Let a$=3\sqrt{2}$ and b $=\frac{1}{5^{\frac{1}{6}}\sqrt{6}}.$ If $x,y\in <!--\; \in \; -->\mathbb{R}$ are such that $3x+2y={\log }_a<!--\; \; -->(18{)}^{\frac{5}{4}}$ and $2x-<!--\; -\; -->y={\log }_b<!--\; \; -->(\sqrt{1080}),$then 4x + 5y is equal to ______. 

The correct answer is 8.

Let us solve the problem step-by-step by simplifying the given logarithmic equations.

Step 1: Simplify the base $a$ and the first logarithmic equation

We are given:
$a=3\sqrt{2}=\sqrt{18}={18}^{1/2}$
The first equation is:
$3x+2y={\log }_a\left({18}^{5/4}\right)$

Substituting $a={18}^{1/2}$ into the equation:
$3x+2y={\log }_{{18}^{1/2}}\left({18}^{5/4}\right)$

Using the property of logarithms, ${\log }_{p^k}\left(p^m\right)=\frac{m}{k}$:
$3x+2y=\frac{5/4}{1/2}=\frac{5}{4}\cdot 2=\frac{5}{2}$
This gives us our first linear equation:
$3x+2y=\frac{5}{2}$     — (Equation 1)

Step 2: Simplify the base $b$ and the second logarithmic equation

We are given:
$b=\frac{1}{5^{1/6}\sqrt{6}}=\frac{1}{5^{1/6}\cdot 6^{1/2}}$

Let us express $b^{-6}$:
$b^{-6}={(5^{1/6}\cdot 6^{1/2})}^6={\left(5^{1/6}\right)}^6\cdot {\left(6^{1/2}\right)}^6=5\cdot 6^3=5\cdot 216=1080$

Therefore, we can write:
$1080=b^{-6}$
Taking the square root on both sides:
$\sqrt{1080}={\left(b^{-6}\right)}^{1/2}=b^{-3}$

The second equation is:
$2x-y={\log }_b\left(\sqrt{1080}\right)$

Substituting $\sqrt{1080}=b^{-3}$:
$2x-y={\log }_b\left(b^{-3}\right)=-3$
This gives us our second linear equation:
$2x-y=-3$     — (Equation 2)

Step 3: Solve the system of linear equations

From Equation 2, we can express $y$ in terms of $x$:
$y=2x+3$

Substitute this expression for $y$ into Equation 1:
$3x+2(2x+3)=\frac{5}{2}$
$3x+4x+6=\frac{5}{2}$
$7x=\frac{5}{2}-6$
$7x=-\frac{7}{2}$
$x=-\frac{1}{2}$

Now, calculate $y$:
$y=2(-\frac{1}{2})+3=-1+3=2$

Step 4: Find the value of $4x+5y$

Substitute the values of $x$ and $y$:
$4x+5y=4(-\frac{1}{2})+5\left(2\right)=-2+10=8$