I3.2 It is given that $\frac{x+y}{2}=\frac{z+x}{3}=\frac{y+z}{4}$ and $x+y+z=36 a$. Find the value of $b$, if $b=x+y$.
$\begin{array}{l}x+y=2 k \ldots \ldots(1) \\ z+x=3 k \ldots \ldots(2) \\ y+z=4 k \ldots \ldots(3) \\ (1)+(2)+(3): 2(x+y+z)=9 k \\ 2(36)\left(\frac{1}{2}\right)=9 k \\ k=4 \\ b=x+y \\ \quad=2 k \\ \quad=2(4)=8\end{array}$