{9}{4}+18\cdot \frac{1}{9}+1$

$=45+2+1$

$=48$

Denominator:

$9b^2+2=9\left(\frac{1}{3}\right)^2+2$

$=1+2$

$=3$

Therefore,

$\sqrt{\frac{20x+38y+18z+1}{9y+9z+2}}=\sqrt{\frac{48}{3}}$

$=\sqrt{16}$

$=4$